distance function on Riemannian manifold $F(x) := dist(x,x^0)$ is geodesic convex?

Question

Assume that $\mathcal{M}$ is a smooth riemannian manifold, define $F(x) := dist(x,x^0)$, where $F:\mathcal{M}\rightarrow \mathbb{R}$, then is it $F$ geodesic convex?

Answer 1

Not necessarily. For example, let $M$ be the unit sphere with its round metric, let $x^0$ be the north pole, and let $x$ and $y$ be two points whose distance from $x^0$ is equal to some constant $r>\pi/2$. If $\gamma:[0,1]\to M$ is a minimizing geodesic from $x$ to $y$, then the distance from $x^0$ to $\gamma(t)$ achieves a strict maximum at $\gamma(1/2)$, which can't happen for a geodesically convex function.

If $M$ is a Cartan-Hadamard manifold (a complete, simply connected Riemannian manifold with nonpositive sectional curvature), then $F$ is geodesically convex. See, for example, Lemma 12.15 in my Introduction to Riemannian Manifolds, 2nd ed., which proves that the squared distance function is strictly geodesically convex. That proof can easily be adapted to show that $F$ is (weakly) geodesically convex.