ScalingStacks

Proof. [01CP]

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Proof.

It is clear that (1)⇒(2)⇒(3)(1)\Rightarrow(2)\Rightarrow(3). The implication (3)⇒(2)(3)\Rightarrow(2) follows from the equality Pθ​(f)−f=Pθ+d​dc​f​(0)P_{\theta}(f)-f=P_{\theta+dd^{c}f}(0). It remains to prove (2)⇒(1)(2)\Rightarrow(1). We may write a given f∈C0​(X)f\in C^{0}(X) as a uniform limit on XX of model functions fjf_{j}, and Pθ​(fj)→Pθ​(f)P_{\theta}(f_{j})\to P_{\theta}(f) uniformly on XX thanks to the Lipschitz property of PθP_{\theta}, see Proposition 2.15. By Theorem 3.1 we thus have

(θ+d​dc​Pθ​(fj))n→(θ+d​dc​Pθ​(f))n\left(\theta+dd^{c}P_{\theta}(f_{j})\right)^{n}\to\left(\theta+dd^{c}P_{\theta}(f)\right)^{n}

in the weak topology of measures. Since Pθ​(fj)−fj→Pθ​(f)−fP_{\theta}(f_{j})-f_{j}\to P_{\theta}(f)-f uniformly on XX and the measures (θ+d​dc​Pθ​(fj))n(\theta+dd^{c}P_{\theta}(f_{j}))^{n} have uniformly bounded (in fact, constant) mass, it follows that

∫(Pθ​(f)−f)​(θ+d​dc​Pθ​(f))n=limj∫(Pθ​(fj)−fj)​(θ+d​dc​Pθ​(fj))n=0.\int\left(P_{\theta}(f)-f\right)\left(\theta+dd^{c}\ P_{\theta}(f)\right)^{n}=\lim_{j}\int\left(P_{\theta}(f_{j})-f_{j}\right)\left(\theta+dd^{c}\ P_{\theta}(f_{j})\right)^{n}=0.

Let us prove the final assertion. Pick θ′∈𝒵1,1​(X)\theta^{\prime}\in\mathcal{Z}^{1,1}(X) such that {θ′}={θ}\{\theta^{\prime}\}=\{\theta\} in N1​(X)N^{1}(X). By the analogue of the d​dcdd^{c}-lemma proved in [BFJ11, Theorem 4.3] there exists g∈𝒟⁡(X)g\in\mathcal{D}(X) such that θ′=θ+d​dc​g\theta^{\prime}=\theta+dd^{c}g. Observe that a function φ\varphi is θ′\theta^{\prime}-psh iff φ+g\varphi+g is θ\theta-psh. As a consequence we get Pθ′​(f)−f=Pθ​(f+g)−(f+g)P_{\theta^{\prime}}(f)-f=P_{\theta}(f+g)-(f+g), hence

∫(Pθ′​(f)−f)​(θ′+d​dc​Pθ′​(f))n=∫(Pθ​(f+g)−(f+g))​(θ+d​dc​Pθ​(f+g))n\int\left(P_{\theta^{\prime}}(f)-f\right)\left(\theta^{\prime}+dd^{c}P_{\theta^{\prime}}(f)\right)^{n}=\int\left(P_{\theta}(f+g)-(f+g)\right)\left(\theta+dd^{c}P_{\theta}(f+g)\right)^{n}

for all f∈C0​(X)f\in C^{0}(X). ∎

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