# Metric projection on closed convex sets in Busemann space

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.

• 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.
– JHM
Jan 8 '21 at 0:46
• @JHM Using the strict convexity of $d^2$ makes sense for uniqueness. It's more the existence part that is evading me. Jan 8 '21 at 18:06
• 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.
– JHM
Jan 9 '21 at 0:53