Upon multiplying by $\varpi^{m}$ with $m\gg 1$, we may assume that $\mathfrak{a}\subset\mathcal{O}_{\mathcal{X}}$ is a vertical ideal sheaf. Pick $J\subset I$ such that $E_{J}$ is non-empty, choose a point $\xi\in E_{J}$ and let $f_{1},\dots,f_{m}$ be generators of $\mathfrak{a}$ at $\xi$. With the notation introduced in the proof of Theorem 3.1 we then have

(3.5)

$$\log|\mathfrak{a}|(\emb_{\mathcal{X}}(s))=\max\left\{-\val_{\mathcal{X},s}(f_{i}),\,i=1,\dots,m\right\}.$$

By (3.4) each function $s\mapsto-\val_{\mathcal{X},s}(f_{i})$ is piecewise affine and convex on $\sigma_{J}$, proving (i). To prove (ii), pick any $x\in X$, set $\xi:=c_{\mathcal{X}}(x)$ and let $J\subset I$ be the set of indices $j\in I$ such that $\xi\in E_{j}$. Arguing similarly with generators of $\mathfrak{a}$, it is enough to show that $|f(x)|\leq|f(p_{\mathcal{X}}(x))|$ for each $f\in\mathcal{O}_{\mathcal{X},\xi}$. Note that the seminorm $f\mapsto|f(x)|$ extends by continuity to $\widehat{\mathcal{O}}_{\mathcal{X},\xi}$ since $\xi=c_{\mathcal{X}}(x)$. Writing in the notation of Remark 3.8 $f=\sum_{\alpha\in\mathbf{N}^{L}}f_{\alpha}z^{\alpha}\in\widehat{\mathcal{O}}_{\mathcal{X},\xi}$ we then have

$$|f(x)|\leq\sup_{f_{\alpha}\neq 0}\prod_{j\in J}|z_{j}(x)|^{\alpha_{j}}$$

by the ultrametric property, using that $|f_{\alpha}(x)|=1$ since each non-zero $f_{\alpha}\in\mathcal{O}_{\mathcal{X},\xi}$ is a unit. On the other hand, if we set $s_{j}:=-\log|z_{j}(x)|$ for $j\in J$ then we have by definition $p_{\mathcal{X}}(x)=\emb_{\mathcal{X}}(s)$, hence

$$\sup_{f_{\alpha}\neq 0}\prod_{j\in J}|z_{j}(x)|^{\alpha_{j}}=|f(p_{\mathcal{X}}(x))|$$

and the result follows. ∎