5.31 Gluing with a small curvature error78243032B1A1

Proof. For fixed $\epsilon >0$, we first glue the local convolutions of $u$ by a regularized maximum, obtaining smooth functions that decrease to $u$ and have curvature error $2\epsilon \omega $; a diagonal sequence then proves the proposition.

Step 1: local regularizations. Fix $\epsilon >0$. By continuity of $\theta $, we can choose finitely many nested coordinate neighborhoods $V_ i\Subset W_ i\Subset U_ i$, with the $V_ i$ covering $X$, and smooth quadratic functions $r_ i$ such that

To do so, freeze the coefficients of $\theta $ at the center of a chart, add a small positive constant form, and shrink the coordinate neighborhood. Then $p_ i=u+r_ i$ is a bounded plurisubharmonic function on $U_ i$; let $p_{i,\delta }$ be its convolution in the coordinates of $U_ i$.

On an overlap, express both $p_ i$ and $p_ j$ in the coordinates of $U_ j$. Their difference is the smooth function $r_ i-r_ j$, so the difference of their convolutions in these coordinates tends uniformly to $r_ i-r_ j$. By corollary 5.30.3, changing the coordinates in which $p_ i$ is convolved introduces another uniformly vanishing error. Thus, on the intersections of the closures of the $W_ i$,

Step 2: gluing. Choose functions $\eta _ i\in C^\infty (U_ i,[-1,0])$ that vanish on $V_ i$ and equal $-1$ near the boundary of $W_ i$, and a constant $C\geq 1$ with $\mathrm{dd}^{\mathrm{c}}\eta _ i\geq -C\omega $. The cover, the functions $r_ i,\eta _ i$ and the constant $C$ may depend on $\epsilon $; they are fixed while $\delta $ tends to zero. Set

Each of these functions satisfies $\mathrm{dd}^{\mathrm{c}}f_{i,\delta }\geq -\theta -2\epsilon \omega $. Every boundary point of $W_ i$ lies in some $V_ j$, and there $\eta _ j=0$ while $\eta _ i=-1$. By (5.31.3), for all small $\delta $,

on a neighborhood of that boundary point. Since the cover is finite, this separation holds uniformly.

Let $\tau (\delta )\downarrow 0$ as $\delta \downarrow 0$, with $\tau (\delta )<\epsilon /(8C)$, and define the value at each point as the regularized maximum $M_{\tau (\delta )}$ of the functions $f_{i,\delta }$ for those $i$ with the point in $W_ i$. A branch disappears only near the boundary of its $W_ i$, where it lies below another branch by more than $2\tau $. By the removal property of $M_\tau $, these local formulas therefore define a global smooth function $\psi _{\epsilon ,\delta }$, and (5.31.1) gives

At every point there is a branch with $\eta _ i=0$, and $p_{i,\delta }\geq p_ i$; hence $\psi _{\epsilon ,\delta }\geq u$. For fixed $\epsilon $, the local convolutions decrease with $\delta $, and the regularized maximum decreases when both its arguments and $\tau $ decrease. Its pointwise limit is

Thus $\psi _{\epsilon ,\delta }\downarrow u$ as $\delta \downarrow 0$, and also in $L^1$.

Step 3: a diagonal sequence. Take $\epsilon _ j\downarrow 0$ and put $v_ j=\psi _{\epsilon _ j,\delta _ j}+2^{-j}$, where $\delta _ j$ is chosen so small that

The second requirement can be met because $v_{j-1}\geq u+2^{-j+1}$. Indeed, as $\delta \downarrow 0$, the continuous functions $\psi _{\epsilon _ j,\delta }+2^{-j}-v_{j-1}$ decrease pointwise to a function bounded above by $-2^{-j}$. Their sets of nonnegative values are decreasing compact sets with empty intersection, so one of them is empty. The resulting sequence $(v_ j)$ decreases, converges to $u$ in $L^1$, and stays above $u$. Its pointwise limit is $(\theta +2\epsilon _ k\omega )$-plurisubharmonic for every $k$, hence $\theta $-plurisubharmonic, and it equals $u$ almost everywhere; by lemma 5.24.3, it equals $u$ everywhere. Taking $\varepsilon _ j=2\epsilon _ j$ proves the proposition. $\square$