existence of hemisphere If $ \beta: (a,b)\rightarrow \mathbb{S}^2 $ is a simple closed curve such that $ \int_a^b\Vert\beta^{\prime}(t) \Vert dt<2\pi$ then there is an open hemisphere (or any rotation of this) containing   the image of $\beta $
thanks!
 A: Here's a constructive proof that feels like it's more complicated than necessary, but I wasn't able to find an easier argument.
Assume without loss of generality that $\beta$ has constant speed $|\dot \beta| = v$ and parametrized on $S^1 = \mathbb R / 2 \pi \mathbb Z$, so that the length condition becomes $v < 1$. Use the usual unit sphere $\mathbb S^2 \subset \mathbb R^3$. 
I will prove that the "centre of mass vector"
$$ X = \int_{\mathbb S^1} \gamma(t) dt \in \mathbb R^3$$
defines a hemisphere $\{ p \in \mathbb S^2 : p \cdot X > 0 \}$ containing $\gamma$. (The integral here is to be understood component-wise in $\mathbb R^3$.)
This involves showing that for every $s$, 
  $$\gamma(s) \cdot X = \int_{s-\pi}^{s+\pi} \gamma(s) \cdot \gamma(t) dt > 0.$$
Since $\gamma$ has speed $v$, $\gamma(t)$ makes an angle of at most $v |t-s|$ with $\gamma(s)$ and so $\gamma(s)\cdot \gamma(t) \ge \cos(v |t-s|)$. Thus
$$ \gamma(s) \cdot X \ge \int_{-\pi}^{\pi} \cos(vt)dt = \frac1v \int_{-v\pi}^{v\pi} \cos(z) dz = \frac2v \sin(v \pi) > 0$$
since $0 < v < 1$.
A: R.A. Horn gave an elementary proof in his article On Fenchel's theorem (American Mathematical Monthly, 78, 1971, 380-381). The argument goes as follow:


*

*Let $\Gamma$ be a curve on $\mathbb{S}^2 \subset \mathbb{R}^3$ of length $<2 \pi$. Let $P,Q \in \Gamma$ cutting $\Gamma$ in two pieces $\Gamma_1, \Gamma_2$ of equal length, ie. $$\mathrm{lg}(\Gamma_1)= \mathrm{lg}(\Gamma_2)= \frac{1}{2} \mathrm{lg}(\Gamma).$$ Thanks to a rotation, we may suppose without loss of generality that $P$ and $Q$ are symmetric with respect to the North pole $N$.

*Suppose that $\Gamma_1$ intersects the equator through a point $R$; let $R'$ and $\Gamma_2'$ denote respectively the images of $R$ and $\Gamma_1$ through the symmetry with respect to $N$. Let $\Gamma':= \Gamma_1 \cup \Gamma_2'$. Notice that $$\mathrm{lg}(\Gamma')= \mathrm{lg}(\Gamma).$$

*Because $\Gamma'$ contains two arcs joining the antipodal points $R$ and $R'$, we deduce $$\mathrm{lg}(\Gamma)= \mathrm{lg}(\Gamma') \geq 2d(R,R')=2 \pi.$$

*Therefore, $\Gamma_1$ cannot intersect the equator; of course, the same thing happens for $\Gamma_2$. Finally, we deduce that $\Gamma$ is contained in the upper hemisphere (after the previous rotation).
