Log-modulus of a holomorphic function is subharmonic

Remarks

Proof idea: let $a$ be a point with $f(a) \ne 0$. On a small disc about $a$ there is a holomorphic $g$ with $f = e^{g}$, so $u = \operatorname{Re} g$ there. The real part of a holomorphic function is harmonic, so $u$ is harmonic near $a$. At a zero $a$ of $f$, $u(a) = -\infty$, and $-\infty$ is at most any mean of $u$, so the sub-mean inequality holds at $a$. The function $u$ is continuous as a map into $[-\infty, \infty)$, so it is upper semicontinuous.

Example: for $f(z) = z$, $u(z) = \log\abs{z}$ is harmonic off $0$ and equals $-\infty$ at $0$. Its mean over the circle of radius $r$ about $0$ is $\log r > -\infty = u(0)$. Composing with the convex increasing function $t \mapsto e^{pt}$ shows that $\abs{f}^p$ is subharmonic for every $p > 0$, by convex functions of subharmonic functions. The function $\log\abs{f}$ is not harmonic on all of $\Omega$ when $f$ has zeros. Its Laplacian in the sense of distributions is $2\pi \sum_a m_a \delta_a$, where $a$ runs over the zeros of $f$ and $m_a$ is the multiplicity of $a$.

Going deeper

For a student

This is the main bridge from complex analysis to potential theory. Applied to $\log\abs{f}$, the maximum principle for subharmonic functions gives the maximum modulus principle, and the convexity results for subharmonic functions give Hadamard's three-circle theorem. Jensen's formula is the mean value formula for $\log\abs{f}$, with the zeros of $f$ as the correction term.