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Let $(X,d)$ be a uniquely geodesic metric space, i.e. for every $x,y\in X$, there is a unique continuous function $f:I=[0,\ell]\to X$ such that $$f(0)=x,f(\ell)=y, d(f(r),f(s))=|r-s|\ \forall r,s\in I,$$ where $\ell=d(x,y)$. The image of $f$ is called the geodesic path joining $x,y$. The geodesic midpoint of $x,y$ is $f(\frac{1}{2}\ell)$.

Let $A,B,C$ be three points in $X$. Let $M$ be the geodesic midpoint of $A,B$. My question is: Is it always true that $$d(M,C)\le \max\{d(A,C),d(B,C)\}?$$

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A triangular graph is not uniquely geodesic. A graph is uniquely geodesic if and only if it is a tree. – t.b. Nov 7 '11 at 16:45
You are right. I deleted my comment. – TCL Nov 7 '11 at 16:48
up vote 1 down vote accepted

The assertion is false. Let $A=(a,0,-a),B=(0,a,-a), C=(0,0,1)$ on $\mathbb{S}^2$, where $a=1/\sqrt{2}$. Let $X$ be the closed geodesic convex hull of $\{A,B,C\}$. Then $X$ is uniquely geodesic. The geodesic midpoint of $A,B$ in $X$ is $M=\frac{1}{\sqrt{6}}(1,1,-2)$, and $d(M,C)=\cos^{-1}(-2/\sqrt{6})>\frac{3\pi}{4}=d(A,C)=d(B,C)$.

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$S^2$ is not uniquely geodesic: any two antipodal points can be connected by two different paths of the same length. – mjqxxxx Nov 7 '11 at 19:34
You are right. I edited my answer so the space is uniquely geodesic. – TCL Nov 7 '11 at 20:01
It is presumable that that property is related to the (Alexandrov ??) curvature of the space. What do you think? – Valerio Capraro Nov 7 '11 at 20:42
@Valerio: Of course, convexity of the metric is plenty enough (so Busemann-non-positively curved), but I'm not sure about the converse – t.b. Nov 7 '11 at 21:04

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