ScalingStacks

Verified tagged author-source HTML · 1904.03696v1 · cited publication edition alignment unverified.

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}fm​κm=κimaxm∈{0,…,d}⁡{∥fm⋅Tm∥ϕ}=inf∑m∈{0,…,d}fm​κm=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. ∎

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