Let $\Sigma$ be a convex surface with strictly positive curvature in $\mathbb{R}^3$ and $\Gamma \subset \Sigma$ be a closed simple space curve with parameterization $\gamma : \mathbb{S}^1 \rightarrow \mathbb{R}^3$. Assume that a functional $\phi: \mathbb{S}^1 \times \mathbb{S}^1 \rightarrow \mathbb{R}^+$ defined as $\phi(u\_1, u\_2) := \Vert \gamma(u\_2) - \gamma(u\_1) \Vert^2$ has a local minimum at $(u\_1, u\_2)$. I would like to show that tangent vectors $T(u\_1)$ and $T(u\_2)$ must then be collinear.


It is easy to prove that the statement holds true when $\Sigma$ is a sphere and that its higher dimensional analogy for curves lying on hyperspheres in spaces of dimension strictly larger that 3 is false.



Closest points of curves on convex surfaces
I consider the question negatively answered by the counter examples provided in comments of @PietroMajer and @fedja. Thank you.


