ScalingStacks

Proof. [028F]

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Proof.

Let

M=−minζ∈ℙn⁡u⁡(ζ)≥0.M=-\min_{\zeta\in{\mathbb{P}}^{n}}u(\zeta)\geq 0.

We work first in an affine chart {zj=1}≡ℂn\{z_{j}=1\}\equiv{\mathbb{C}}^{n}. Let Xj=X∩{zj=1}X_{j}=X\cap\{z_{j}=1\} and let ρj≥0\rho_{j}\geq 0 be the potential of ω\omega in this chart with ρj​(0)=0\rho_{j}(0)=0. Then φ+ρj\varphi+\rho_{j} is psh on XjX_{j} and since u≤0u\leq 0,

φ+ρj+M<u+ρj+M≤11+ε​u+ρj+M​on​Xj.\varphi+\rho_{j}+M<u+\rho_{j}+M\leq\frac{1}{1+\varepsilon}\,u+\rho_{j}+M\;{\rm on}\;X_{j}.

Note that (1+ε)−1​u+ρj+M≥0(1+\varepsilon)^{-1}u+\rho_{j}+M\geq 0 is a continuous psh exhaustion function on ℂn{\mathbb{C}}^{n}. Theorem A yields a psh function ψ~\widetilde{\psi} on ℂn{\mathbb{C}}^{n} so that

ψ~<c1+ε​u+c​ρj+c​M​on​ℂn,ψ~=φ+ρj+M​on​Xj.\widetilde{\psi}<\frac{c}{1+\varepsilon}\,u+c\rho_{j}+cM\;{\rm on}\;{\mathbb{C}}^{n}\;,\;\;\widetilde{\psi}=\varphi+\rho_{j}+M\;{\rm on}\;X_{j}.

The function ψj=ψ~−c​ρj−c​M\psi_{j}=\widetilde{\psi}-c\rho_{j}-cM extends uniquely to a c​ωc\omega-psh function on ℙn{\mathbb{P}}^{n} which verifies

ψj≤c1+ε​u​on​ℙn.\psi_{j}\leq\frac{c}{1+\varepsilon}\,u\;\;{\rm on}\;{\mathbb{P}}^{n}.

Moreover on X∩{zj=1}X\cap\{z_{j}=1\} we have

ψj=φ−(c−1)​ρj−(c−1)​M=φ+(c−1)​θj−(c−1)​M,\psi_{j}=\varphi-(c-1)\rho_{j}-(c-1)M=\varphi+(c-1)\theta_{j}-(c-1)M,

where

θj​(z)=log⁡|zj||z0|2+…+|zn|2.\theta_{j}(z)=\log\frac{|z_{j}|}{\sqrt{|z_{0}|^{2}+\ldots+|z_{n}|^{2}}}\;.

Hence ψj=−∞\psi_{j}=-\infty on X∩{zj=0}X\cap\{z_{j}=0\}.

We finally let ψ=max⁡{ψ0,…,ψn}\psi=\max\{\psi_{0},\ldots,\psi_{n}\}. This is a c​ωc\omega-psh function on ℙn{\mathbb{P}}^{n} which verifies the desired conclusions, since θ=max⁡{θ0,…,θn}\theta=\max\{\theta_{0},\ldots,\theta_{n}\}. ∎

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