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I am looking for a proof or, better, a reference to a proof of the following known fact.

Let $(M,g)$ be a smooth Riemannian manifold with boundary. Assume the sectional curvature of $M$ is at least $\kappa$. Then if $M$ is an Alexandrov space with curvature bounded below then it is locally geodesically convex, namely any shortest path between any two sufficiently close points is a geodesic in $M$.

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    $\begingroup$ In Alexandrov geometry any shortest path is a geodesic by definition. Are you using a Riemannian definition of a geodesic (as the solution to a certain ODE)? $\endgroup$
    – Igor Belegradek
    Commented Jul 20, 2020 at 14:25
  • $\begingroup$ @IgorBelegradek: yes, this is what I meant. $\endgroup$
    – asv
    Commented Jul 20, 2020 at 14:27
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    $\begingroup$ Does the usual proof that a shortest path is a geodesic (in a manifold without boundary) fail? Where? $\endgroup$
    – Igor Belegradek
    Commented Jul 20, 2020 at 14:29
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    $\begingroup$ Manifolds with boundary which are not Alexandrov spaces do not have to be locally geodesically convex because shortest path might be contained in the boundary and thus be a geodesic there rather then in the manifold. Probably more general situation is possible: shortest path might consist of many pieces of geodesics on the manifold and on its boundary. $\endgroup$
    – asv
    Commented Jul 20, 2020 at 14:33
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    $\begingroup$ Why don't the references I gave in your other question mathoverflow.net/questions/365457/… provide an answer? All you need to do is to embed $M$ into a complete Riemannian manifold of the same dimension and apply theorem 1.5 of arxiv.org/abs/1902.00942. For example, double $M$ along the boundary and extend the Riemannian metric on $M$ to a Riemannian metric on the double. $\endgroup$
    – Igor Belegradek
    Commented Jul 20, 2020 at 19:16

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