As in the proof of Proposition 4.99, it is enough to prove equation (5.76). By replacing $\psi$ by $\psi-m_{\sigma}$, we can assume without loss of generality that $m_{\sigma}=0$. By the continuity of the metric, the function $\psi$ can be extended to a continuous function ${\overline{\psi}}_{\sigma}$ on $N_{\sigma}$. Fix $u_{0}\in N(\sigma)_{\mathbb{R}}$, write $s={\overline{\psi}}_{\sigma}(u_{0})$ and let $u\in N_{\mathbb{R}}$ such that $\pi_{\sigma}(u)=u_{0}$. By definition

$$(\pi_{\sigma})_{\ast}(\psi)(u_{0})=\sup_{p\in\mathbb{R}\sigma}\psi(u+p).$$

It is clear that $\sup_{p\in\mathbb{R}\sigma}\psi(u+p)\geq s$. Suppose that $\sup_{p\in\mathbb{R}\sigma}\psi(u+p)>s$. Let $q\in\mathbb{R}\sigma$ such that $\psi(u+q)>s$ and let $\varepsilon=(\psi(u+q)-s)/2$. By the definition of the topology of $N_{\sigma}$, there exists a $p\in\mathbb{R}\sigma$ such that

(5.78)

$$s-\varepsilon<\psi(u+p+\sigma)<s+\varepsilon.$$

Since $\sigma$ is a cone of maximal dimension in $\mathbb{R}\sigma$, there exists a point $r\in(q+\sigma)\cap(p+\sigma)$. By the right inequality of equation (5.78) $\psi(u+r)<\psi(u+q)$. By concavity of $\psi$ this implies that

(5.79)

$$\lim_{\lambda\to\infty}\psi(u+r+\lambda(r-q))=-\infty.$$

Since, by construction $u+r+\mathbb{R}_{\geq 0}(r-q)$ is contained in $u+p+\sigma$, equation (5.79) contradicts the left inequality of equation (5.78). Hence $\sup_{p\in\mathbb{R}\sigma}\psi(u+p)=s$, which proves equation (5.76). ∎