By Corollary 5.17, we have $\Psi=\operatorname{rec}(\psi_{\infty})$. By equation (3.49), we have $\operatorname{rec}(\psi_{\infty})=\lim_{\lambda\to\infty}\lambda^{-1}\psi_{\infty}(\lambda(u,v))$. Statement (2) follows readily from this and from the expression for $\psi_{\infty}$ in Lemma 8.17.

The function $\Psi$ is strictly concave on $\Sigma$, because ${\mathcal{O}}_{\mathbb{P}(E)}(1)$ is an ample line bundle. Hence $\Sigma=\Pi(\Psi)$ and this is the fan described in statement (1).

Let $(e_{1}^{\vee},\dots,e_{n}^{\vee},f_{1}^{\vee},\dots,f_{r}^{\vee})$ be the dual basis of $M$ induced by the basis of $N$. By Proposition 3.64 and statement (2), we have

$$\Delta=\operatorname{conv}\bigg(0,(a_{0}e^{\vee}_{k})_{1\leq k\leq n},(f^{\vee}_{\ell})_{1\leq\ell\leq r},(a_{\ell}e^{\vee}_{k}+f^{\vee}_{\ell})_{\stackrel{{\scriptstyle 1\leq k\leq n}}{{\scriptscriptstyle 1\leq\ell\leq r}}}\bigg).$$

Statement (3) follows readily from this.

For the first part of statement (4), it suffices to compute the Legendre-Fenchel dual of $\psi_{\infty}$ at a point $(x,y)$ in the interior of the polytope. Lemma 8.17 shows that $\psi_{\infty}$ is strictly concave. Hence, by Theorem 3.52(3), $\nabla\psi_{\infty}$ is a homeomorphism between $N_{\mathbb{R}}$ and $\Delta^{\circ}$. Thus, there exist a unique $(u,v)\in N_{\mathbb{R}}$ such that, for $i=1,\dots,n$ and $j=1,\dots,r$,

$$x_{i}=\frac{\partial\psi_{\infty}}{\partial u_{i}}(u,v),\quad y_{j}=\frac{\partial\psi_{\infty}}{\partial v_{j}}(u,v).$$

We use the conventions $x_{0}=L(y)-\sum_{i=1}^{n}x_{i}$, $y_{0}=1-\sum_{j=1}^{r}y_{j}$, and $u_{0}=v_{0}=0$ as before, and also $\eta=\sum_{i=0}^{n}\operatorname{e}^{-2u_{i}}$ and $\psi=\psi_{\infty}$, so that $-2\psi=\log\big(\sum_{j=0}^{r}\operatorname{e}^{-2v_{j}}\eta^{a_{j}}\big)$. Computing the gradient of $\psi$, we obtain, for $i=1,\dots,n$ and $j=1,\dots,r$,

$$x_{i}\operatorname{e}^{-2\psi}=\Big(\sum_{j=0}^{r}a_{j}\eta^{a_{j}-1}\operatorname{e}^{-2v_{j}}\Big)\operatorname{e}^{-2u_{i}},\quad y_{j}\operatorname{e}^{-2\psi}=\eta^{a_{j}}\operatorname{e}^{-2v_{j}}.$$

Combining these expressions, we obtain, for $i=0,\dots,n$ and $j=0,\dots,r$,

$$\frac{x_{i}}{L(y)}=\frac{\operatorname{e}^{-2u_{i}}}{\eta},\quad y_{j}=\frac{\operatorname{e}^{-2v_{j}+2\psi}}{\eta^{a_{j}}}.$$

From the case $i=0$ we deduce $\eta=L(y)/x_{0}$ and from the case $j=0$ it results $2\psi=\log(y_{0})+a_{0}\log(x_{0}/L(y))$. From this, one can verify

$$u_{i}=\frac{1}{2}\log\Big(\frac{x_{0}}{x_{i}}\Big),\quad v_{j}=\frac{1}{2}\log\Big(\frac{y_{0}}{y_{j}}\Big)+\frac{a_{0}-a_{j}}{2}\log\Big(\frac{x_{0}}{L(y)}\Big).$$

From Theorem 3.52(4), we have $\psi^{\vee}(x,y)=\langle x,u\rangle+\langle y,v\rangle-\psi(u,v)$. Inserting the expressions above for $\psi$, $u_{i}$ and $v_{j}$ in terms of $x,y$, we obtain the stated formula.

For $v\neq\infty$, we have $\psi_{v}=\Psi$. The last statement follows from Example 3.16. ∎