12.15 Varying the probability measure8B161DDAE77F

Proof. Since $D_\nu (\gamma )=\int _ X(\nu /\mu )^{1/n}(\mu /\gamma ^ n)^{1/n}\gamma ^ n$, the inequalities $m\mu \leq \nu \leq M\mu $ give

for every admissible metric $\gamma $. The numerators of the quotients are nonnegative and do not involve the measure. Dividing and taking infima proves the assertion. $\square$

Proof. For a smooth potential $v$ with $\omega +\mathrm{dd}^{\mathrm{c}}v>0$, put

Then $\mu _ v$ is a smooth positive probability measure, and $v$ solves the normalized equation for $\mu _ v$ with scalar $M_ v$. Uniqueness of the scalar and theorem 12.2.3 give $\mathcal V_{\mu _ v}(\omega )=M_ v$. Hence, if the volume does not depend on the measure, neither does the mass. Conversely, suppose that all these masses equal one constant $M$. For an arbitrary smooth positive probability measure $\mu $, integrating the normalized equation gives $c_\omega (\mu )=M$, so $\mathcal V_\mu (\omega )=M$ for every $\mu $. $\square$

Proof. For each $\mu \in \mathscr M$, the smooth normalized equation realizes $\mathcal V_\mu (\omega )$ as the mass $\int _ X(\omega +\mathrm{dd}^{\mathrm{c}}v)^ n$ of a smooth potential $v$ with $\omega +\mathrm{dd}^{\mathrm{c}}v>0$. Conversely, every such potential realizes its mass as $\mathcal V_{\mu _ v}(\omega )$ for the measure $\mu _ v$ in the proof of proposition 12.15.2. Thus the set of volumes $\mathcal V_\mu (\omega )$, $\mu \in \mathscr M$, coincides with the set of these smooth masses.

It remains to show that the smooth masses have the same extrema as the masses of bounded potentials, which define $v_\pm (\omega )$. If $u\in \mathrm{PSH}(X,\omega )\cap L^\infty $, theorem 5.32.1 gives uniformly bounded smooth strictly $\omega $-plurisubharmonic functions $u_ j$ decreasing to $u$. Their Monge–Ampère measures converge by the monotone continuity of bounded products, and so do their total masses, since $X$ is compact. Every bounded mass is therefore a limit of smooth strictly positive masses. Conversely, smooth potentials are bounded. This proves the first two identities; no uniform bound on the masses is needed.

For the last formula, take $n$th roots and interchange the infima over $\mu $ and over $\gamma $ in theorem 12.2.3. For a fixed Gauduchon metric $\gamma $, Hölder's inequality and $\int _ X\mu =1$ give

with equality for the smooth probability measure $\mu =\gamma ^ n/\int _ X\gamma ^ n$. The numerator is positive and does not depend on $\mu $, so the infimum of the quotient over $\mu $ is obtained by maximizing the denominator. This gives (12.15.5). For $n=1$ both denominators equal one, and the same argument applies. $\square$

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 $. $\square$