$|f|^{\alpha}$ subharmonic for $\alpha > 0$

Question

Wikipedia claims that given a holomorphic function $f$ then $\log(|f(z)|)$ being subharmonic implies that the function $\varphi_{\alpha}(z) := |f(z)|^{\alpha}$ is subharmonic for every $\alpha > 0$. Why is this true?

Answer 1

This is a special case of the following, applied to $|f(z)|^\alpha = \exp(\alpha \log (|f(z)|))$:

If $v: D \to \Bbb R \cup \{ -\infty \}$ is subharmonic in the domain $D$ and $h: \Bbb R \to \Bbb R$ is increasing and convex, then $V = h \circ v$ is subharmonic in $D$.

Remark: If $v(z_0) = -\infty$ then $V(z_0)$ is defined as $V(z_0) = \lim_{t \to -\infty} h(t) = \inf \{ h(t) \mid t \in \Bbb R \}$.

Proof: This follows easily from Jensen's inequality for integrals: If $v(z_0) > -\infty$ and $\overline {B_r(z_0)} \subset D$ then $$ V(z_0) = h(v(z_0)) \le h \left(\frac{1}{2\pi r}\int_0^{2\pi} v(z_0 + re^{it}) \, dt \right) \\ \le \frac{1}{2\pi r}\int_0^{2\pi} h(v(z_0 + re^{it})) \, dt = \frac{1}{2\pi r}\int_0^{2\pi} V(z_0 + re^{it}) \, dt \, . $$

If $v(z_0) = -\infty$ then $V(z_0) \le V(z_0 + re^{it})$ and the integral inequality holds obviously.