Lemma 8.10.2FFBF9CD42640

Proof. First let $u$ be smooth admissible. The first-cone inequality gives $\mu _ u\geq 0$, since $T$ is conformal to $\omega ^{n-1}$, and $\mathrm{dd}^{\mathrm{c}}T=0$ gives $\mu _ u(X)=A$. Applying distributional Stokes to $\mathrm{dd}^{\mathrm{c}}(u^2)$ yields the energy identity

Together with the $L^2$ bound this controls $\| u\| _{H^1}$. For bounded admissible $u$, choose smooth admissible $u_ j\downarrow u$ by 8.6.6, discard a finite initial segment, and take $\| u_ j\| _\infty \leq M+1$ as in 8.9.1. Bounded convergence gives strong $L^2$ convergence. The energy bound gives weak $H^1$ compactness, and every weak cluster point agrees with $u$ in $L^2$; thus the entire sequence converges weakly to $u$ in $H^1$. This proves the Sobolev bound. The trace measures converge weakly to $\mu _ u$, preserving positivity and total mass.

For any smooth real $\phi $, the first-order Stokes identity is

It remains valid for $u\in H^1$ by smooth Sobolev approximation. Both terms are bounded by $C\| u\| _{H^1}\| \phi \| _{H^1}$; the background term $\int \phi \omega \wedge T$ is bounded by $C\| \phi \| _{H^1}$. The preceding capacity lemma gives the bound for $\mu _ u$. In particular the derivative $\mathrm dT$ has not been discarded.

For a smooth admissible approximant, write the eigenvalues of $\omega +\mathrm{dd}^{\mathrm{c}}u_ j$ in an $\omega $-unitary frame as $\lambda _ i$. Since $m\geq 2$,

Consequently every coefficient of this form is dominated by its trace. On a fixed coordinate chart this bounds the coefficient variations of $\mathrm{dd}^{\mathrm{c}}u_ j$ by $C(\mu _{u_ j}+\omega ^ n)$. Passing to the weak measure limits of 8.9.1 gives the same domination by $C(\mu _ u+\omega ^ n)$. Finally $\int _ E\omega ^ n\leq \operatorname {Cap}_1(E)$ and $\operatorname {Cap}_1(E)\leq \| 1\| _{H^1}^2$ absorb the smooth term into the claimed square-root estimate. $\square$