ScalingStacks

Proof. Since $\lvert\mathord{\cdot}\rvert_{k}$ is discrete, for any $\epsilon>0$, there exists $\boldsymbol{\delta}$ with $|\boldsymbol{\delta}|\leq\epsilon$ such that the elements $\{\alpha(\lVert T_{i}\rVert_{\phi(\boldsymbol{\delta})})\}_{i\in\{0,\dots,d\}}$ are $\mathbb{Q}$-independent in $\mathbb{R}/H(k,\lvert\mathord{\cdot}\rvert)$. By Proposition 5.5, for any $n\in\mathbb{N}$ and any $s_{n}=\sum_{\lvert J\rvert=n}f_{J}\cdot\boldsymbol{T}^{J}\in V_{n}(\mathscr{O}(1))$,

$$\mathrm{e}^{-n\epsilon}\Big\lVert\sum_{\lvert J\rvert=n}f_{J}\cdot\boldsymbol{T}^{J}\Big\rVert_{n\phi(\boldsymbol{\delta})}\leq\Big\lVert\sum_{\lvert J\rvert=n}f_{J}\cdot\boldsymbol{T}^{J}\Big\rVert_{n\phi}\leq\mathrm{e}^{n\epsilon}\Big\lVert\sum_{\lvert J\rvert=n}f_{J}\cdot\boldsymbol{T}^{J}\Big\rVert_{n\phi(\boldsymbol{\delta})}.$$

By Proposition 5.3, one has

$$\max_{\lvert J\rvert=n}\Big\{\mathrm{e}^{-n\epsilon}\lvert f_{J}\rvert\prod_{i\in\{0,\dots,d\}}(\mathrm{e}^{\delta_{i}}r_{i})^{j_{i}}\Big\}\leq\Big\lVert\sum_{\lvert J\rvert=n}f_{J}\cdot\boldsymbol{T}^{J}\Big\rVert_{n\phi}\leq\max_{\lvert J\rvert=n}\Big\{\mathrm{e}^{n\epsilon}\lvert f_{J}\rvert\prod_{i\in\{0,\dots,d\}}(\mathrm{e}^{\delta_{i}}r_{i})^{j_{i}}\Big\}.$$

Fix $n$ and let $\epsilon\to 0$, one gets

$$\Big\lVert\sum_{\lvert J\rvert=n}f_{J}\cdot\boldsymbol{T}^{J}\Big\rVert_{n\phi}=\max_{\lvert J\rvert=n}\Big\{\lvert f_{J}\rvert\prod_{i\in\{0,\dots,d\}}r_{i}^{j_{i}}\Big\}.$$

So one has

$$\Big\vvvert\sum_{\lvert J\rvert<\infty}f_{J}\cdot\boldsymbol{T}^{J}\Big\vvvert_{n\phi}=\sup_{n\in\mathbb{N}}\max_{\lvert J\rvert=n}\Big\{\lvert f_{J}\rvert\prod_{i\in\{0,\dots,d\}}r_{i}^{j_{i}}\Big\}=\max_{\lvert J\rvert<\infty}\Big\{\lvert f_{J}\rvert\prod_{i\in\{0,\dots,d\}}r_{i}^{j_{i}}\Big\}.$$

Hence $\vvvert\mathord{\cdot}\vvvert_{\phi}$ is a Gauss norm of multi-radius $\boldsymbol{r}$ on $V_{{\scriptscriptstyle\bullet}}(\mathscr{O}(1))$. ∎