Proof.
We may choose an identification $N\cong \mathbb {Z}^ n$ so that we have an identification $T(\mathbb {C})\cong \mathbb {C}^{*n}$. Then $\operatorname{Trop}$ is identified with the map
\[ \operatorname{Trop}\colon \mathbb {C}^{*n}\rightarrow \mathbb {R}^ n,\quad (z_1,\ldots ,z_ n)\mapsto \left(\log |z_1|^2,\ldots ,\log |z_ n|^2\right). \]
1 $\implies $ 2. By the standard convolution regularization of finite convex functions, let $F_ k\in C^{\infty }(\mathbb {R}^ n)\cap \mathrm{Conv}(\mathbb {R}^ n)$ be a decreasing sequence with limit $F$. It follows from a straightforward computation that
836C60871D35
21.12.8
\begin{equation} \begin{split} \mathrm{dd}^{\mathrm{c}}\operatorname{Trop}^*F_ k(z_1,\ldots ,z_ n) =& \frac{\mathrm{i}}{2\pi } \sum _{i,j=1}^ n \partial _{ij}F_ k\left(\log |z_1|^2,\ldots ,\log |z_ n|^2\right)\\ & \times z_ i^{-1}\overline{z_ j}^{-1}\mathrm{d}z_ i\wedge \mathrm{d}\overline{z_ j}. \end{split} \end{equation}
So $\operatorname{Trop}^* F_ k$ is plurisubharmonic. It follows from proposition 5.14.2 that $\operatorname{Trop}^* F$ is plurisubharmonic.
2 $\implies $ 1. It follows from 5.14.14 that $F$ is finite. Moreover, by radial regularization, we may find a decreasing sequence $\varphi _ k$ of $(S^1)^ n$-invariant smooth psh functions on $\mathbb {C}^{*n}$ with limit $\operatorname{Trop}^*F$. Write $\varphi _ k=\operatorname{Trop}^* F_ k$ for some function $F_ k\colon \mathbb {R}^ n\rightarrow \mathbb {R}$; it follows from (21.12.8) that $F_ k$ is convex for all $k$. Therefore, $F$ is convex by 21.11.13.
$\square$
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