Theorem 14.24.198EC31947E4C

of theorem 14.24.1. By corollary 14.19.29, $u,v$ are continuous. In dimension one, $\mathrm{dd}^{\mathrm{c}}w=0$, so the maximum principle makes $w$ constant. Assume $n\geq 2$.

Suppose that $w=u-v$ is nonconstant. Let $a=\min _ Xw$, $K=\{ w=a\} $, and $\mu =F\omega ^ n$. By lemma 14.24.11, $\mu (K)=0$. Choose $B_0>\| 1\| _{L^2(\omega ^ n)}$ and let $\varepsilon >0$ be the constant from proposition 14.24.13. Since $\mu (K)=0$, the compact set $K_0=K\cap \{ F\geq \varepsilon /2\} $ has zero $\omega ^ n$-volume. Fix $A>1$ with $\varepsilon A>\max _ XF$. By outer regularity of the smooth volume measure, choose an open neighborhood $O$ of $K_0$ so small that

Take a smooth $0\leq \chi \leq 1$ equal to one near $K_0$ and supported in $O$, and set $g=1+(A-1)\chi $. If $K_0=\varnothing $, take $\chi =0$. Then $g\geq 1$, $g=A$ on $K_0$, and $\| g\| _{L^2(\omega ^ n)}<B_0$.

By proposition 14.24.13, there is a smooth strictly $\omega $-plurisubharmonic $q$ with $\omega _ q^ n=G\omega ^ n\geq \varepsilon g\omega ^ n$. On $K_0$ we have $G\geq \varepsilon A>F$; on $K\setminus K_0$ we have $F<\varepsilon /2<\varepsilon \leq G$. Thus $G>F$ throughout the compact set $K$. By compactness and continuity, we can choose $0<\kappa <1$ such that $F<\kappa G$ on $K$; put $W=\{ F<\kappa G\} $. The complement of this open neighborhood of $K$ is compact, and $w-a$ has a strictly positive minimum there if it is nonempty. Hence some $\eta >0$ satisfies $\{ w<a+\eta \} \subset W$.

Proposition 14.24.2 asserts that $\mu (\{ w<a+\eta \} \cap \{ F>\kappa G\} )>0$. By the preceding inclusion, however, this set is empty. This contradiction shows that $w$ is constant. If $\sup _ Xu=\sup _ Xv$, the constant is zero. $\square$