Proposition 15.18.3145A010B2A5AF

Proof. Where $u$ is smooth and strictly $\vartheta $-plurisubharmonic, it provides its own quadratic separation, so the argument of proposition 15.18.25 applies around every point of $O$. A first application gives trace growth of order $\sqrt{\log j}$; the resulting gradient convergence then allows the variant of lemma 15.18.8 with a gradient bound, which removes this growth.

For $p$ in a fixed compact subset of $O$, take a uniformly sized ball $B_ r(p)\Subset O$ and set

with $\kappa >0$ small, so $\vartheta +\mathrm{dd}^{\mathrm{c}}q_ p\geq \varepsilon \beta $. If $e_ j=Cj^{-a}$ and $v_{j,p}=u_ j-q_ p$, then

With $A_ j=D\log j$ and $D\kappa r^2>b+1$, the same maximum argument as above gives

The constants are uniform on compact subsets of $O$.

Put $T_ j=\vartheta +\mathrm{dd}^{\mathrm{c}}u_ j$ and $w_ j=u_ j-u$. On slightly larger compact sets, (15.18.33) bounds $|\Delta w_ j|$ by $C\sqrt{\log j}$. Poisson estimates and Sobolev embedding give

Interpolating with $\lVert w_ j\rVert _\infty \leq Cj^{-a}$ shows $\nabla w_ j\to 0$ uniformly on smaller compact sets. The gradients of $v_{j,p}$ are therefore uniformly bounded. Repeat the maximum argument with $\log \operatorname {tr}_\beta T_ j-A_ jv_{j,p}$, now using the variant of lemma 15.18.8 with a gradient bound. Its interior trace bound is now independent of $A_ j$, while the boundary term is suppressed as before. Evaluation at $p$ gives

proving (15.18.32). $\square$