Theorem 12.13.1.B5C9469E9BB3 For $s<t$, the full Gauduchon infimum is attained by a balanced metric and
\[ \mathcal V_\mu (\theta _ s)=\frac{nW}{H(t-s)},\qquad \mathcal V_\mu (\theta _ t)=\frac{nW}{H(0)}, \]
where the last value is zero if $H(0)=+\infty $. Every minimizing Gauduchon metric $g_ s$ at $s$ has the same subtraction cost
503F3EDAE05A
12.13.2
\begin{equation} \frac{\int _ Xw\wedge g_ s^{n-1}}{D_\mu (g_ s)} =\mathcal C(s) :=n^{-(n-1)/n}W^{1/n} H(t-s)^{-(n+1)/n}J(t-s). \end{equation}
For every selection of these metrics, the improper integral satisfies
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12.13.3
\begin{equation} \int _0^ t\mathcal C(s)\, \mathrm{d}s =\lambda _\mu (\theta _0)-\lambda _\mu (\theta _ t)<\infty . \end{equation}
Proof.
Put $\epsilon =t-s$ and $L=a+\epsilon $. Apply theorem 14.2.1 on the closed positive base to obtain
\[ A_\epsilon =\alpha +\mathrm{dd}^{\mathrm{c}}\phi _\epsilon >0,\qquad A_\epsilon ^ m=\frac{W}{H(\epsilon )}\frac{\sigma }{L}. \]
The target has total mass $W$, so closed Stokes fixes its scalar. The construction of theorem 12.6.32, now without a normalization of $W$ or a nonconstancy requirement on $a$, gives
\[ \Omega _\epsilon =A_\epsilon +Lw,\qquad \Omega _\epsilon ^ n=\frac{nW}{H(\epsilon )}\mu , \qquad \gamma _ s=L^{-1/m}\Omega _\epsilon . \]
Indeed,
\[ \gamma _ s^ m=L^{-1}A_\epsilon ^ m +mA_\epsilon ^{m-1}\wedge w \]
is closed: its first summand has top degree on the base, and its second has closed factors. Thus $\gamma _ s$ is balanced. Gauduchon Stokes and the matrix arithmetic–geometric mean inequality give
\[ \int _ X\theta _ s\wedge \xi ^ m \geq \left(\frac{nW}{H(\epsilon )}\right)^{1/n}D_\mu (\xi ) \]
for every Gauduchon metric $\xi $. The metric $\gamma _ s$ attains equality, with numerator $nW$ and denominator $(nW/H)^{m/n}H$. This proves the full volume formula.
For every fixed test, its quotient at $\theta _ t+\epsilon w$ decreases to its quotient at $\theta _ t$. The two infima, over $\epsilon >0$ and over the tests, commute, as in theorem 12.11.1. Hence the endpoint value is the limit of the strict values. Monotone convergence of $H(\epsilon )$ gives the asserted value, including when the integral is infinite.
The cost numerator for the displayed balanced metric is $WJ/H$. Dividing by its denominator gives (12.13.2). Also $\mathrm{d}H(t-s)/\mathrm{d}s=J(t-s)$, so differentiation of the volume formula gives $-\mathrm{d}\lambda _\mu (\theta _ s)/\mathrm{d}s=\mathcal C(s)$. By proposition 12.3.2 this is the cost of every minimizing metric, regardless of the regularity of the selection. On each interval $[0,r]$ with $r<t$, integrate this identity. Since $\mathcal C\geq 0$ and the root volume has the finite endpoint limit just proved, letting $r\uparrow t$ gives (12.13.3).
$\square$
Pome
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