ScalingStacks

Proof. [032U]

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Proof.

By Stokes theorem, ∫Xd​dc​φ∧T∧ωn−p−1=0\int_{X}dd^{c}\varphi\wedge T\wedge\omega^{n-p-1}=0, hence

‖ωφ∧T‖:=∫Xωφ∧T∧ωn−p−1=∫XT∧ωn−p:=‖T‖.||\omega_{\varphi}\wedge T||:=\int_{X}\omega_{\varphi}\wedge T\wedge\omega^{n-p-1}=\int_{X}T\wedge\omega^{n-p}:=||T||.

Consider now ψ∈L1​(T)\psi\in L^{1}(T). Since TT has measure coefficients, this simply means that ψ\psi is integrable with respect to the total variation of these measures. Assume first ψ≤0\psi\leq 0, φ≥0\varphi\geq 0 and φ,ψ\varphi,\psi are smooth. Then

‖ψ‖L1​(T∧ωφ):=∫X(−ψ)​T∧ωφ∧ωn−p−1=‖ψ‖L1​(T)+∫X(−ψ)​T∧d​dc​φ∧ωn−p−1.||\psi||_{L^{1}(T\wedge\omega_{\varphi})}:=\int_{X}(-\psi)T\wedge\omega_{\varphi}\wedge\omega^{n-p-1}=||\psi||_{L^{1}(T)}+\int_{X}(-\psi)T\wedge dd^{c}\varphi\wedge\omega^{n-p-1}.

Now it follows from Stokes theorem that

∫X(−ψ)​T∧d​dc​φ∧ωn−p−1=∫Xφ​T∧(−d​dc​ψ)∧ωn−p−1\int_{X}(-\psi)T\wedge dd^{c}\varphi\wedge\omega^{n-p-1}=\int_{X}\varphi T\wedge(-dd^{c}\psi)\wedge\omega^{n-p-1}
≤∫Xφ​T∧ωn−p≤supXφ​∫XT∧ωn−p,\leq\int_{X}\varphi T\wedge\omega^{n-p}\leq\sup_{X}\varphi\int_{X}T\wedge\omega^{n-p},

where the forelast inequality follows from φ​T∧ωn−p≥0\varphi T\wedge\omega^{n-p}\geq 0 and −d​dc​ψ≤ω-dd^{c}\psi\leq\omega. This yields

‖ψ‖L1​(T∧ωφ)≤‖ψ‖L1​(T)+supXφ​‖T‖.||\psi||_{L^{1}(T\wedge\omega_{\varphi})}\leq||\psi||_{L^{1}(T)}+\sup_{X}\varphi||T||.

The general case follows by regularizing φ,ψ\varphi,\psi, observing that ωφ=ωφ′\omega_{\varphi}=\omega_{\varphi^{\prime}} where φ′=φ−infXφ≥0\varphi^{\prime}=\varphi-\inf_{X}\varphi\geq 0, and decomposing ψ=ψ′+supXψ\psi=\psi^{\prime}+\sup_{X}\psi with ψ′=ψ−supXψ≤0\psi^{\prime}=\psi-\sup_{X}\psi\leq 0. ∎

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