If $V$ has an orthogonal basis with respect to $\|\raisebox{1.72218pt}{.}\|$, then the assertion follows from Lemma 3.3.

In general, by Proposition 1.3, for each $n\in\mathbb{Z}_{>0}$, we choose a basis

$$(e_{n,0},e_{n,1},\ldots,e_{n,r})$$

of $V$ such that

$$(1-1/n)\max\{|c_{0}|\|e_{n,0}\|,\ldots,|c_{r}|\|e_{n,r}\|\}\leq\|c_{0}e_{n,0}+\cdots+c_{r}e_{n,r}\|$$

for all $c_{0},\ldots,c_{r}\in k$. If we set

$$\|c_{0}e_{n,0}+\cdots+c_{r}e_{n,r}\|_{n}:=\max\{|c_{0}|\|e_{n,0}\|,\ldots,|c_{r}|\|e_{n,r}\|\}$$

for $c_{0},\ldots,c_{r}\in k$. Then $(1-1/n)\|\raisebox{1.72218pt}{.}\|_{n}\leq\|\raisebox{1.72218pt}{.}\|\leq\|\raisebox{1.72218pt}{.}\|_{n}$, so that

$$(1-1/n)|\raisebox{1.72218pt}{.}|_{(V,\|\raisebox{1.20552pt}{.}\|_{n})}^{\mathrm{quot}}(x)\leq|\raisebox{1.72218pt}{.}|_{(V,\|\raisebox{1.20552pt}{.}\|)}^{\mathrm{quot}}(x)\leq|\raisebox{1.72218pt}{.}|_{(V,\|\raisebox{1.20552pt}{.}\|_{n})}^{\mathrm{quot}}(x)$$

for all $x\in X^{\mathrm{an}}$. Let $\omega$ be a local basis of $L$ over an open set $U$. Then the above inequalities imply that

$$\log(1-1/n)\leq\log\left(|\omega|_{(V,\|\raisebox{1.20552pt}{.}\|)}^{\mathrm{quot}}(x)\right)-\log\left(|\omega|_{(V,\|\raisebox{1.20552pt}{.}\|_{n})}^{\mathrm{quot}}(x)\right)\leq 0$$

for all $x\in U^{\mathrm{an}}$, which shows that the sequence $\left\{\log\left(|\omega|_{(V,\|\raisebox{1.20552pt}{.}\|_{n})}^{\mathrm{quot}}(x)\right)\right\}_{n=1}^{\infty}$ converges to $\log\left(|\omega|_{(V,\|\raisebox{1.20552pt}{.}\|)}^{\mathrm{quot}}(x)\right)$ uniformly on $U^{\mathrm{an}}$. Thus, by the previous observation, $\log\left(|\omega|_{(V,\|\raisebox{1.20552pt}{.}\|)}^{\mathrm{quot}}(x)\right)$ is continuous on $U^{\mathrm{an}}$. ∎