Theorem 12.15.4B6CA5669E402

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. $\square$