The logarithm of the modulus of a holomorphic function is subharmonic

Let $D\subseteq \mathrm{\Omega }$ be a disc on which $f$ has no zeros. By [L3], there is a holomorphic function $L$ on $D$ with $\exp L=f$. Writing $L=\alpha +i\beta$, one has $\alpha =\log \mathrm{\mid }f\mathrm{\mid }$ on $D$. Since $L$ is holomorphic and smooth by [L4], the Cauchy-Riemann equations imply $\mathrm{\Delta }\alpha =0$, so [L1] makes $\log \mathrm{\mid }f\mathrm{\mid }$ subharmonic on every zero-free disc.

Fix a zero $a$ of $f$, and let $m={\mathrm{ord}}_a(f)$. By [L2], on a small disc about $a$ one has $f(z)=(z-a{)}^{m}g(z)$ with $g(a)\mathrm{\ne }0$. Shrinking if necessary, $g$ has no zeros there, so step 1.1 makes $\log \mathrm{\mid }g\mathrm{\mid }$ harmonic and hence subharmonic on that disc.

On the punctured disc around $a$, [step 2.1, algebra] $u(z)=m\log \mathrm{\mid }z-a\mathrm{\mid }+\log \mathrm{\mid }g(z)\mathrm{\mid }\mathrm{.}$ The function $\log \mathrm{\mid }z-a\mathrm{\mid }$ is harmonic on the punctured disc, and at the center $a$ its value is $-\mathrm{\infty }$ while every circle average is finite; hence it is subharmonic there. Therefore the right-hand side is subharmonic on the whole disc, agreeing with $u$ away from $a$ and with $u(a)=-\mathrm{\infty }$ at the center.

Every point of $\mathrm{\Omega }$ lies either on a zero-free disc covered by step 1.1 or on a zero-containing disc covered by step 3.1. So $u$ is subharmonic throughout $\mathrm{\Omega }$.