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Suppose $Y$ is a proper, uniquely geodesic metric space. In such a space, is any local geodesic in fact a geodesic? Here the terms "geodesic" and "local geodesic" are taken in the metric sense: a geodesic is a globally length-minimizing path, whereas a local geodesic is a locally length-minimizing path.

I suspect the answer is, in general, negative (so that there may exist local geodesics which are not geodesics) but I haven't been able to come up with an example to see this. Also, if a counterexample exists, can it be taken to be a Riemannian manifold?

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  • $\begingroup$ Cross-posted at Mathoverflow here; see that post for a counter-example. $\endgroup$ Jan 28 at 22:38

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Not an answer, but too long for a comment:

Given a complete Riemannian manifold $M$ such that every two points are joined by a unique minimizing geodesic it follows that for a fixed $p \in M$ every point $q \in M$ different from $p$ is a regular point for the distance function $d(p,.)$. As a consequence $M$ is diffeomorphic to $\mathbb R^n$. So if there is a counterexample among Riemannian manifolds it is obtained via some metric on $\mathbb R^n$.

A good source for this (not found online) is Groves "critical point theory for distance functions". The principle ideas can be extracted from http://www-personal.umich.edu/~wangzuoq/635W12/Notes/Lec%2037.pdf, in particular Lemma 2.1.

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