ScalingStacks

Proof. Since the $T^{n}$-action on $(\mathbb{C}^{*})^{n}$ is holomorphic, $\Phi(\zeta_{1}e^{i\theta_{1}},\ldots\zeta_{n}e^{i\theta_{n}})$ is psh in $\zeta$ for any choice of $\theta_{i}$, so the average function $\bar{\Phi}$ is also psh. Any $T^{n}$-invariant psh function must be convex in the log coordinates, because of the formula

$$\sqrt{-1}\partial\bar{\partial}\bar{\Phi}=\frac{1}{4}\sum\frac{\partial^{2}\bar{\Phi}}{\partial x_{i}\partial x_{j}}\sqrt{-1}d\log\zeta_{i}\wedge d\overline{\log\zeta_{j}}\geq 0.$$

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