Let $\gamma\colon [0,1] \to \mathbb R^2$ (continuous) be a simple closed plane curve and $\,\mathcal C$ its image. Let $x,y$ be some functions such that $\gamma(t)=(x(t),y(t))$. $\gamma$ is said differentiable if $x,y$ are differentiable, $\gamma'(0)=\gamma'(1)$ and $\gamma'(t)$ is never the zero vector.

$\gamma$ is said strictly convex if for any two points on $\,\mathcal C$, the line segment that joins them stays entirely inside the interior of $\,\mathcal C$ (as defined by the Jordan curve theorem), except for the endpoints.

The tangent line to $\gamma$ at $t \in [0,1[$ is the set $$ \{ \gamma(t)+r\gamma'(t) : r \in \mathbb R\} .$$

Is this true?

If $\gamma$ is strictly convex, then any line crosses $\mathcal C$ in at most two points. If $\gamma$ is also differentiable, then a line $s$ crosses $\mathcal C$ in just one point iff $s$ is tangent to $\gamma$. No interior point of $\mathcal C$ are on a tangent line.

  • $\begingroup$ To answer the first part of your question, imagine a construction where a line $l$ crosses $\mathcal{C}$ at points $a$, $b$, and $c$. WLOG, let $c$ be between $a$ and $b$ on $l$. Let $c_{\epsilon}$ and $c_{-\epsilon}$ be displaced from $c$ by small arc lengths $\epsilon$ and $-\epsilon$ along $\mathcal{C}$, respectively (where $c_{-\epsilon}$ is between $a$ and $c$, and $c_{\epsilon}$ is between $c$ and $b$). What about the line segments connecting $a$ and $c_{-\epsilon}$, $c_{-\epsilon}$ and $c_{\epsilon}$, and $c_{\epsilon}$ and $b$? Are they all on the interior of $\mathcal{C}$? $\endgroup$ – Michael Lee Oct 21 '16 at 15:46
  • $\begingroup$ Maybe I should modify my definition... $\endgroup$ – OliverN Oct 21 '16 at 16:04
  • $\begingroup$ See the edit. Now I can say that if $c$ is between $a$ and $b$, then it is in the interior of $\mathcal C$, impossible. Is it correct? $\endgroup$ – OliverN Oct 21 '16 at 16:08
  • $\begingroup$ That looks absolutely correct. The second claim should also be true by boundary arguments. $\endgroup$ – Fimpellizieri Oct 21 '16 at 16:22
  • $\begingroup$ Making the second claim more precise; since $\gamma'(t)$ is never zero, one can always use the implicit function theorem to write $\gamma$ as the graph of a function, locally. Now, it's easy to show that if a line touches a graph at a point and the derivatives do not agree on that point, then the line crosses the graph. Can you see how this solves the problem? $\endgroup$ – Fimpellizieri Oct 21 '16 at 16:26

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