ScalingStacks

Proof. [00KN]

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Proof.

By assumption, there exists a familly of norms {∥⋅∥n}n∈ℕ\{\lVert\mathord{\cdot}\rVert_{n}\}_{n\in\mathbb{N}} such that uniformly for x∈Xanx\in X^{\mathrm{an}},

limn→∞1n​FS​(∥⋅∥n)​(∗)​(x)=|∗|ϕ​(x),\quad\lim_{\begin{subarray}{c}n\to\infty\end{subarray}}\frac{1}{n}\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n})(\ast)(x)=\lvert\ast\rvert_{\phi}(x),

so for any ϵ>0\epsilon>0, there exists N0∈ℕN_{0}\in\mathbb{N} such that for any n≥N0n\geq N_{0}

dist⁡(n​ϕ,FS⁡(∥⋅∥n))≤n​ϵ,\dist(n\phi,\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n}))\leq n\epsilon,

hence by Lemma 3.9 and 3.10,

dist⁡(FS⁡(∥⋅∥n​ϕ),FS⁡(∥⋅∥FS⁡(∥⋅∥n)))≤n​ϵ.\dist(\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n\phi}),\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n})}))\leq n\epsilon.

By Proposition 3.11, one has OPENFS⁡(∥⋅∥FS⁡(∥⋅∥n)))=FS⁡(∥⋅∥n)\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n})}))=\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n}), so

dist⁡(FS⁡(∥⋅∥n​ϕ),FS⁡(∥⋅∥n))≤n​ϵ,\dist(\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n\phi}),\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n}))\leq n\epsilon,

and

dist⁡(1n​FS​(∥⋅∥n​ϕ),1n​FS​(∥⋅∥n))≤ϵ.\dist(\frac{1}{n}\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n\phi}),\frac{1}{n}\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n}))\leq\epsilon.

Taking limit for n→∞n\to\infty and then for ϵ→0\epsilon\to 0, one has

dist(𝒫(⦀⋅⦀ϕ,ϕ)=0,\dist(\mathcal{P}(\vvvert\mathord{\cdot}\vvvert_{\phi},\phi)=0,

so the two metrics are equal. ∎

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