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user6976

Hi. I simply wanted to add to the above that this is, I think, specific to a product of two trees, it is not true for at least three. Take R^3$\mathbb{R}^3$ with the l^1$l^1$ metric and inside it the plane x+y+z=1$x+y+z=1$. It is convex (because Euclidean segments are geodesics also for l^1$l^1$), not strongly convex, and not median: it contains the points (1,0,0), (0,1,0), (0,0,1)$(1,0,0), (0,1,0), (0,0,1)$ but not their median point (0,0,0)$(0,0,0)$.

Hi. I simply wanted to add to the above that this is, I think, specific to a product of two trees, it is not true for at least three. Take R^3 with the l^1 metric and inside it the plane x+y+z=1. It is convex (because Euclidean segments are geodesics also for l^1), not strongly convex, and not median: it contains the points (1,0,0), (0,1,0), (0,0,1) but not their median point (0,0,0).

I simply wanted to add to the above that this is, I think, specific to a product of two trees, it is not true for at least three. Take $\mathbb{R}^3$ with the $l^1$ metric and inside it the plane $x+y+z=1$. It is convex (because Euclidean segments are geodesics also for $l^1$), not strongly convex, and not median: it contains the points $(1,0,0), (0,1,0), (0,0,1)$ but not their median point $(0,0,0)$.

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Hi. I simply wanted to add to the above that this is, I think, specific to a product of two trees, it is not true for at least three. Take R^3 with the l^1 metric and inside it the plane x+y+z=1. It is convex (because Euclidean segments are geodesics also for l^1), not strongly convex, and not median: it contains the points (1,0,0), (0,1,0), (0,0,1) but not their median point (0,0,0).