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I am looking for a proof of the following statement:

Let $X$ be a complete Busemann space. For any point $x\in X$ and any nonempty closed convex set $A\subseteq X$, there is a unique $a\in A$ such that $d(x,a) = d(x,A)$.

Some context:

By a Busemann space, I mean a geodesic metric space such that for any two affinely parametrized geodesics, $\gamma_0$ and $\gamma_1$, the map $(s,t)\mapsto d(\gamma_0(s),\gamma_1(t))$ is convex. In Papadopoulos' Metric Spaces, Convexity and Nonpositive Curvature, the above statement is given as Proposition 8.4.8. However, I found the proof rather unconvincing as it addresses uniqueness but not existence (existence is not clarified earlier in the text). In Bridson and Haefliger's Metric Spaces of Non-positive Curvature there is a nice proof for CAT(0) spaces, but it doesn't seem to directly generalize to the weaker curvature condition of Busemann spaces.

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  • $\begingroup$ If I'm not mistaken, one can argue that $d^2/2$ is strictly convex if $d$ is convex. And with strict convexity, for every $x$ we deduce the uniqueness of $argmin_{a\in A} d^2(x,a)/2$, from which uniqueness of $argmin_{a\in A} d(x,a)$ follows. The strict convexity of quadratic distance $d^2/2$ versus the convexity of $d$ is basically why $L^2$ optimal transport is so regular and $L^1$ optimal transport more difficult. $\endgroup$
    – JHM
    Commented Jan 8, 2021 at 0:46
  • $\begingroup$ @JHM Using the strict convexity of $d^2$ makes sense for uniqueness. It's more the existence part that is evading me. $\endgroup$
    – Logan Fox
    Commented Jan 8, 2021 at 18:06
  • $\begingroup$ The infimum $\inf_{a\in A} d(x,a)$ obviously exists since $d\geq 0$. If $d$ is proper, then every minimizing sequence will be contained in a compact ball, and there will exist a convergent subsequence. Without $d$ proper, there is another possibility, using fact that convex hulls of finite subsets is always compact in nonpositive curvature (convex hull of finite subset is the continuous image of a compact simplex). If $d$ nonproper and $A$ noncompact, then i think minimum possibly does not exist. $\endgroup$
    – JHM
    Commented Jan 9, 2021 at 0:53

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It turns out this statement is not true without additional conditions on the space. See this counterexample for a closed convex set in as strictly convex Banach space which does not admit a metric projection from the origin (every strictly convex Banach space is Busemann).

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