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Positive powers of the modulus of a holomorphic function are subharmonic

Statement

Let $f$ be holomorphic on a complex domain $Ω$ , not identically zero on any connected component, and let $p > 0$ . Then the function $z ⟼ ∣ f (z) ∣^{p}$ is subharmonic on $Ω$ .

Facts & Assumptions

Given: A holomorphic function $f$ on a complex domain $Ω$ , not identically zero on any connected component, and a real number $p > 0$ .

[L1]

The function $\log ∣ f ∣$ , with value $- \infty$ at the zeros of $f$ , is subharmonic (The logarithm of the modulus of a holomorphic function is subharmonic).

Proof

technique · direct

1.1

Put $u = \log ∣ f ∣$ . By [L1], for every closed disc $\overline{D (a, r)} \subseteq Ω$ , [L1, algebra] $u (a) \leq \frac{1}{2 π} \int_{0}^{2 π} u (a + r e^{i t}) d t .$ Exponentiating and using Jensen's inequality for the convex increasing map $x \mapsto e^{p x}$ gives $∣ f (a) ∣^{p} = e^{p u (a)} \leq \frac{1}{2 π} \int_{0}^{2 π} e^{p u (a + r e^{i t})} d t = \frac{1}{2 π} \int_{0}^{2 π} ∣ f (a + r e^{i t}) ∣^{p} d t .$

L1algebra

2.1

The function $∣ f ∣^{p}$ is continuous, hence upper semicontinuous, and is not identically zero on a connected component because $f$ is not identically zero there. Step 1.1 is exactly the submean inequality, so $∣ f ∣^{p}$ is subharmonic.

step 1.1given∎

Depends on

The logarithm of the modulus of a holomorphic function is subharmonic

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

5 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Harold P. Boas, Class Notes Math 618: Complex Variables II, Spring 2016 (standard reference, not scraped)