Since the condition of being semipositive is closed, it is enough to check it in the open set $X_{0}^{{\text{\rm an}}}$. We choose an integral basis of $M=N^{\vee}$. This determines isomorphisms

$$X_{0}^{{\text{\rm an}}}\simeq(\mathbb{C}^{\times})^{n},\quad X_{0}(\mathbb{R}_{\geq 0})\simeq(\mathbb{R}_{>0})^{n},\quad N_{\mathbb{C}}\simeq\mathbb{C}^{n},\quad N_{\mathbb{R}}\simeq\mathbb{R}^{n}.$$

Let $z_{1},\dots,z_{n}$ be the coordinates of $X_{0}^{{\text{\rm an}}}$ and $u_{1},\dots,u_{n}$ the coordinates of $N_{\mathbb{R}}$ determined by these isomorphisms. With these coordinates the map

$${\operatorname{val}}\colon X_{0}^{{\text{\rm an}}}\to N_{\mathbb{R}}$$

is given by

$${\operatorname{val}}(z_{1},\dots,z_{n})=\frac{-1}{2}(\log(z_{1}\bar{z}_{1}),\dots,\log(z_{n}\bar{z}_{n})).$$

As usual, we denote ${\overline{L}}=(L,\|\cdot\|)$. Set $g=g_{{\overline{L}},s}=\log\|s\|$. Then, the integral valued first Chern class is given by

(5.30)

$$\frac{1}{2\pi i}c_{1}(\overline{L})=\frac{1}{\pi i}\partial\bar{\partial}g=\frac{-i}{\pi}\sum_{k,l}\frac{\partial^{2}g}{\partial z_{k}\partial\bar{z}_{l}}\,\text{\rm d}z_{k}\land\,\text{\rm d}\bar{z}_{l}.$$

The standard orientation of the unit disk $\mathbb{D}\subset\mathbb{C}$ is given by $\,\text{\rm d}x\land\,\text{\rm d}y=(i/2)\,\text{\rm d}z\land\,\text{\rm d}\bar{z}$. Hence, the metric of ${\overline{L}}$ is semipositive if and only if the matrix $G=(\frac{\partial^{2}g}{\partial z_{k}\partial\bar{z}_{l}})_{k,l}$ is semi-negative definite. Since

(5.31)

$$\frac{\partial^{2}g}{\partial z_{k}\partial\bar{z}_{l}}=\frac{1}{4z_{k}\bar{z}_{l}}\frac{\partial^{2}\psi}{\partial u_{k}\partial\bar{u}_{l}},$$

if we write $\operatorname{Hess}(\psi)=(\frac{\partial^{2}\psi}{\partial u_{k}\partial\bar{u}_{l}})_{k,l}$ and $Z=\operatorname{diag}((2z_{1})^{-1},\dots,(2z_{n})^{-1})$, then $G=\bar{Z}^{t}\operatorname{Hess}(\psi)Z$. Therefore $G$ is semi-negative definite if and only if $\operatorname{Hess}(\psi)$ is semi-negative definite, hence, if and only if $\psi$ is concave. ∎