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. ∎

Since the measure $c_{1}({\overline{L}})^{n}\land\delta_{X_{\Sigma}}$ is given by a smooth volume form and $X^{{\text{\rm an}}}_{\Sigma}\setminus X^{{\text{\rm an}}}_{0}$ is a set of Lebesgue measure zero, the measure $c_{1}({\overline{L}})^{n}\land\delta_{X_{\Sigma}}$ is determined by its restriction to the dense open subset $X_{0}^{{\text{\rm an}}}$. Thus, to prove equation (5.34) it is enough to show that

(5.35)

$${\operatorname{val}}_{\ast}(c_{1}({\overline{L}})^{n}\land\delta_{X_{\Sigma}}|_{X^{{\text{\rm an}}}_{0}})=n!\mathcal{M}_{M}(\psi).$$

We use the coordinate system of the proof of Proposition 5.29. We denote by ${\widetilde{{\operatorname{\mathbf{e}}}}}\colon N_{\mathbb{C}}\to X_{0}(\mathbb{C})$ the map induced by the morphism $\mathbb{C}\to\mathbb{C}^{\times}$ given by $z\mapsto\exp(-z)$. We write $u_{k}+iv_{k}$ for the complex coordinates of $N_{\mathbb{C}}$. Then

(5.36)

$${\widetilde{{\operatorname{\mathbf{e}}}}}^{\ast}\left(\frac{\,\text{\rm d}z_{k}\land\,\text{\rm d}\bar{z}_{k}}{z_{k}\bar{z}_{k}}\right)=(-2i)\,\text{\rm d}u_{k}\land\,\text{\rm d}v_{k}.$$

Using now equations (5.30), (5.31) and (5.36), we obtain that,

	$\displaystyle\frac{1}{(2\pi i)^{n}}{\widetilde{{\operatorname{\mathbf{e}}}}}^{\ast}c_{1}({\overline{L}})^{n}$	$\displaystyle={\widetilde{{\operatorname{\mathbf{e}}}}}^{\ast}\left(\frac{1}{(i\pi)^{n}}n!\det G\,\text{\rm d}z_{1}\land\,\text{\rm d}\bar{z}_{1}\land\dots\land\,\text{\rm d}z_{n}\land\,\text{\rm d}\bar{z}_{n}\right)$
		$\displaystyle=\frac{(-1)^{n}}{(2\pi)^{n}}n!\det\operatorname{Hess}(\psi)\,\text{\rm d}u_{1}\land\,\text{\rm d}v_{1}\land\dots\land\,\text{\rm d}u_{n}\land\,\text{\rm d}u_{n}.$

Since the map ${\operatorname{val}}$ is the composition of ${\widetilde{{\operatorname{\mathbf{e}}}}}^{-1}$ with the projection $N_{\mathbb{C}}\to N_{\mathbb{R}}$, integrating with respect to the variables $v_{1},\dots,v_{n}$ in the domain $[0,2\pi]^{n}$, taking into account the natural orientation of $\mathbb{C}^{n}$ and the orientation of $N_{\mathbb{R}}$ given by the coordinate system, and the fact that the normalization factor $1/(2\pi i)^{n}$ is implicit in the current $\delta_{X_{\Sigma}}$, we obtain

$${\operatorname{val}}_{\ast}(c_{1}({\overline{L}})^{n}\land\delta_{X_{\Sigma}}|_{X_{0}^{{\text{\rm an}}}})=(-1)^{n}n!\det\operatorname{Hess}(\psi)\,\text{\rm d}u_{1}\land\dots\land\,\text{\rm d}u_{n}.$$

Thus equation (5.35) follows from Proposition 3.94. Finally, the last statement follows from the fact that, in a compact Abelian group there is a unique Haar measure with fixed total volume. ∎

Let $\|\cdot\|^{\prime}$ be any toric smooth metric. Then $\|\cdot\|_{\mathbb{S}}$ extends to a smooth metric if and only if $\log(\|s\|_{\mathbb{S}}/\|s\|^{\prime})$ can be extended to a smooth function on $X_{\Sigma}^{{\text{\rm an}}}$. But we have

$$\log(\|s\|_{\mathbb{S}}/\|s\|^{\prime})=\int_{\mathbb{S}^{{\text{\rm an}}}}\log(\|s(t\cdot p)\|/\|s(t\cdot p)\|^{\prime})\,\text{\rm d}\mu_{\operatorname{Haar}}(t)$$

and the right-hand side can be extended to a smooth function on the whole $X_{\Sigma}^{{\text{\rm an}}}$. Clearly the metric $\|\cdot\|_{\mathbb{S}}$ is toric. Moreover

$$c_{1}(L,\|\cdot\|_{\mathbb{S}})=\int_{\mathbb{S}^{{\text{\rm an}}}}t^{\ast}c_{1}(L,\|\cdot\|)\,\text{\rm d}\mu_{\operatorname{Haar}}(t).$$

Therefore, if $(L,\|\cdot\|)$ is semipositive, then $(L,\|\cdot\|_{\mathbb{S}})$ is semipositive. ∎