Is the distance function convex over geodesically convex sets?

Question

Let $M$ be a complete Riemannian manifold and $y \in B \subset M$, where $B$ is a geodesically convex baIl. Define $f(x):=d(x,y)^2$ for $x \in B$. Is $f$ geodesically convex?

Answer 1

Not necessarily. Let $M = S^2 \subset \mathbb{R}^3$ be the sphere with its usual metric and $B$ its northern hemisphere. Let $y \in B$ be a point that is close to the equator, and let $\gamma$ be a great circle arc that is a slight tilt of the equator that achieves its maximum $z$-coordinate on the other side of the sphere from $y$. Then the (smooth) function $f \circ \gamma \colon t \mapsto d(\gamma(t), y)^2$ achieves a maximum, so cannot be convex. Therefore, $f$ is not geodesically convex.

More precisely, let $\theta, \phi > 0$ be small positive constants such that $\theta + \phi < \frac{\pi}{2}$. Let $y = (0, \cos \theta, \sin \theta)$ and $\gamma(t) = (\cos t, -\cos \phi \sin t, \sin \phi \sin t)$ for $t \in (0, \pi)$. Then, one can check that $$d(\gamma(t), y)^2 = \arccos(-\cos(\theta + \phi) \sin t).$$ Furthermore, this function achieves a maximum at $t = \frac{\pi}{2}$; hence, it is not convex (alternatively, compute second derivatives).