Let $(X,d)$ be a metric space. A barycenter of a Borel probability measure $\mu$ on $X$ is a minimizer of the function \begin{equation} \begin{split} f \colon X & \to \mathbb{R}\\ x &\mapsto \frac 12 \int_X d(x,\cdot)^2 \mathop{}\!\mathrm{d}\mu~. \end{split} \end{equation} NB: A barycenter is also sometimes called a center of mass, a Fréchet mean or a Karcher mean.

In general, neither existence nor uniqueness of barycenters hold.

I am interested in the case where $X$ is a possibly incomplete simply connected Riemannian manifold (without boundary) of negative sectional curvature, and moreover is uniquely geodesic (any two points are connected by a unique minimizing geodesic). It is not hard to show that uniqueness of barycenters holds, because $f$ is a strictly convex function. I wonder if existence holds, especially in the case where $\mu$ is finitely supported (I would like to define the barycenter of any finite set of points).

Question: Does existence hold in this case?

Here are a couple more observations. Note that I could have merely assumed $X$ to be a metric space -- a (possibly incomplete) uniquely geodesic (locally compact) $\mathrm{CAT}(0)$ metric space, but maybe the techniques of Riemannian geometry can be useful. In the setting of metric spaces, the papers that I have looked at always assume completeness. Karcher's original proof of existence in the setting of Riemannian manifolds also assumes completeness, but maybe it can be adapted. It consists in observing that if the support of $\mu$ is contained in some closed convex ball $B = \overline{B(x_0,r)}$, then the gradient of $f$ which is given by \begin{equation} \mathrm{grad} f (x) = -\int_X \exp_x^{-1}(\cdot) \mathop{}\!\mathrm{d}\mu \end{equation} is pointing outwards on the boundary of $B$, therefore a minimum of $f$ on $B$ must be attained in the interior. In my situation, the problem is that a closed ball is not necessarily compact. I thought I could adapt this argument by replacing a convex ball by a compact convex set, but I am unsure whether it is true that any finite set of points is contained in some compact convex set in this class of manifolds...

Edit: Misha pointed out to me in the comments below that Teichmüller space with the Weil-Petersson metric is exhausted by compact convex sets. S. Wolpert proves it in "Geodesic length functions and the Nielsen problem". This solves my problem in this particular situation (according to the discussion above). I wonder that a uniquely geodesic simply connected Riemannian manifold of negative sectional curvature is always exhausted by compact convex sets.

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    $\begingroup$ Just to be sure, by "uniquely geodesic" you imply both existence and uniqueness of geodesics, right? $\endgroup$ – Benoît Kloeckner Jul 29 '17 at 8:40
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    $\begingroup$ The naturally occurring examples of such incomplete metrics that I know also satisfy a stronger condition that the manifold is also exhausted by compact convex subsets. (The most interesting one is the Weil-Peterson metric.) Is this satisfied by by chance? If so, then you will have a cheap proof of existence of a barycenter for compactly supported measures. (For CAT(0) spaces, the naturally occurring examples are also some Bruhat-Tits buildings.) What is the context in which incomplete examples appear in your case? $\endgroup$ – Misha Jul 29 '17 at 8:52
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    $\begingroup$ Actually, you just need that your vectorfield $\mathrm{grad}f$ is complete to prove existence of a centre of mass. In the case of a complete manifold (and simply connected, negatively curved), you can prove that this vectorfield is complete and that any flow converges to its unique zero (which is, as you know the barycenter). The thing is that if it is complete, you can prove a priori that any flow converges to a zero, and as you said, you already know that there is a unique zero. Completeness should be given by conditions such as the ones Misha is talking about. $\endgroup$ – M. Dus Jul 29 '17 at 9:29
  • $\begingroup$ @BenoîtKloeckner: yes $\endgroup$ – seub Jul 29 '17 at 14:42
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    $\begingroup$ @M. Dus: Ok, I think I see. It is probably the same proof (based on the observation that the gradient is outward pointing on the boundary) $\endgroup$ – seub Jul 29 '17 at 15:07

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