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In practice, it is important to consider a variant of plurisubharmonic functions. We will fix a complex manifold $X$.
Definition 5.17.1.3DC2C56EBB31 Let $\theta $ be a closed real smooth $(1,1)$-form on $X$.
A $\theta $-plurisubharmonic function or $\theta $-psh function for short on $X$ is a function $\varphi \colon X\rightarrow [-\infty ,\infty )$ such that for each $x\in X$ and each open neighborhood $U$ of $x$ in $X$ satisfying the condition that $\theta =\mathrm{dd}^{\mathrm{c}}g$ for some smooth function $g$ on $U$, we have $g+\varphi |_ U\in \mathrm{PSH}(U)$. The set of $\theta $-psh functions on $X$ is denoted by $\mathrm{PSH}(X,\theta )$.
A quasi-plurisubharmonic function on $X$ is a function $\varphi \colon X\rightarrow [-\infty ,\infty )$ such that there exists a smooth closed real $(1,1)$-form $\theta '$ on $X$ such that $\varphi \in \mathrm{PSH}(X,\theta ')$. The set of quasi-plurisubharmonic functions on $X$ is denoted by $\mathrm{QPSH}(X)$.
There is a natural preorder on $\mathrm{QPSH}(X)$ defined as follows:
See Definition 16.7.1.
It is sometimes convenient to include the constant function $-\infty $ as well. Then we always write $-\infty \preceq \varphi $ for any $\varphi \in \mathrm{QPSH}(X)$. Formally, we write $-\infty \sim -\infty $ as well.
When $X$ is not compact, one can still define similar notions, but the generalization is not unique, and we shall not consider them here.
Proposition 5.17.3.EC4B3805B84DAssume that $X$ is compact. Let $\theta $ be a closed real smooth $(1,1)$-form on $X$. Then for any $a,b\in \mathbb {R}$, $a\leq b$, the set
\[ \left\{ \varphi \in \mathrm{PSH}(X,\theta ): \sup _ X \varphi \in [a,b]\right\} \]
is compact with respect to the $L^1$-topology. Moreover, $\varphi \mapsto \sup _ X \varphi $ is $L^1$-continuous for $\varphi \in \mathrm{PSH}(X,\theta )$.
Qualification and proof. The compactness assertion needs connectedness of $X$, or a separate normalization on every connected component. Indeed, take the disjoint union of two compact connected complex manifolds, put $\theta=0$, and let $u_j$ equal $0$ on the first component and $-j$ on the second. Then $\sup_Xu_j=0$, but the $L^1$ norms tend to infinity, so this sequence has no convergent subsequence. Thus a proof for the stated disconnected case would be impossible.
For connected $X$ of positive dimension, normalized global plurisubharmonic compactness proves the assertion, including continuity of $u\mapsto\sup_Xu$. In dimension zero, connected $X$ is a point and the assertion is compactness of $[a,b]$. More generally a compact manifold has finitely many connected components. Requiring $a_l\leq\sup_{X_l}u\leq b_l$ on every component gives a finite product of these compact sets, hence a compact set in $L^1(X)$. The global supremum is then the maximum of finitely many continuous component suprema and is continuous.
This is an immediate consequence of
[Proposition 8.5, Exercise 1.20, GZ17].
Corollary 5.17.5.B5E91AB35965 Assume that $X$ is compact. Let $\theta $ be a closed real smooth $(1,1)$-form on $X$, and $(\varphi _ i)_{i\in I}$ be a decreasing net in $\mathrm{PSH}(X,\theta )$ such that $\sup _ X \varphi _ i$ is uniformly bounded from below. Then $\varphi =\inf _{i\in I} \varphi _ i$ is a $\theta $-psh function and
88BE4BEDE25D
5.17.6
\begin{equation} \adjustlimits \lim _{i\in I}\sup _ X \varphi _ i=\sup _ X \varphi ,\quad \varphi _ i\xrightarrow {L^1}\varphi . \end{equation}
Proof.
We may assume that $X$ is connected.
It suffices to show that $\varphi \not\equiv -\infty $. From this, it follows that $\varphi \in \mathrm{PSH}(X,\theta )$ and $\varphi _ i\to \varphi $ in $L^1$ by proposition 5.14.2. The $L^1$-continuity of $\eta \mapsto \sup _ X\eta $ on $\mathrm{PSH}(X,\theta )$ from proposition 5.17.3 then gives the first equality in (5.17.6).
Fix $i_0\in I$ and replace $I$ by the cofinal tail $I_{\geq i_0}$. This does not change the infimum. On this tail we have
\[ \sup _ X\varphi _ i\leq \sup _ X\varphi _{i_0}, \]
while the suprema are bounded from below by assumption. Hence proposition 5.17.3 shows that the net $(\varphi _ i)_{i\in I}$ is relatively compact with respect to the $L^1$-topology. We may therefore take a subnet $(\varphi _ j)_{j\in J}$ converging in $L^1$ to some $\psi \in \mathrm{PSH}(X,\theta )$. Since $(\varphi _ j)_{j\in J}$ is a Willard subnet of $(\varphi _ i)_{i\in I}$, we know that it is also decreasing. Therefore, $\psi =\inf _{j\in J} \varphi _ j$ due to proposition 5.14.2 and proposition 5.14.6.
Fix $i_0\in I$. For all sufficiently large $j\in J$, the corresponding index of the subnet dominates $i_0$, hence
\[ \varphi _ j\leq \varphi _{i_0}. \]
Therefore, $\psi \leq \varphi _{i_0}$. As $i_0$ was arbitrary, $\psi \leq \varphi $ everywhere. In particular, $\varphi $ is not identically $-\infty $.
$\square$
Proposition 5.17.7.A1D8E7565DF1 Assume that $X$ is compact. Let $\theta $ be a closed real smooth $(1,1)$-form on $X$ and $E$ be a prime divisor over $X$. Then
\[ \sup \Bigl \{ \nu (\varphi ,E): \varphi \in \mathrm{PSH}(X,\theta ) \Bigr \} <\infty . \]
Proof.
It follows from the proof of [Exercise 2.7, GZ17] that $\nu (\bullet ,E)$ is upper semicontinuous with respect to the $L^1$-topology on $\mathrm{PSH}(X,\theta )$. Since adding a constant does not change Lelong numbers, we may restrict to the compact set $\{ \varphi \in \mathrm{PSH}(X,\theta ):\sup _ X\varphi =0\} $ from proposition 5.17.3. The desired upper bound follows.
$\square$
Proposition 5.17.8.AF2A3F35959C Let $\pi \colon Y\rightarrow X$ be a proper bimeromorphic morphism from a compact Kähler manifold $Y$. Let $\theta $ be a closed real smooth $(1,1)$-form on $X$. Then the pull-back gives a bijection
\[ \pi ^*\colon \mathrm{PSH}(X,\theta )\xlongrightarrow {\sim }\mathrm{PSH}(Y,\pi ^*\theta ). \]
Proof. Injectivity holds because $\pi$ is an isomorphism over a dense open subset, and plurisubharmonic representatives are determined by their values almost everywhere. For surjectivity, descent across a smooth-target modification proves that every $\pi^*\theta$-plurisubharmonic function on $Y$ descends to a $\theta$-plurisubharmonic function on $X$. Locally, the descended function is defined off the exceptional image using the inverse map, extended by its upper semicontinuous regularization, and the removable-singularity argument in that proposition proves positivity and equality of its pullback with the original function. Thus every function on the right is a pullback, uniquely.
This follows from a more general result theorem
9.10.5.
Since we shall constantly need the balayage argument of proposition 5.14.21, we explain explicitly the global version of it.
Proposition 5.17.9.5F9E17A436E9 Fix a closed smooth real $(1,1)$-form $\theta $ on $X$. Let $\varphi \in \mathrm{PSH}(X,\theta )$, and $B\Subset B' \subseteq X$ be two holomorphic balls. Assume that $\varphi $ is locally bounded on $B'$. Then there is a unique $\psi \in \mathrm{PSH}(X,\theta )$ so that $\psi $ is bounded on an open neighborhood of $\overline{B}$, and
\[ \begin{cases} \left( \theta |_ B+\mathrm{dd}^{\mathrm{c}}\psi |_ B \right)^ n = 0,\\ \psi = \varphi \quad \textup{on } X\setminus B. \end{cases} \]
Moreover, $\psi \geq \varphi $.
We shall call $\psi $ the
balayage of $\varphi $ on $B$.
Proof.
By the Dolbeault–Grothendieck lemma, after possibly shrinking $B'$, we may write $\theta =\mathrm{dd}^{\mathrm{c}}g$ for some bounded function $g\in C^{\infty }(B')$. We can then use $g$ to translate $\theta $-psh functions to psh functions, and our assertion follows immediately from proposition 5.14.21.
$\square$
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