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 $J$.

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    $\begingroup$ If you consider the lines $\gamma_1(t)=(t,0)$ and $\gamma_2(t)=(t,t)$ in the plane (which is a CAT(0) space) you see that your function $f$ equals $f(t)=t$ which is not strictly convex. I guess for strict convexity you need strictly negative curvature. $\endgroup$ – Teri Oct 27 '14 at 23:14
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    $\begingroup$ Yes but then $f(0)=0$. I stated the conjecture that if $f>0$ then $f$ is constant or strictly convex. $\endgroup$ – user35593 Oct 28 '14 at 5:42
  • $\begingroup$ I think you are right about your conjecture. Maybe its best to consider spaces of strictly negative curvature and CAT(0) spaces separately. $\endgroup$ – user35593 Oct 28 '14 at 5:47
  • $\begingroup$ Ach, I overlooked the $f>0$ assumption, sorry. By 'my conjecture' I assume you mean the statement 'strict convexity requires strictly negative curvature'. $\endgroup$ – Teri Oct 28 '14 at 14:54

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 $u_4\in[u_2u_3]$.

  • $\begingroup$ Yes but then $a_1,\dots,a_m$ do all lie on the same geodesic (The geodesic connecting $a_1$ and $a_2$). My claim excludes this case. $\endgroup$ – user35593 Oct 28 '14 at 5:42
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    $\begingroup$ I did not see it, but the same construction gives non-uniqueness for $m=3$ and $n=4$; I will update my answer. $\endgroup$ – Anton Petrunin Oct 29 '14 at 1:04

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