# A question on Hilbert geometries as metric-measure spaces

Recall that a Hilbert geometry is the interior of a convex body $K \subset \mathbb{R}^n$ provided with the metric $$d(x,y) = \frac{1}{2} \ln\left(\frac{|x-b|}{|y-b|}\frac{|y-a|}{|x-a|}\right) ,$$ where $a$ and $b$ are the points of intersection of the boundary of $K$ and the straight line determined by $x$ and $y$ with the provision that $x$ lie in the segment $ay$ (and $y$ lie in the segment $xb$).

Question. Given two metric balls $B_1$ and $B_2$ of finite Hausdorff measure $\nu > 0$ in a Hilbert geometry $(K,d_K)$, is it true that the Hausdorff measure of $(B_1 + B_2)/2$, defined as the set formed by the midpoints for the Hilbert metric of all line segments having one endpoint in $B_1$ and another endpoint in $B_2$, is never greater than $\nu$?

The idea is to see if Hilbert geometries behave as Hadamard manifolds in some sense.

• I have a feeling that the answer is "yes", but I am not so well-versed in this area to qualify my answer. My hunch is, however, you might need to qualify the notion of "midpoint" (I'm presuming it is the argmin in terms of $d^2(x,a)+d^2(x,b)$)... – Suvrit Apr 30 '14 at 14:22
• I thought about the metric midpoint (for the Hilbert metric) of the line segment. The distances to $a$ and $b$ are infinite. – alvarezpaiva Apr 30 '14 at 15:04