Proposition 12.15.687CB9377C0BB

Proof. Suppose that a Gauduchon metric $\gamma $ attains the infimum in (12.15.5), and put $m=v_-(\omega )^{1/n}$. The numerator of its quotient is strictly positive, so $m>0$. For the measure $\mu _\gamma =\gamma ^ n/\int _ X\gamma ^ n$, equality holds in the Hölder inequality above, so the test $\gamma $ gives $\lambda _{\mu _\gamma }(\omega )\leq m$. Theorem 12.15.4 gives the opposite inequality. Hence the solution of the smooth equation for $\mu _\gamma $ has mass $m^ n=v_-(\omega )$, which gives a minimizing smooth potential. For $n\geq 2$, $\gamma $ also minimizes the quotient for $\mu _\gamma $, so by theorem 12.2.3 the solution form is conformal to $\gamma $. Since both have volume forms proportional to $\mu _\gamma $, the conformal factor is constant, and the solution metric is Gauduchon.

Conversely, let $v$ be a smooth potential minimizing the mass. For every smooth real $f$, the potentials $v\pm tf$ are admissible for sufficiently small $t>0$. The vanishing of the first variation of the mass therefore gives

Thus $\Omega =\omega +\mathrm{dd}^{\mathrm{c}}v$ is Gauduchon. Taking $\gamma =\Omega $ in the quotient and using Stokes' theorem for the Gauduchon metric $\Omega $ gives

so the infimum in (12.15.5) is attained. In dimension one all metrics are Gauduchon and all masses equal $\int _ X\omega $. $\square$