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Proof of Lemma 3.6 . [015L]

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Proof of Lemma 3.6.

The direct implication follows from the continuity of Log𝒱\operatorname{Log}_{\mathcal{V}}. For the reverse implication, assume that limt→0(Log𝒱)∗​μt=μ0\lim_{t\to 0}(\operatorname{Log}_{\mathcal{V}})_{*}\mu_{t}=\mu_{0} and consider the following three subsets of C0​(𝒱)C^{0}({\mathcal{V}}): A1A_{1} is the set of functions of the form Log𝒱∗​φ\operatorname{Log}_{\mathcal{V}}^{*}\varphi, where φ∈C0​(Δ​(𝒳))\varphi\in C^{0}(\Delta({\mathcal{X}})); A2A_{2} is the set of functions of the form π∗​g\pi^{*}g, where g∈C0​(𝔻r)g\in C^{0}({\mathbb{D}}_{r}); and A3=Cc0​(𝒱∖Δ⁡(𝒳))A_{3}=C^{0}_{c}({\mathcal{V}}\setminus\Delta({\mathcal{X}})) together with the constant function 1. Then the real vector space A⊂C0​(𝒱)A\subset C^{0}({\mathcal{V}}) spanned by functions of the form f1​f2​f3f_{1}f_{2}f_{3}, with fi∈Aif_{i}\in A_{i} is easily seen to be an ℝ{\mathbb{R}}-algebra that separates points and contains all constant functions. By the Stone-Weierstrass Theorem, AA is dense in C0​(𝒱)C^{0}({\mathcal{V}}), so it suffices to prove that lim∫⁡f​μt=∫f​μ0\lim\int f\mu_{t}=\int f\mu_{0} for f∈Af\in A. By linearity, we may assume f=f1​f2​f3f=f_{1}f_{2}f_{3} with fi∈Aif_{i}\in A_{i}. We may further assume f3=1f_{3}=1. Write f1=Log𝒱∗​φf_{1}=\operatorname{Log}_{\mathcal{V}}^{*}\varphi and f2=π∗​gf_{2}=\pi^{*}g. Then

limt→0∫Xtf​μt=limt→0g⁡(t)​∫Xtφ∘Log𝒱⁡μt=limt→0g⁡(t)​∫Δ⁡(𝒳)φ​(Log𝒱)∗​μt=g⁡(0)​∫Δ⁡(𝒳)φ​μ0=∫f​μ0,\lim_{t\to 0}\int_{X_{t}}f\mu_{t}=\lim_{t\to 0}g(t)\int_{X_{t}}\varphi\circ\operatorname{Log}_{\mathcal{V}}\mu_{t}\\ =\lim_{t\to 0}g(t)\int_{\Delta({\mathcal{X}})}\varphi\ (\operatorname{Log}_{\mathcal{V}})_{*}\mu_{t}=g(0)\int_{\Delta({\mathcal{X}})}\varphi\mu_{0}=\int f\mu_{0},

which completes the proof. ∎

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