PomeProposition 5.10.3.1A6F10A07BA5 Let $\Omega \subset \mathbb {C}^ n$ be a domain, $n\geq 1$, and let $g=(g_1,\ldots ,g_ r)$, $r\geq 1$, be a tuple of holomorphic functions on $\Omega $, not all identically zero. Then the function $\log \sum _ j|g_ j|^2$, with the value $-\infty $ at the common zeros of the $g_ j$, is plurisubharmonic on $\Omega $. In particular, for every $a\in \mathbb {C}^ n$ the function $\log |z-a|^2$ is plurisubharmonic on $\mathbb {C}^ n$.
Proposition 5.10.31A6F10A07BA5
Proof. By the Cauchy–Schwarz inequality, with equality for $\lambda $ proportional to $(\overline{g_1(z)},\ldots ,\overline{g_ r(z)})$, we have
\[ \log \sum _ j|g_ j(z)|^2 =\sup _{\sum _ j|\lambda _ j|^2=1}\log \left|\sum _ j\lambda _ jg_ j(z)\right|^2 \]
for every $z\in \Omega $, both sides being $-\infty $ at a common zero. The supremum does not change if we omit the $\lambda $ for which $\sum _ j\lambda _ jg_ j$ vanishes identically: at a point $z$ that is not a common zero, the maximizing $\lambda $ gives $\sum _ j\lambda _ jg_ j(z)\neq 0$, and at a common zero all terms equal $-\infty $. Some $g_ j$ is not identically zero, so the remaining family is nonempty. Its members are plurisubharmonic by proposition 5.5.3, theorem 5.9.1, and they are locally uniformly bounded above by the left-hand side. Their supremum $\log \sum _ j|g_ j|^2$ is already upper semicontinuous: it is continuous as a map to $[-\infty ,\infty )$, because $\sum _ j|g_ j|^2$ is continuous and its logarithm tends to $-\infty $ at a common zero. Hence it equals its regularization, which is plurisubharmonic by theorem 5.10.2. The particular case is the tuple $(z_1-a_1,\ldots ,z_ n-a_ n)$. $\square$
Comments
Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.