Lemma 8.11.13E64937A67BD2

Proof. First take smooth potentials, put $\sigma _ i=\omega +\mathrm{dd}^{\mathrm{c}}u_ i$, and choose the fixed local correction $\tau =\mathrm{dd}^{\mathrm{c}}\rho \geq 2\omega $. Then $\beta _ i=\mathrm{dd}^{\mathrm{c}}(u_ i+\rho )=\sigma _ i+(\tau -\omega )$ lies in the same local cone. Since $\tau -\omega \geq 0$, mixed positivity gives

Here $Q_ p\wedge \omega ^{m-p}=\sigma _1\cdots \sigma _ p\omega ^{n-p}$. The last bound follows from smooth mixed-trace domination at $s=m$ for the bounded corrected potentials and a normalized bounded sum with a fixed strictly plurisubharmonic function. Positivity of $Q_ p$ bounds each of its coefficient variations on the compact coordinate set by a fixed multiple of its trace measure.

For bounded potentials choose the smooth decreasing global approximations already established in the chapter. The joint convergence in 8.11.12 identifies their positive-current limit with the previously constructed $Q_ p$. On a compact neighborhood of $K$ inside $U$, lower semicontinuity on open sets preserves trace and coefficient variation estimates. For an arbitrary open $G\supset E$ intersect with a fixed relatively compact neighborhood of $K$ before taking this limit; its capacity is at most $c_ m(G)$. Therefore every such trace or variation measure $\mu $ satisfies $\mu (E)\leq Cc_ m(G)$. Infimize over $G$ to obtain the asserted outer-open bound. The constants of the smooth estimates are independent of the approximation index. The argument does not multiply a rough current of positive order or use an unproved rough diagonal polynomial expansion. $\square$