00KS Proof. For j∈{0,…,dn}j\in\{0,\dots,d_{n}\}, let λj∈κ^(x)\lambda_{j}\in\widehat{\kappa}(x) be such that sn,j(x)=λj⋅en(x)s_{n,j}(x)=\lambda_{j}\cdot e_{n}(x). By construction, |en(x)|FS(∥⋅∥n)=infμj∈κ^(x),∑μj⋅sn,j(x)=en(x)‖∑j∈{0,…,dn}μj⋅sn,j‖=infμj∈κ^(x),∑μj⋅sn,j(x)=en(x)maxj{|μj|κ^(x)⋅∥sn,j∥}=infκj∈κ^(x),∑κj=1maxj{|κj⋅λj−1|κ^(x)⋅∥sn,j∥}=infκj∈κ^(x),∑κj=1maxj{|κj|κ^(x)⋅|λj|κ^(x)−1⋅∥sn,j∥}=minj∈{0,…,dn}{|λj|κ^(x)−1⋅∥sn,j∥n}.\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. ∎