(1) Fix $l\in{L(x)}\setminus\{0\}$. For $\epsilon>0$, let $(e_{1},\ldots,e_{n})$ be an $e^{-\epsilon}$-orthogonal basis of $H^{0}(X,L)$ with respect to $\|\raisebox{1.72218pt}{.}\|_{h}$. There is $s\in H^{0}(X,L)\otimes_{k}\hat{\kappa}(x)$ such that $s(x)=l$ and $\|s\|_{h,\hat{\kappa}(x)}\leq e^{\epsilon}|l|^{\mathrm{quot}}_{h}(x)$. We set $s=a_{1}e_{1}+\cdots+a_{n}e_{n}$ ($a_{1},\ldots,a_{n}\in\hat{\kappa}(x)$). Then, by Proposition 1.9,

	$\displaystyle\|s\|_{h,\hat{\kappa}(x)}$	$\displaystyle\geq e^{-\epsilon}\max\{|a_{1}|_{x}\|e_{1}\|_{h},\ldots,|a_{n}|_{x}\|e_{n}\|_{h}\}$
		$\displaystyle\geq e^{-\epsilon}\max\{|a_{1}|_{x}|e_{1}|_{h}(x),\ldots,|a_{n}|_{x}|e_{n}|_{h}(x)\}\geq e^{-\epsilon}|l|_{h}(x),$

so that $|l|_{h}(x)\leq e^{2\epsilon}|l|^{\mathrm{quot}}_{h}(x)$, and hence the assertion follows because $\epsilon$ is an arbitrary positive number.

(2) By (1), we have $\|\raisebox{1.72218pt}{.}\|_{h}\leq\|\raisebox{1.72218pt}{.}\|_{h}^{\mathrm{quot}}$. On the other hand, as $|s|^{\mathrm{quot}}_{h}(x)\leq\|s\|_{h}$ for $s\in H^{0}(X,L)$, we have $\|s\|^{\mathrm{quot}}_{h}\leq\|s\|_{h}$.

(3) For $\epsilon>0$, there are $s\in H^{0}(X,L)\otimes_{k}\hat{\kappa}(x)$ and $s^{\prime}\in H^{0}(X,L^{\prime})\otimes_{k}\hat{\kappa}(x)$ such that

$$s(x)=l,\ s^{\prime}(x)=l^{\prime},\ \|s\|_{h,\hat{\kappa}(x)}\leq e^{\epsilon}|l|_{h}^{\mathrm{quot}}(x)\ \text{and}\ \|s^{\prime}\|_{h^{\prime},\hat{\kappa}(x)}\leq e^{\epsilon}|l^{\prime}|_{h^{\prime}}^{\mathrm{quot}}(x).$$

Here let us see that $\|s\cdot s^{\prime}\|_{h\otimes h^{\prime},\hat{\kappa}(x)}\leq e^{2\epsilon}\|s\|_{h,\hat{\kappa}(x)}\|s^{\prime}\|_{h^{\prime},\hat{\kappa}(x)}$. Let $(s_{1},\ldots,s_{m})$ and $(s^{\prime}_{1},\ldots,s^{\prime}_{m^{\prime}})$ be $e^{-\epsilon}$-orthogonal bases of $H^{0}(X,L)$ and $H^{0}(X,L^{\prime})$, respectively. If we set $s=t_{1}s_{1}+\cdots+t_{m}s_{m}$ and $s^{\prime}=t^{\prime}_{1}s^{\prime}_{1}+\cdots+t^{\prime}_{m^{\prime}}s^{\prime}_{m^{\prime}}$ ($t_{1},\ldots,t_{m},t^{\prime}_{1},\ldots,t^{\prime}_{m^{\prime}}\in\hat{\kappa}(x)$), then

$$s\cdot s^{\prime}=\sum_{i,j}t_{i}t^{\prime}_{j}s_{i}\cdot s^{\prime}_{j}.$$

Thus,

	$\displaystyle\|s\cdot s^{\prime}\|_{h\otimes h^{\prime},\hat{\kappa}(x)}$	$\displaystyle\leq\max_{i,j}\left\{|t_{i}|_{x}|t^{\prime}_{j}|_{x}\|s_{i}\cdot s^{\prime}_{j}\|_{h\otimes h^{\prime}}\right\}\leq\max_{i,j}\left\{|t_{i}|_{x}|t^{\prime}_{j}|_{x}\|s_{i}\|_{h}\|s^{\prime}_{j}\|_{h^{\prime}}\right\}$
		$\displaystyle\leq\max_{i}\left\{|t_{i}|_{x}\|s_{i}\|_{h}\right\}\max_{j}\left\{|t^{\prime}_{j}|_{x}\|s^{\prime}_{j}\|_{h^{\prime}}\right\}$
		$\displaystyle\leq e^{2\epsilon}\|s\|_{h,\hat{\kappa}(x)}\|s^{\prime}\|_{h^{\prime},\hat{\kappa}(x)}.$

Therefore, we have $(s\cdot s^{\prime})(x)=l\cdot l^{\prime}$ and

$$|l\cdot l^{\prime}|_{h\otimes h^{\prime}}^{\mathrm{quot}}(x)\leq\|s\cdot s^{\prime}\|_{h\otimes h^{\prime},\hat{\kappa}(x)}\leq e^{2\epsilon}\|s\|_{h,\hat{\kappa}(x)}\|s^{\prime}\|_{h^{\prime},\hat{\kappa}(x)}\leq e^{4\epsilon}|l|_{h}^{\mathrm{quot}}(x)|l^{\prime}|_{h^{\prime}}^{\mathrm{quot}}(x),$$

as required. ∎