Show that $\log|f|$ is a plurisubharmonic function

Question

$\Omega \subset C^n$. $f \in O(\Omega)$. Show that $\log|f|$ is a plurisubharmonic function.

I have tried two methods.

The first one is calculating Hessain matrix of $log|f|$, but it is too hard.

The second one is assuming $\log|f|$ not a PSH function. Then $g(t)=\log|f(Z_0+t\alpha )|$ is not subharmonic, which means $g(0)>\frac{1}{2\pi } \int_0^{2\pi } \log |f(Z_0+\epsilon e^{i \theta} \alpha)| d \theta$. However, I cannot get a contradiction by this.

This is actually an example on my textbook without proof. Thanks in advance.

Answer 1

By definition, as you seem to realize, you need to show that the map $\lambda\mapsto\log|f(z+\lambda w)|$ is subharmonic in the open subset of the plane where it is defined.

So you only need the one-variable case:

If $V\subset\Bbb C$ is open and $f\in H(V)$ then $u=\log|f|$ is subharmonic in $V$.

And this is trivial. First, $u$ is certainly usc, since it's a continuous $[-\infty,\infty)$-valued function. So we need to show that if $z\in V$ there exists $\rho>0$ such that $$u(z)\le\frac1{2\pi}\int_0^{2\pi}u(z+re^{it})\,dt\quad(0<r<\rho).$$ And this is trivial. It's clear if $f(z)=0$, since then $u(z)=-\infty$. If $f(z)\ne0$ choose $r>0$ so $f$ has no zero in $D(z,r)$. Then in $B(z,r)$ we have $u=\Re \log f$, so we have equality above (since the real part of a holomorphic function is harmonic).