We set $c_{i}=\|e_{i}\|$ for $i=0,\ldots,r$. Without loss of generality, we may assume that $j=0$, that is, we need to show that

$$|\tilde{e}_{0}|_{\overline{V}}^{\mathrm{quot}}(x)=\frac{1}{\max\{1/c_{0},|a_{10}|_{x}/c_{1},\ldots,|a_{r0}|_{x}/c_{r}\}}.$$

Since

$$\ker(\pi_{x}:V\otimes_{k}\hat{\kappa}(x)\to L\otimes_{\mathscr{O}_{X}}\hat{\kappa}(x))=\langle e_{1}-a_{10}(x)e_{0},\ldots,e_{r}-a_{r0}(x)e_{0}\rangle$$

for $x\in U^{\mathrm{an}}_{0}$, we have

$$|\tilde{e}_{0}|_{\overline{V}}^{\mathrm{quot}}(x)=\inf\left.\left\{f(\lambda_{1},\ldots,\lambda_{r})\ \right|\ (\lambda_{1},\ldots,\lambda_{r})\in\hat{\kappa}(x)^{r}\right\},$$

where $f(\lambda_{1},\ldots,\lambda_{r}):=\big\|e_{0}+\sum_{i=1}^{r}\lambda_{i}(e_{i}-a_{i0}(x)e_{0})\big\|_{\hat{\kappa}(x)}$. Note that

$$f(\lambda_{1},\ldots,\lambda_{r})=\max\left\{c_{0}\left|1-\sum\nolimits_{i=1}^{r}\lambda_{i}a_{i0}(x)\right|_{x},c_{1}|\lambda_{1}|_{x},\ldots,c_{r}|\lambda_{r}|_{x}\right\}.$$

As

$$\max\{\alpha_{0},\ldots,\alpha_{r}\}\max\{\beta_{0},\ldots,\beta_{r}\}\geq\max\{\alpha_{0}\beta_{0},\ldots,\alpha_{r}\beta_{r}\}$$

for $\alpha_{0},\ldots,\alpha_{r},\beta_{0},\ldots,\beta_{r}\in\mathbb{R}_{\geq 0}$, we have

$$f(\lambda_{1},\ldots,\lambda_{r})\cdot\max\{1/c_{0},|a_{10}(x)|_{x}/c_{1},\ldots,|a_{r0}(x)|_{x}/c_{r}\}\\ \geq\max\left\{\left|1-\sum\nolimits_{i=1}^{r}\lambda_{i}a_{i0}(x)\right|_{x},|\lambda_{1}a_{10}(x)|_{x},\ldots,|\lambda_{r}a_{r0}(x)|_{x}\right\}\\ \geq\left|1-\sum\nolimits_{i=1}^{r}\lambda_{i}a_{i0}(x)+\sum\nolimits_{i=1}^{r}\lambda_{i}a_{i0}(x)\right|_{x}=1.$$

Therefore, we obtain

$$\inf\left.\left\{f(\lambda_{1},\ldots,\lambda_{r})\ \right|\ (\lambda_{1},\ldots,\lambda_{r})\in\hat{\kappa}(x)^{n}\right\}\\ \geq\frac{1}{\max\{1/c_{0},|a_{10}(x)|_{x}/c_{1},\ldots,|a_{r0}(x)|_{x}/c_{r}\}}.$$

We need to see that

$$f(\eta_{1},\ldots,\eta_{r})=\frac{1}{\max\{1/c_{0},|a_{10}(x)|_{x}/c_{1},\ldots,|a_{r0}(x)|_{x}/c_{r}\}}.$$

for some $\eta_{1},\ldots,\eta_{r}\in\hat{\kappa}(x)$. As $f(0,\ldots,0)=c_{0}$, the assertion holds if

$$\max\{1/c_{0},|a_{10}(x)|_{x}/c_{1},\ldots,|a_{r0}(x)|_{x}/c_{r}\}=1/c_{0}.$$

Next we assume that

$$\max\{1/c_{0},|a_{10}(x)|_{x}/c_{1},\ldots,|a_{r0}(x)|_{x}/c_{r}\}=|a_{i0}(x)|_{x}/c_{i}$$

for some $i$. Clearly $a_{i0}(x)\not=0$. If we set

$$\eta_{j}=\begin{cases}0&\text{if $j\not=i$},\\ 1/a_{i0}(x)&\text{if $j=i$},\end{cases}$$

then $f(\eta_{1},\ldots,\eta_{n})=c_{i}/|a_{i0}(x)|_{x}$, as required. ∎