I have the following problem.
Suppose $\alpha:[0,1]\rightarrow \mathbb R^2$ is regular, closed and convex curve.
For which $c\in \mathbb R$ can the following function:
$\kappa:[0,1]\rightarrow\mathbb R$ ; $\kappa(t)= ct^4(1-t)^4$ can be it's curvature (meaning signed curvature , because curvature is always positive by definition)?
My (attempt of) solution:
- Because $\alpha$ is convex we know that $\alpha\prime\prime\geq0$ $ \forall t\in[0,1]$
- Because $\alpha$ is closed then there exists $T>0$ such that $\alpha(t) = \alpha(t+T)$ $ \forall t\in[0,1]$.
- $c\neq0$ because otherwise we have that $\kappa=0$ therefore, $\alpha$ is part of straight line and can't be closed. contradiction.
- I've read that convex curves curvature is $\geq0$ but I can't find a proper proof for this.
Therefore, in conclusion I get that for any $c>0$ $\kappa(t)$ can be it's curvature.
Did I'm miss something?
Ok. So I've managed to combine a formal answer for this question.
$\gamma$ is convex i.e. simple curve by definition therefore, by Hopf Umlaufsatz theorem taking in account that $\gamma$ is simple and closed we get that the value of $\int_\gamma \kappa(t) dt = +/- 2\pi$.
Therefore by simple computation (and taking in account that $\kappa > 0$ because it is convex) we get that $c=1260\pi$