Lemma 5.28.13.D263A93C1D69 Let $h\leq a$ be locally bounded $\theta $-plurisubharmonic functions, and let $\chi $ be a smooth function on a neighborhood of $(-\infty ,0]$ with $0\leq \chi '\leq 1$ and $\chi ''\geq 0$. Then $w=a+\chi (h-a)$ is $\theta $-plurisubharmonic and
\[ \theta _ w=(1-\chi '(h-a))\theta _ a+\chi '(h-a)\theta _ h +\chi ''(h-a)\, \mathrm{d}(h-a)\wedge \mathrm d^{\mathrm c}(h-a). \]
In particular, $\theta _ w^ n\geq \chi '(h-a)^ n\theta _ h^ n$. The positivity of $\theta _ w$ and this inequality also hold for convex functions $\chi $ of class $C^{1,1}$ on compact intervals with $0\leq \chi '\leq 1$.
Proof.
The function $G(x,y)=x+\chi (y-x)$, for $y\leq x$, is convex and nondecreasing in each variable: its first derivatives are $1-\chi '$ and $\chi '$, and its Hessian is
\[ \chi ''\begin{pmatrix} 1
& -1
\\ -1
& 1
\end{pmatrix}\geq 0. \]
For smooth potentials, the chain rule gives the identity. For bounded potentials, add a common quadratic correction and convolve. Convolution preserves the order of the potentials, and since $G(x+r,y+r)=r+G(x,y)$, adding the same correction to both arguments adds it to $G$. The bounded chain rule (5.27.9) and its products with the gradient term give the identity in the limit. Freezing the smooth background as in the proof of proposition 5.28.2 preserves the exact lower bound. Every term on the right is positive, and the positivity of bounded mixed products gives the last assertion.
For a convex weight of class $C^{1,1}$, extend it convexly across zero with derivative in $[0,1]$, and smooth it in one variable. The smooth convex weights and their first derivatives converge uniformly on compact intervals, and the derivatives stay in $[0,1]$. On each smaller coordinate ball, the corresponding potentials converge uniformly to $w$. The continuity of bounded products gives convergence of their top powers, while the weights on the right-hand side converge uniformly against the fixed measure $\theta _ h^ n$. Passing to the limit proves the positivity and the product inequality with the original smooth background.
$\square$
Pome
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