# Geodesic convexity on Riemannian manifolds with boundary

Let $M$ be a smooth, connected and bounded manifold and let $f:M \rightarrow \mathbb{R}$ be a smooth function.

The conditions to ensure that $f$ is a geodesically convex on $M$ is to ensure firstly that $M$ is geodesically convex manifold. To do that Hopf-Rinow theorem states that $M$ must be a complete topological space. Secondly, the covariant derivative of $f$, $D^2 f$ is non-negative (or equivalently its geodesic hessian matrix $H_g f$ is semi-definite positive).

Assume that $D^2 f$ is non-negative on the topological closure of $M$.

My question is : Is $f$ a geodesically convex function on $\overline{M}$?

Thanks in advance?

• First, there is no reason for $\bar M$ to still be a Riemannian manifold; metric completion loses smoothness most of the time. Second, even if $\bar M$ happens to be smooth, there is no reason for the extension of $f$ to $\bar M$ to be so. In particular, the symbol $D^2 f$ might make no sense. Third, in order to extend $f$ to a continuous function on $\bar M$ you will have to assume it uniformly continuous; assuming $f$ to have compact support seems reasonable. In any case, you will have to make so many assumptions on $M$ and $f$, that the problem becomes irrelevant. – Alex M. Jul 5 at 9:54
• Also, minor nitpick: there is no such thing as "topological completion" because completion is not a property related to the underlying topological structure. In your case, you probably mean "metric completion", in which case your manifold must be Riemannian. In general, completeness is studied in the context of uniform structures (even though the most general setting is that of Cauchy structures). – Alex M. Jul 5 at 9:57
• Dear @ Alex M. Thank for your remarks. – jaogye Jul 6 at 11:42