Let $(M,g), dim(M)=n,$ be a complete Riemannian manifold and let $f(x):=d(x,p)^2, p\in M$ being fixed. I'm interested in conditions where $f$ is strictly or weakly geodesically convex. I see the following:
- Lemma 12.15 from John Lee's "Introduction to Riemannian manifolds" shows that when $(M,g)$ is Cartan-Hadamard, i.e. simply connected with non-positive sectional curvature, $f$ is strictly geodesically convex, i.e. for every geodesic $\gamma\subset M, f\circ \gamma$ is strictly convex, in the sense that its second derivative is strictly positive.
One of the major ingredients in this proof is the Hessian comparison theorem for manifolds with $sec\le 0.$
It seems to me that within a normal neighborhood $U$ of $p,f$ is strictly geodesically convex? Sorry if it's a trivial question to many, but it's been a while I've studied Riemannian geometry (I switched careers and am coming back to research) and thus I'm a bit rusty on some stuff. The reason I think $f$ is strictly geodesically convex in a normal neighborhood is that w.r.t. a normal chart given by normal coordinates $(x_1\dots x_n)$, $f(x)$ can be written as $\sum_{j=1}^{n}x_j^2,$ which has $2.Id_n$ as its Hessian, so $r(x)$ is strictly geodesically convex, but only within $U.$. But then I'm a bit confused: why do we need the Hessian comparison theorems to conclude for the Cartan-Hadamard manifolds that $f$ is strictly convex? Won't the fact that $exp_p$ is a global diffeomorphism (Cartan-Hadamard theorem) alone prove that we can take the normal neighborhood to be all of $M,$ and thus on all of the tangent space $T_pM, f(x)=\sum_{j=1}^{n}x_j^2,$ which is clearly geodesically convex. So where does my argument fail? Please correct me if my argument is wrong?
Is there any other more general results that deals with convexity of the squared distance function?