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8.3. Continuity [01C3]

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8.3. Continuity

Finally we show that φ\varphi is continuous. For this we use capacity estimates in the spirit of Kołodziej [Koł98, Koł03]; see also [EGZ09]. The following result (and its proof) is a translation of [EGZ09, Lemma 2.3].

Lemma 8.3.

Let φ,ψ∈ℰ1​(X,ω)\varphi,\psi\in\mathcal{E}^{1}(X,\omega) with ψ≤0\psi\leq 0. Then

Capω{φ<ψ}≤t−n∫{φ<(1−t)ψ+t}MA(φ)\Capa_{\omega}\{\varphi<\psi\}\leq t^{-n}\int_{\left\{\varphi<(1-t)\psi+t\right\}}\MA(\varphi)

for 0<t<10<t<1.

Proof.

Fix u∈PSH⁡(X,ω)u\in\PSH(X,\omega) with 0≤u≤10\leq u\leq 1 and set ψt:=(1−t)​ψ+t​u\psi_{t}:=(1-t)\psi+tu. We have

{φ<ψ}⊆{φ<ψt}⊆{φ<(1−t)ψ+t}.\{\varphi<\psi\}\subseteq\{\varphi<\psi_{t}\}\subseteq\{\varphi<(1-t)\psi+t\}.

since ψ≤0\psi\leq 0. Now MA⁡(ψt)≥tn​MA⁡(u)\MA(\psi_{t})\geq t^{n}\MA(u) by (3.1), so

tn∫{φ<ψ}MA(u)≤∫{φ<ψ}MA(ψt)≤∫{φ<ψt}MA(ψt)≤∫{φ<ψt}MA(φ)≤∫{φ<(1−t)ψ+t}MA(φ),t^{n}\int_{\{\varphi<\psi\}}\MA(u)\leq\int_{\{\varphi<\psi\}}\MA(\psi_{t})\leq\int_{\{\varphi<\psi_{t}\}}\MA(\psi_{t})\\ \leq\int_{\{\varphi<\psi_{t}\}}\MA(\varphi)\leq\int_{\{\varphi<(1-t)\psi+t\}}\MA(\varphi),

where the third inequality follows from the comparison principle (6.7). Taking the supremum over uu completes the proof. ∎

As a consequence, we get the following version of the ’domination principle’, sufficient for our purpose.

Lemma 8.4.

Let φ∈PSH⁡(X,ω)∩C0​(X)\varphi\in\PSH(X,\omega)\cap C^{0}(X) and ψ∈ℰ1​(X,ω)\psi\in\mathcal{E}^{1}(X,\omega). Assume that ν:=MA⁡(ψ)\nu:=\MA(\psi) is supported in the dual complex Δ𝒳\Delta_{\mathcal{X}} of some SNC model 𝒳\mathcal{X}, and that φ≤ψ\varphi\leq\psi ν\nu-a.e. Then φ≤ψ\varphi\leq\psi on XX.

Proof.

Upon translating by a constant we may assume that 0≥φ≥−C0\geq\varphi\geq-C. Let ε>0\varepsilon>0. If we choose 0<t≪10<t\ll 1 such that t⁡(C+1)≤ε/2t(C+1)\leq\varepsilon/2 then we have

ν{ψ+ε<(1−t)φ+t}≤ν{ψ+ε/2<φ}=0.\nu\{\psi+\varepsilon<(1-t)\varphi+t\}\leq\nu\{\psi+\varepsilon/2<\varphi\}=0.

By Lemma 8.3 it follows that

Capω{ψ+ε<φ}≤t−nν{ψ+ε<(1−t)φ+t}=0\Capa_{\omega}\{\psi+\varepsilon<\varphi\}\leq t^{-n}\nu\{\psi+\varepsilon<(1-t)\varphi+t\}=0

(since MA⁡(ψ+ε)=ν\MA(\psi+\varepsilon)=\nu). But {ψ+ε<φ}\{\psi+\varepsilon<\varphi\} is open by continuity of φ\varphi, hence empty by Lemma 4.2. We have thus proved that φ≤ψ+ε\varphi\leq\psi+\varepsilon on XX for all ε>0\varepsilon>0, and the result follows. ∎

Now let φ∈ℰ1​(X,ω)\varphi\in\mathcal{E}^{1}(X,\omega) be a solution to MA⁡(φ)=μ\MA(\varphi)=\mu, with μ\mu supported in a dual complex Δ𝒳\Delta_{\mathcal{X}}. We may normalize φ\varphi by supXφ=−1\sup_{X}\varphi=-1. Let (φj)j(\varphi_{j})_{j} be a decreasing net of ω\omega-psh model functions converging to φ\varphi. We are going to show that φj→φ\varphi_{j}\to\varphi uniformly on XX, which will in particular imply that φ\varphi is continuous.

By Theorem 2.10 we have supXφj→supXφ\sup_{X}\varphi_{j}\to\sup_{X}\varphi, so we may assume φj≤0\varphi_{j}\leq 0 for all jj. Fix ε>0\varepsilon>0. Since φ\varphi is continuous on Δ𝒳\Delta_{\mathcal{X}}, the monotone convergence φj→φ\varphi_{j}\to\varphi is uniform on Δ𝒳\Delta_{\mathcal{X}} by Dini’s lemma. We thus have φj≤φ+ε\varphi_{j}\leq\varphi+\varepsilon μ\mu-a.e. for j≫1j\gg 1, and Lemma 8.4 yields φj≤φ+ε\varphi_{j}\leq\varphi+\varepsilon on XX, which concludes the proof.

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