Proposition 9.9.3D6705783DDAD

Proof. Let $u$ be locally plurisubharmonic on the compact connected space $X$. It is upper semicontinuous, so it attains its maximum $M\in [-\infty ,\infty )$. If $M=-\infty $, then $u\equiv -\infty $. Suppose that $M$ is finite, and let $x$ be a point with $u(x)=M$. We show that $u=M$ near $x$. If $x$ is an isolated point, there is nothing to prove. Otherwise, let $X_1$ be a branch of $(X,x)$ of positive dimension, and take a parametrization of $(X_1,x)$, with its connected covering locus $X_{1,\delta }$, dense in the representative $X_1$. By proposition 9.5.2, applied to the subgerm consisting of the singular locus of $X$ and the points of $X_1$ outside $X_{1,\delta }$, there is a nonconstant holomorphic curve $\gamma $ with $\gamma (0)=x$ and $\gamma (t)\in X_{1,\delta }\cap X_{\mathrm{reg}}$ for $t\neq 0$. The function $u\circ \gamma $ is subharmonic by lemma 9.9.2(2), and it attains its finite maximum $M$ at $0$; by the maximum principle (proposition 3.4.7) it equals $M$ near $0$. Thus $u$ attains the value $M$ at points of $X_{1,\delta }$.

By lemma 9.9.2(2) and (3), applied to the inclusion of the connected manifold $X_{1,\delta }$, the restriction of $u$ to $X_{1,\delta }$ is plurisubharmonic or identically $-\infty $, hence plurisubharmonic, since it takes the value $M$. On a coordinate ball around a point where $u=M$, it is subharmonic by theorem 5.9.1, and it attains its maximum $M$ at the center, so it equals $M$ on this ball by proposition 3.4.7. Thus the set where $u=M$ is open in $X_{1,\delta }$; it is closed by upper semicontinuity and nonempty, so $u=M$ on $X_{1,\delta }$. Since $X_{1,\delta }$ is dense in $X_1$, upper semicontinuity gives $u\geq M$, hence $u=M$, on all of $X_1$. Doing this for each branch, we find that $u=M$ on a neighborhood of $x$.

Hence the set $\{ u=M\} $ is open. It is also closed, because $\{ u<M\} $ is open for an upper semicontinuous function bounded above by $M$. Since $X$ is connected, $u\equiv M$. $\square$