Fenchel's theorem

In differential geometry, Fenchel's theorem is an inequality on the total absolute curvature of a closed smooth space curve, stating that it is always at least $$2\pi$$. Equivalently, the average curvature is at least $$ 2 \pi/L$$, where $$L$$ is the length of the curve. The only curves of this type whose total absolute curvature equals $$2\pi$$ and whose average curvature equals $$ 2 \pi/L$$ are the plane convex curves. The theorem is named after Werner Fenchel, who published it in 1929.

The Fenchel theorem is enhanced by the Fáry–Milnor theorem, which says that if a closed smooth simple space curve is nontrivially knotted, then the total absolute curvature is greater than $4&pi;$.

Proof
Given a closed smooth curve $$\alpha:[0,L]\to\mathbb{R}^3$$ with unit speed, the velocity $$\gamma=\dot\alpha:[0,L]\to\mathbb{S}^2$$ is also a closed smooth curve (called tangent indicatrix). The total absolute curvature is its length $$l(\gamma)$$.

The curve $$\gamma$$ does not lie in an open hemisphere. If so, then there is $$v\in\mathbb{S}^2$$ such that $$\gamma\cdot v>0$$, so $$\textstyle0=(\alpha(1)-\alpha(0))\cdot v=\int_0^L\gamma(t)\cdot v\,\mathrm{d}t>0$$, a contradiction. This also shows that if $$\gamma$$ lies in a closed hemisphere, then $$\gamma\cdot v\equiv0$$, so $$\alpha$$ is a plane curve.

Consider a point $$\gamma(T)$$ such that curves $$\gamma([0,T])$$ and $$\gamma([T,L])$$ have the same length. By rotating the sphere, we may assume $$\gamma(0)$$ and $$\gamma(T)$$ are symmetric about the axis through the poles. By the previous paragraph, at least one of the two curves $$\gamma([0,T])$$ and $$\gamma([T,L])$$ intersects with the equator at some point $$p$$. We denote this curve by $$\gamma_0$$. Then $$l(\gamma)=2l(\gamma_0)$$.

We reflect $$\gamma_0$$ across the plane through $$\gamma(0)$$, $$\gamma(T)$$, and the north pole, forming a closed curve $$\gamma_1$$ containing antipodal points $$\pm p$$, with length $$l(\gamma_1)=2l(\gamma_0)$$. A curve connecting $$\pm p$$ has length at least $$\pi$$, which is the length of the great semicircle between $$\pm p$$. So $$l(\gamma_1)\ge2\pi$$, and if equality holds then $$\gamma_0$$ does not cross the equator.

Therefore, $$l(\gamma)=2l(\gamma_0)=l(\gamma_1)\ge2\pi$$, and if equality holds then $$\gamma$$ lies in a closed hemisphere, and thus $$\alpha$$ is a plane curve.