First we consider the case $n=1$. Fix $l\in{L(x)}\setminus\{0\}$. For $\epsilon>0$, there is $s\in V\otimes_{k}\hat{\kappa}(x)$ such that $\tilde{s}(x)=l$ and $\|s\|_{\hat{\kappa}(x)}\leq e^{\epsilon}|l|_{h}(x)$.

Note that $\|\tilde{e}\|_{h}\leq\|e\|$ for all $e\in V$. Let $(e_{1},\ldots,e_{r})$ be an $e^{-\epsilon}$-orthogonal basis of $V$ with respect to $\|\raisebox{1.72218pt}{.}\|$. If we set $s=a_{1}e_{1}+\cdots+a_{r}e_{r}$ ($a_{1},\ldots,a_{r}\in\hat{\kappa}(x)$), then, by Proposition 1.9,

	$\displaystyle\|\tilde{s}\|_{h,\hat{\kappa}(x)}$	$\displaystyle\leq\max\{|a_{1}|_{x}\|\tilde{e}_{1}\|_{h},\ldots,|a_{r}|_{x}\|\tilde{e}_{r}\|_{h}\}$
		$\displaystyle\leq\max\{|a_{1}|_{x}\|e_{1}\|,\ldots,|a_{r}|_{x}\|e_{r}\|\}$
		$\displaystyle\leq e^{\epsilon}\|s\|_{\hat{\kappa}(x)},$

so that

$$|l|_{h}^{\mathrm{quot}}(x)\leq\|\tilde{s}\|_{h,\hat{\kappa}(x)}\leq e^{\epsilon}\|s\|_{\hat{\kappa}(x)}\leq e^{2\epsilon}|l|_{h}(x),$$

and hence $|l|_{h}^{\mathrm{quot}}(x)\leq|l|_{h}(x)$ by taking $\epsilon\to 0$. Thus the assertion for $n=1$ follows from (1) in Lemma 3.5.

In general, by using (3) in Lemma 3.5,

$$|l^{n}|_{h^{n}}(x)=\left(|l|_{h}(x)\right)^{n}=\left(|l|_{h}^{\mathrm{quot}}(x)\right)^{n}\geq|l^{n}|_{h^{n}}^{\mathrm{quot}}(x),$$

and hence we have the assertion by (1) in Lemma 3.5. ∎