Write $\|\cdot\|=\|\cdot\|_{\mathcal{X},\mathcal{L},e}$ for short. Let $\mathcal{U}=\operatorname{Spec}({\mathcal{A}})\ni{\operatorname{red}}(p)$ be an open affine trivializing set of $\mathcal{L}$ and $\sigma$ be a generator of $\mathcal{L}|_{{\mathcal{U}}}$. Then $s^{\otimes e}=\lambda\sigma$ with $\lambda$ in the fraction field of ${\mathcal{A}}$. We have that $\lambda(p)\in H$ and, by definition, $\|s(p)\|=|\lambda(p)|^{1/e}$. If $\lambda(p)=0$, the equation is clearly satisfied. Denote temporarily by $C$ the right-hand side of (2.22). If $\lambda(p)\not=0$,

$$\lambda(p)^{-1}\widetilde{p}^{\ast}s^{\otimes e}=\widetilde{p}^{\ast}\sigma\in{\widetilde{p}}^{*}\mathcal{L}.$$

Hence $\|s(p)\|\geq C$. Moreover, if $a\in H^{\times}$ is such that $a^{-1}\widetilde{p}^{\ast}s^{\otimes e}\in{\widetilde{p}}^{*}\mathcal{L}$, then there is an element $\alpha\in H^{\circ}\setminus\{0\}$ with $a^{-1}\widetilde{p}^{\ast}s^{\otimes e}=\alpha\widetilde{p}^{\ast}\sigma$. Therefore, $a=\lambda(p)/\alpha$ and $|a|^{1/e}=|\lambda(p)|^{1/e}/|\alpha|^{1/e}\geq|\lambda(p)|^{1/e}$. Thus, $\|s(p)\|\leq C$. ∎