For $j\in\{0,\dots,d_{n}\}$, let $\lambda_{j}\in\widehat{\kappa}(x)$ be such that $s_{n,j}(x)=\lambda_{j}\cdot e_{n}(x)$. By construction,

$$\begin{split}\lvert e_{n}(x)\rvert_{\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n})}=&\inf_{\mu_{j}\in\widehat{\kappa}(x),\ \sum\mu_{j}\cdot s_{n,j}(x)=e_{n}(x)}\Big\|\sum_{j\in\{0,\dots,d_{n}\}}\mu_{j}\cdot s_{n,j}\Big\|\\ =&\inf_{\mu_{j}\in\widehat{\kappa}(x),\ \sum\mu_{j}\cdot s_{n,j}(x)=e_{n}(x)}\max_{j}\Big\{\lvert\mu_{j}\rvert_{\widehat{\kappa}(x)}\cdot\lVert s_{n,j}\rVert\Big\}\\ =&\inf_{\kappa_{j}\in\widehat{\kappa}(x),\ \sum\kappa_{j}=1}\max_{j}\Big\{\lvert\kappa_{j}\cdot\lambda_{j}^{-1}\rvert_{\widehat{\kappa}(x)}\cdot\lVert s_{n,j}\rVert\Big\}\\ =&\inf_{\kappa_{j}\in\widehat{\kappa}(x),\ \sum\kappa_{j}=1}\max_{j}\Big\{\lvert\kappa_{j}\rvert_{\widehat{\kappa}(x)}\cdot\lvert\lambda_{j}\rvert_{\widehat{\kappa}(x)}^{-1}\cdot\lVert s_{n,j}\rVert\Big\}\\ =&\min_{j\in\{0,\dots,d_{n}\}}\Big\{\lvert\lambda_{j}\rvert_{\widehat{\kappa}(x)}^{-1}\cdot\lVert s_{n,j}\rVert_{n}\Big\}.\end{split}$$

The last equality is obtained with Lemma 3.13. ∎