CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26

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A plane harmonic function bounded above or below is constant

Statement

A harmonic function on $C$ that is bounded above, or bounded below, is constant.

Facts & Assumptions

Given: A harmonic function $u : C \to R$ .

[L1]

The plane $C$ is star-shaped and therefore homologically simply connected, so every harmonic function on it has a harmonic conjugate (Star-shaped plane domains are homologically simply connected, Harmonic conjugates exist on homologically simply connected plane domains, Homologically simply connected complex domains).

[L2]

Every bounded entire function is constant (Liouville's theorem: every bounded entire function is constant).

[L3]

The complex exponential is entire, satisfies $∣ \exp (x + i y) ∣ = e^{x}$ , and obeys the usual product and chain rules; a holomorphic function with zero derivative on a domain is constant (The complex exponential is entire and its complex derivative is itself, $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ , Linearity, product, reciprocal, and quotient rules for complex derivatives, A holomorphic function with zero derivative on a domain is constant).

Proof

technique · direct

1.1L1L2L3

Suppose first that $u$ is bounded above. By [L1], choose a harmonic conjugate $v$ and set $F : = u + i v$ , a holomorphic function on $C$ . Then $\exp (F)$ is entire by [L3], and its modulus is $e^{u}$ , which is bounded because $u$ is. So [L2] makes $\exp (F)$ constant.

2.1step 1.1L3

Differentiating the identity $\exp (F) \equiv c$ gives $0 = (\exp (F))^{'} = \exp (F) F^{'}$ . Since an exponential value is never $0$ , [L3] gives $F^{'} = 0$ , and [L3] again makes $F$ constant. Therefore $u = Re F$ is constant.

3.1step 2.1algebra∎

If instead $u$ is bounded below, then $- u$ is harmonic and bounded above, so step 2.1 applied to $- u$ makes $- u$ , and therefore $u$ , constant.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

48 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

Jeremy Orloff, MIT 18.04 Topic 5: Introduction to Harmonic Functions (standard reference, not scraped)
Sigurdur Helgason, MIT 18.112 Lecture 16: Harmonic Functions (standard reference, not scraped)