Let $X$ be a hadamard space and $\gamma\_1, \gamma\_2 \colon \mathbb{R}\rightarrow X$ be two geodesics. Part 2 of Coroallary 2.5 in
<http://www.iam.uni-bonn.de/fileadmin/WT/Inhalt/people/Karl-Theodor_Sturm/papers/paper41.pdf>
states that $f(t):=d(\gamma\_1(t),\gamma\_2(t))$ is convex. I wonder under what conditions $f$ is even strictly convex. My conjecture is that if $f>0$ then $f$ is strictly convex or constant. However I dont know how to prove nor where to find such a result. My motivation is that I want to prove uniqueness of a minimizer of a functional $J\colon X^n\rightarrow \mathbb{R}$ of the following form
$$J(u)=\sum\_{i=1}^m d^2(a\_i,u\_i)+\sum\_{(i,j)\in E} d(u\_i,u\_j),$$
where $1\leq m<n$, $a\_1,\dots,a\_m \in X$, , $E\subset \{1,\dots,n\}^2$. I want to prove that if the graph corresponding to $E$ is connected and $a\_1,\dots,a\_m$ do not lie on the same geodesic then there exist a unique minimizer.


Edit::The conjecture about $J$ is wrong, just assume that $i$ is a node with exactly two neighbours $j\_1$, $j\_2$. Then $u\_i$ can be choosen anywhere on the geodesic between $u\_{j\_1}$ and $u\_{j\_2}$ without changing the value of $J$.



Geodesic comparison in Hadamard space
The minimum of $J$ is not unique $m=3$, $n=4$, $E=\{(1,2),(1,3),(2,4),(3,4)\}$ and $X=\mathbb R^2$.


Assume that $a\_1$, $a\_2$ and $a\_3$ are vertices of huge equilateral triangle.
Then at the minimum of  

$$d^2(a\_1,u\_1)+d^2(a\_2,u\_2)+d^2(a\_3,u\_3)+\\+d(u\_1,u\_2)+d(u\_1,u\_3)+d(u\_2,u\_3)$$
the points $u\_i$ lie are vertices of slighlty smaller equilateral triangle insde $\triangle a\_1a\_2a\_3$.


Now, 
$$J(u)= \\
=d^2(a\_1,u\_1)+d^2(a\_2,u\_2)+d^2(a\_3,u\_3)+\\+d(u\_1,u\_2)+d(u\_1,u\_3)+d(u\_2,u\_4)+d(u\_3,u\_4)\ge\\
\ge d^2(a\_1,u\_1)+d^2(a\_2,u\_2)+d^2(a\_3,u\_3)+\\+d(u\_1,u\_2)+d(u\_1,u\_3)+d(u\_2,u\_3)$$
and equality holds if $u\_4\in[u\_2u\_3]$.
So, you should take $u\_1$, $u\_2$ and $u\_3$ as above and any $u\_4\in[u\_2u\_3]$. 


