Proposition 5.17.3EC4B3805B84D

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.