PomeLemma 20.3.65.72230836D2EE There are $U_ L\sim U$ with $\sup U_ L=0$, $\mathrm{dd}^{\mathrm{c}}U_ L\geq -\epsilon _ Lw$ and $\epsilon _ L\downarrow 0$, such that $U_ L\to 0$ in $L^1(\mathbb P^1)$. Each has zero Lelong number at $p$ and is locally bounded off $p$.
Lemma 20.3.6572230836D2EE
Proof. For $t\leq 0$ put $\chi (t)=1-(1-t)^\alpha $. Then
\[ \chi '(t)=\alpha (1-t)^{\alpha -1}>0,\qquad \chi ''(t)=\alpha (1-\alpha )(1-t)^{\alpha -2}>0. \]
For $L>0$ set $\epsilon _ L=\chi '(-L)$, $C_ L=-\epsilon _ LL-\chi (-L)$, and define
\[ \chi _ L(t)= \begin{cases} \chi (t)+C_ L,& t<-L,\\ \epsilon _ Lt,& -L\leq t\leq 0. \end{cases} \]
The values and first derivatives match, and $\chi _ L$ is convex and increasing with derivative at most $\epsilon _ L$. Off $p$, the chain rule gives $\mathrm{dd}^{\mathrm{c}}\chi _ L(G_ p)\geq -\epsilon _ Lw$. Canonical extension across $p$ retains this bound by the removable-singularity theorem, so $U_ L=\chi _ L(G_ p)$ has the required curvature. Near $p$ it differs from $U$ by $C_ L$, and elsewhere both are bounded. Its growth is sublogarithmic, hence its Lelong number is zero.
Writing $x=1+L$, we have $C_ L=(1-\alpha )x^\alpha +\alpha x^{\alpha -1}-1\geq 0$. Also $\chi (t)\leq \epsilon _ Lt$ for $-L\leq t\leq 0$, by integrating $\chi '\geq \epsilon _ L$ on this interval. It follows that $U\leq U_ L\leq 0$. Since $\epsilon _ L\to 0$, the functions tend to zero off $p$, and dominated convergence gives the $L^1$ limit. $\square$
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