Theorem 12.2.32DB664A6EFDF

Proof. For any Hermitian metric $\gamma $, the arithmetic–geometric mean inequality for the eigenvalues of $\Omega _\theta $ relative to $\gamma $ gives the pointwise inequality

If $\gamma $ is Gauduchon, integration and Stokes' theorem replace $\Omega _\theta $ in the numerator by $\theta $. Every test quotient is therefore at least $c_\theta (\mu )^{1/n}$.

For $n\geq 2$, take the Gauduchon metric $\gamma =\mathrm{e}^ g\Omega _\theta $ conformal to $\Omega _\theta $, given by theorem 11.4.2, where $g$ is a smooth real function. Since all eigenvalues of $\Omega _\theta $ relative to $\gamma $ are equal at every point, equality holds pointwise and hence in the quotient. This proves the formula and attainment. Conversely, if a Gauduchon metric attains the infimum, the nonnegative smooth difference of the two sides of the pointwise inequality has zero integral. It therefore vanishes everywhere, and the equality case of the arithmetic–geometric mean inequality makes $\gamma $ conformal to $\Omega _\theta $. Since the Gauduchon metric in a conformal class is unique up to a constant factor, this proves the last assertion.

For $n=1$, we integrate the equation and use $\int _ X\mathrm{dd}^{\mathrm{c}}u_\theta =0$. This gives $c_\theta (\mu )=\int _ X\theta $, which equals every quotient in the definition. $\square$