Write $\psi=\psi_{{\overline{L}},s}$ for short. First we consider the Archimedean case. We have that $q=|p_{r}|^{2}\prod_{i}(z-\xi_{i})^{\ell_{i}}$. By Proposition 8.1,

$$\psi(u)=-\frac{1}{2}\log(q(\operatorname{e}^{-2u}))=-\log|p_{r}|-\frac{1}{2}\sum_{i}\ell_{i}\log|\operatorname{e}^{-2u}-\xi_{i}|,$$

which proves (1). Hence,

$$\psi^{\prime}(u)=\sum_{i}\ell_{i}\frac{1}{1-\xi_{i}\operatorname{e}^{2u}}\quad\text{and}\quad\psi^{\prime\prime}(u)=\sum_{i}2\ell_{i}\frac{\xi_{i}\operatorname{e}^{2u}}{(1-\xi_{i}\operatorname{e}^{2u})^{2}}.$$

The Monge-Ampère measure of $\psi$ is given by $-\psi^{\prime\prime}\,\text{\rm d}u$, and so the above proves (2). To prove (3) we apply Lemma 8.6. We have that $\operatorname{stab}(\psi)=[0,m_{r}]$, $\psi^{\vee}(0)=0$, and $\psi^{\vee}(m_{r})=\log|p_{r}|$. Thus,

(8.9)

$$\operatorname{h}^{\operatorname{tor}}_{{\overline{L}}}(Y)=m_{r}\log|p_{r}|+\int_{-\infty}^{\infty}(m_{r}-\psi^{\prime})\psi^{\prime}\,\text{\rm d}u.$$

We have $\displaystyle m_{r}-\psi^{\prime}(u)=\sum_{i}\ell_{i}\bigg(1-\frac{1}{1-\xi_{i}\operatorname{e}^{2u}}\bigg)=-\sum_{i}\ell_{i}\frac{\xi_{i}\operatorname{e}^{2u}}{1-\xi_{i}\operatorname{e}^{2u}}$. Hence,

$$(m_{r}-\psi^{\prime}(u))\psi^{\prime}(u)=-\bigg(\sum_{i}\ell_{i}\frac{\xi_{i}\operatorname{e}^{2u}}{1-\xi_{i}\operatorname{e}^{2u}}\bigg)\bigg(\sum_{j}\ell_{j}\frac{1}{1-\xi_{j}\operatorname{e}^{2u}}\bigg)\\ =-\sum_{i}\ell_{i}^{2}\frac{\xi_{i}\operatorname{e}^{2u}}{(1-\xi_{i}\operatorname{e}^{2u})^{2}}-\sum_{i\neq j}\ell_{i}\ell_{j}\frac{\xi_{i}\operatorname{e}^{2u}}{(1-\xi_{i}\operatorname{e}^{2u})(1-\xi_{j}\operatorname{e}^{2u})}.$$

Moreover $\displaystyle\int_{-\infty}^{\infty}\frac{\xi_{i}\operatorname{e}^{2u}}{(1-\xi_{i}\operatorname{e}^{2u})^{2}}\,\text{\rm d}u=\bigg[\frac{1}{2(1-\xi_{i}\operatorname{e}^{2u})}\bigg]^{\infty}_{-\infty}=-\frac{1}{2}$ and

$$\int_{-\infty}^{\infty}\frac{\xi_{i}\operatorname{e}^{2u}}{(1-\xi_{i}\operatorname{e}^{2u})(1-\xi_{j}\operatorname{e}^{2u})}\,\text{\rm d}u=\bigg[\frac{\xi_{i}}{2(\xi_{i}-\xi_{j})}(\log(1-\xi_{j}\operatorname{e}^{2u}))-\log(1-\xi_{i}\operatorname{e}^{2u})\bigg]^{\infty}_{-\infty}\\ =\frac{\xi_{i}}{2(\xi_{i}-\xi_{j})}(\log(-\xi_{i})-\log(-\xi_{j})),$$

for the principal determination of $\log$. These calculations together with equation (8.9) imply that

$$\operatorname{h}^{\operatorname{tor}}_{{\overline{L}}}(Y)=m_{r}\log|p_{r}|+\frac{1}{2}\sum_{i}\ell_{i}^{2}+\frac{1}{2}\sum_{i\neq j}\ell_{i}\ell_{j}\frac{\xi_{i}}{\xi_{i}-\xi_{j}}(\log(-\xi_{i})-\log(-\xi_{j}))\\ =m_{r}\log|p_{r}|+\frac{1}{2}\sum_{i}\ell_{i}^{2}+\frac{1}{2}\sum_{i<j}\ell_{i}\ell_{j}\frac{\xi_{i}+\xi_{j}}{\xi_{i}-\xi_{j}}(\log(-\xi_{i})-\log(-\xi_{j})),$$

which proves (3).

Next we consider the non-Archimedean case. Let $U\subset{\overline{K}}^{\times}$ be a sufficiently small open subset and $\zeta\in U$. For short, write $v_{i}={\operatorname{val}}_{{\overline{K}}}(\xi_{i})$. By Proposition 8.1, the genericity of $\zeta$, and the condition $m_{i}\not=m_{j}$ for $i\not=j$, imply

$$\psi({\operatorname{val}}_{{\overline{K}}}(\zeta))=\min_{i}\{0,m_{i}{\operatorname{val}}_{{\overline{K}}}(\zeta)+{\operatorname{val}}_{K}(p_{i})\}={\operatorname{val}}_{{\overline{K}}}(q(\zeta)).$$

By the factorization of $q$,

$${\operatorname{val}}_{{\overline{K}}}(q(\zeta))={\operatorname{val}}_{K}(p_{r})+\sum_{i}\ell_{i}{\operatorname{val}}_{{\overline{K}}}(\zeta-\xi_{i})={\operatorname{val}}_{K}(p_{r})+\sum_{i}\ell_{i}\min\{{\operatorname{val}}_{{\overline{K}}}(\zeta),v_{i}\}.$$

The image of ${\operatorname{val}}_{{\overline{K}}}\colon{\overline{K}}^{\times}\to\mathbb{R}$ is a dense subset. We deduce that, $u\in\mathbb{R}$,

$$\psi(u)={\operatorname{val}}_{K}(p_{r})+\sum_{i}\ell_{i}\min\{u,{\operatorname{val}}_{K}(\xi_{i})\},$$

which proves (4). The gradient of this function is, for $u\in\mathbb{R}$,

$$\partial\psi(u)=\begin{cases}\Big[\sum_{j:v_{j}>v_{i}}\ell_{j},\sum_{j:v_{j}\geq v_{i}}\ell_{j}\Big]&\text{ if }u=v_{i}\text{ for some }i,\\ \sum_{j:v_{j}>x}\ell_{j}&\text{ otherwise.}\end{cases}$$

Hence, the associated Monge-Ampère measure is $\sum_{i}\ell_{i}\delta_{v_{i}},$ which proves (5). The derivative of $\psi$ in the sense of (8.5) is, for $u\in\mathbb{R}$,

$$\psi^{\prime}(u)=\begin{cases}\sum_{j:v_{j}>v_{i}}\ell_{j}+\frac{1}{2}\sum_{j:v_{j}=v_{i}}\ell_{j}&\text{ if }u=v_{i}\text{ for some }i,\\ \sum_{j:v_{j}>x}\ell_{j}&\text{ otherwise.}\end{cases}$$

Moreover, $\operatorname{stab}(\psi)=[0,m_{r}]$, $\psi^{\vee}(0)=0$ and $\psi^{\vee}(m_{r})=-{\operatorname{val}}_{K}(p_{r})$. By Lemma 8.6

(8.10)

$$\operatorname{h}^{\operatorname{tor}}_{{\overline{L}}}(Y)=-m_{r}\lambda_{K}{\operatorname{val}}_{K}(p_{r})+\lambda_{K}\int_{-\infty}^{\infty}(m_{r}-\psi^{\prime})\psi^{\prime}\,\text{\rm d}u.$$

If we write

$$f_{i}(u)=\begin{cases}0,&\text{ if }x\leq v_{i}\\ \ell_{i},&\text{ if }x>v_{i},\end{cases}$$

then, we have that, almost everywhere $\psi^{\prime}(u)=\sum_{i}\ell_{i}-f_{i}(u)$ and $m_{r}-\psi^{\prime}(u)=\sum_{i}f_{i}$. Therefore

(8.11)

$$\int_{-\infty}^{\infty}(m_{r}-\psi^{\prime})\psi^{\prime}\,\text{\rm d}u=\sum_{i,j}\int_{-\infty}^{\infty}f_{i}(\ell_{j}-f_{j})\,\text{\rm d}u=\sum_{i,j}\ell_{i}\ell_{j}\max\{0,v_{j}-v_{i}\}.$$

Thus, joining together (8.10), (8.11) and the relation $\log(|\zeta|)=-\lambda_{K}{\operatorname{val}}_{K}(\zeta)$ we deduce

$$\operatorname{h}^{\operatorname{tor}}_{{\overline{L}}}(Y)=m_{r}\log|p_{r}|+\sum_{i,j}\ell_{i}\ell_{j}\max\{0,\log(|\xi_{i}|/|\xi_{j}|)\},$$

finishing the proof of the theorem. ∎