First let us see that $|s|_{h}(x)\leq|s|_{\mathscr{L}}(x)$ for $s\in H$. Let $\omega_{\xi}$ be a local basis of $\mathscr{L}$ at $\xi=r_{\mathscr{X}}(x)$. If we set $s=s_{\xi}\omega_{\xi}$, then

$$|s|_{\mathscr{L}}(x)=|s_{\xi}|_{x}.$$

As $s_{\xi}^{-1}s\in\mathscr{L}_{\xi}$ and $\mathscr{H}\otimes_{\mathfrak{o}_{k}}\mathscr{O}_{\mathscr{X},\xi}\to\mathscr{L}_{\xi}$ is surjective, there are $l_{1},\ldots,l_{r}\in\mathscr{H}$ and $a_{1},\ldots,a_{r}\in\mathscr{O}_{\mathscr{X},\xi}$ such that $s_{\xi}^{-1}s=a_{1}l_{1}+\cdots+a_{r}l_{r}$. Therefore,

	$\displaystyle\left|s_{\xi}^{-1}s\right|_{h}(x)$	$\displaystyle\leq\max\left\{|a_{1}l_{1}|_{h}(x),\ldots,|a_{r}l_{r}|_{h}(x)\right\}$
		$\displaystyle=\max\left\{|a_{1}|_{x}|l_{1}|_{h}(x),\ldots,|a_{r}|_{x}|l_{r}|_{h}(x)\right\}\leq 1,$

so that $|s|_{h}(x)\leq|s_{\xi}|_{x}=|s|_{\mathscr{L}}(x)$, as required.

Next let us see that $|l|_{\mathscr{L}}(x)\leq\|l\|_{\hat{\kappa}(x)}$ for all $l\in H\otimes\hat{\kappa}(x)$. By Proposition 1.9, $(e_{1},\ldots,e_{r})$ is an orthonormal basis of $H\otimes\hat{\kappa}(x)$ with respect to $\|\raisebox{1.72218pt}{.}\|_{\hat{\kappa}(x)}$. Thus, if we set $l=a_{1}e_{1}+\cdots+a_{r}e_{r}$ ($a_{1},\ldots,a_{r}\in\hat{\kappa}(x)$), then

	$\displaystyle|l|_{\mathscr{L}}(x)$	$\displaystyle\leq\max\{|a_{1}|_{x}|e_{1}|_{\mathscr{L}}(x),\ldots,|a_{r}|_{x}|e_{r}|_{\mathscr{L}}(x)\}$
		$\displaystyle\leq\max\{|a_{1}|_{x},\ldots,|a_{r}|_{x}\}=\|l\|_{\hat{\kappa}(x)}.$

Finally let us see that $|s|_{\mathscr{L}}(x)\leq|s|_{h}(x)$ for $s\in H$. For $\epsilon>0$, we choose $l\in H\otimes\hat{\kappa}(x)$ such that $l(x)=s(x)$ and $\|l\|_{\hat{\kappa}(x)}\leq e^{\epsilon}|s|_{h}(x)$. Then, by the previous observation,

$$|s|_{\mathscr{L}}(x)=|l|_{\mathscr{L}}(x)\leq\|l\|_{\hat{\kappa}(x)}\leq e^{\epsilon}|s|_{h}(x).$$

Thus the assertion follows. ∎