Proposition 5.41.3ED1DA0551379

Proof. The classical local Kiselman principle states that if a plurisubharmonic function on the product of a domain with a strip does not depend on the imaginary part of the strip variable, then its infimum over the real part is plurisubharmonic or identically $-\infty $; the interval of the real part may be unbounded. We use this local result from [SGPT].

Fix $x_0\in X$, choose local holomorphic coordinates centered at $x_0$, and let $\kappa $ be their Euclidean form. Fix $\eta >0$. On a sufficiently small coordinate ball the constant form $\beta =\theta (x_0)$ satisfies

Choose a quadratic function $r$ with $\mathrm{dd}^{\mathrm{c}}r=\beta +\eta \kappa $. Then $U+r\circ \pi $ is plurisubharmonic on the product of the ball with $S_ I$. Its infimum over the parameter is $q+r$, so the local principle shows that $q+r$ is plurisubharmonic or identically $-\infty $ on the ball; in the first case

In a fixed atlas, the Euclidean forms are bounded by a fixed multiple of $\omega $, so letting $\eta \downarrow 0$ proves the exact lower bound $\mathrm{dd}^{\mathrm{c}}q\geq -\theta $. Since a plurisubharmonic function cannot be identically $-\infty $ on a nonempty open set, the local alternative propagates across overlapping balls on the connected manifold $X$. Thus either $q\in \mathrm{PSH}(X,\theta )$ or $q\equiv -\infty $. No regularization is needed, because an infimum of upper semicontinuous functions is upper semicontinuous.

Now suppose that $I$ is bounded. By lemma 5.41.1, the function $t\mapsto u_ t(x)$ is finite and convex for almost every $x$, and a supporting affine function at any point of $I$ bounds it from below on the bounded interval $I$. Hence $q(x)>-\infty $ almost everywhere, which excludes the alternative $q\equiv -\infty $. $\square$