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Let $\Sigma$ be an Alexandrov space of curvature $\geq 1$ without boundary. Let $X$ be the cone over $\Sigma$. $X$ is well known to be non-negatively curved.

Let $f\colon X\to \mathbb{R}$ be a concave function (i.e. its restriction to any shortest path is concave in the usual sense). Assume in addition that $f$ is 1-homogeneous and $f\geq 0$. Does it follow that $f\equiv 0$?

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Yes.

The fact that $f$ is concave and 1-homogeneous implies that $f$ on $\Sigma$ is $(-f)$-concave. That is, $(f(\gamma(t)))''+f(\gamma(t)) \le 0$ for any shortest geodesic $\gamma$ of $\Sigma$. Consider a minimum point $p$ of $f$ on $\Sigma$, then by solving the ODE one obtains for any $q \in \Sigma$, $$ f(q)\le f(p) \cos|pq|_{\Sigma}. $$

Since $f \ge 0$, it is clear that $f \equiv 0$ on $\Sigma$ and hence on $X$.

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