Let $C\subset\mathbb{R}^2$ be a simple closed convex curve with $O$ as its interior point and its regular parametrization is given as $\gamma_C:[0,1]\rightarrow C$ such that $\gamma_C(0)=(1,0)$ and $\gamma_C'(0)=(0,1)$.
Let $L\subset\mathbb{R}^2$ be a simple curve with its regular parametrization given as $\gamma_L:[0,1]\rightarrow L$ such that $\gamma_L(0)=(1,0)$ and $\gamma_L'(0)=(-1,1)$.

Show that there exists $t\in[0,1]$ such that $\gamma_L(t)$ is interior to $C$.

I am trying to prove a lemma in my research regarding topology but I come across this small problem about geometry of a curve. I do not attend a full course on geometry of curves, hence I may lack of ideas to solve this. It is geometrically intuitive but I need an algebraic approach to prove this. I try Taylor's Theorem or Cauchy's Mean Value Theorem but I'm going nowhere.

Please advice me on what I should do. Thank you in advance.

  • $\begingroup$ The claim, as stated, is not true. Do you have any additional information (like the second curve being closed, too)? $\endgroup$ – Thomas May 2 '17 at 16:01
  • $\begingroup$ Can you specify why it is not true? The second curve needs not to be closed. The crucial part is that C has a tangential vector upwards. If it is not upwards, then, it may be not true. Just geometrical intuition. $\endgroup$ – Nur Nordin May 2 '17 at 17:09
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    $\begingroup$ It's not true because the original curve $C$ may be turning to the right. Take $C$ to be the circle $(x-2)^2+y^2=1$ (traversed clockwise), for example. $\endgroup$ – Ted Shifrin May 2 '17 at 20:14
  • $\begingroup$ But in that example, O is not interior to C. But actually I get what you mean. Yeah, I need to add another assumption. Can I say that C is convex? $\endgroup$ – Nur Nordin May 2 '17 at 22:15
  • $\begingroup$ The closed curve bisects, near it's starting point, any sufficiently small ball centered at that point into two components. In order to ensure that what you claim is true you somehow have to ensure that the component 'to the left' is contained in what you refer to as the interior region, because that's the direction into which the second curve is pointing at its starting point. Convexity of $C$ will obviously allow you to conclude this. $\endgroup$ – Thomas May 3 '17 at 5:10

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