00LX Claim 5.2. For any i∈{0,…,d}i\in\{0,\dots,d\}, one has ∥Ti∥ϕ=ri=|Ti|ϕ(x(𝒓)).\lVert T_{i}\rVert_{\phi}=r_{i}=\lvert T_{i}\rvert_{\phi}(x(\boldsymbol{r})). In other words, the maximum of the function |Ti|ϕ(x)\lvert T_{i}\rvert_{\phi}(x) on (ℙkd)an(\mathbb{P}^{d}_{k})^{\mathrm{an}} is rir_{i}, and the maximum values of these d+1d+1 functions can be attained at the same point x(𝒓)x(\boldsymbol{r}).
00LY Proof. By the orthogonality of the basis {Ti}i∈{0,…,d}\{T_{i}\}_{i\in\{0,\dots,d\}}, we can compute |Ti|ϕ(x(𝒓))=inf(∑m∈{0,…,d}fm⋅Tm)(x(𝒓))=(Ti)(x(𝒓))(f0,…,fd)∈kd+1∥∑m∈{0,…,d}fm⋅Tm∥ϕ=inf∑m∈{0,…,d}fmκm=κimaxm∈{0,…,d}{∥fm⋅Tm∥ϕ}=inf∑m∈{0,…,d}fmκm=κimaxm∈{0,…,d}{|fm||κm|}=inf∑m∈{0,…,d}fm(κm/κi)=1maxm∈{0,…,d}{|fm||κm/κi|⋅|κi|}=|κi|=ri.\begin{split}\lvert T_{i}\rvert_{\phi}(x(\boldsymbol{r}))&=\inf_{\begin{subarray}{c}(\sum_{m\in\{0,\dots,d\}}f_{m}\cdot T_{m})(x(\boldsymbol{r}))=(T_{i})(x(\boldsymbol{r}))\\ (f_{0},\dots,f_{d})\in k^{d+1}\end{subarray}}\Big\lVert\sum_{m\in\{0,\dots,d\}}f_{m}\cdot T_{m}\Big\rVert_{\phi}\\ &=\inf\limits_{\sum_{m\in\{0,\dots,d\}}f_{m}\kappa_{m}=\kappa_{i}}\max_{m\in\{0,\dots,d\}}\Big\{\lVert f_{m}\cdot T_{m}\rVert_{\phi}\Big\}\\ &=\inf\limits_{\sum_{m\in\{0,\dots,d\}}f_{m}\kappa_{m}=\kappa_{i}}\max_{m\in\{0,\dots,d\}}\Big\{\lvert f_{m}\rvert\lvert\kappa_{m}\rvert\Big\}\\ &=\inf\limits_{\sum_{m\in\{0,\dots,d\}}f_{m}(\kappa_{m}/\kappa_{i})=1}\max_{m\in\{0,\dots,d\}}\Big\{\lvert f_{m}\rvert\lvert\kappa_{m}/\kappa_{i}\rvert\cdot\lvert\kappa_{i}\rvert\Big\}\\ &=\lvert\kappa_{i}\rvert=r_{i}.\end{split} The last equality is obtained by Lemma 3.13. ∎