Geodesic line with endpoints in interior of Riemannian manifold or Alexandrov space

Let $$X$$ be a finite dimensional Alexandrov space with curvature bounded below and non-empty boundary. Let $$\gamma$$ be a shortest geodesic path in $$X$$ whose endpoints belong to the interior of $$X$$.

Is it true that $$\gamma$$ is contained in the interior of $$X$$? Is that true at least under the assumption that $$X$$ is smooth Riemannian manifold with smooth boundary (thus $$X$$ must be locally geodesically convex)?

• The case when $X$ is a smooth compact domain in a Riemannian manifold follows from elementary arguments once you know that $X$ is geodesically convex (which follows from the assumption that $X$ is an Alexandrov space as I mention in the comments to Michor's answer). Namely, if $[x,y]$ is a segment in $X$ with $x,y$ in the interior, then the segments from $x$ to points of a small neighborhood of $y$ fill a neighborhood of $(x,y)$, and since by convexity of $X$ all such segments must lie in $X$, no point of $(x,y)$ lies in $\partial X$. – Igor Belegradek Jul 14 '20 at 2:29

• I wish to add that a smooth compact domain in a Riemannian manifold must be convex if the intrinsic metric on the domain is $CD(K,\infty)$, see Theorem 1.5 in arxiv.org/abs/1902.00942. This applies to every finite-dimensional Alexandrov space by arxiv.org/abs/1003.5948. – Igor Belegradek Jul 14 '20 at 1:59