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Let $X$ be a Hadamard space. Any two points $x$ and $y$ of $X$ have a unique midpoint $m = m(x,y)$.

Given $x$ and $y$ any two points of $X$, is it always possible to find an affine function $f : X \rightarrow \mathbb{R}$ with $f(x) \neq f(y)$ ? (where by "affine" I mean $\forall x',y' : f(m(x',y')) = \frac{f(x') + f(y')}{2})$ ?

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    $\begingroup$ Is this definition of affine equivalent to the standard one (that a function is affine if its restriction to every geodesic is affine)? If yes, then the answer is given by arxiv.org/abs/math/0511583 "Spaces with many affine functions" by Hitzelberger-Lytchak. $\endgroup$
    – Igor Belegradek
    Commented Dec 19, 2012 at 23:58
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    $\begingroup$ An example (the first example?) of a Hadamard space without this property is $\mathbb H^2$. $\endgroup$
    – Will Sawin
    Commented Dec 20, 2012 at 0:51
  • $\begingroup$ I'd say that the tripod is even an easier example. $\endgroup$
    – Misha
    Commented Dec 20, 2012 at 0:53
  • $\begingroup$ What's the tripod? $\endgroup$
    – Will Sawin
    Commented Dec 20, 2012 at 3:26
  • $\begingroup$ Tripod is the metric graph obtained by gluing three segments at a common vertex. $\endgroup$
    – Misha
    Commented Dec 20, 2012 at 4:02

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