TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-26

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Plane harmonic functions satisfy the mean-value property

Statement

Every plane harmonic function satisfies both the circle and disc mean-value properties of The circle and disc mean-value properties.

Facts & Assumptions

Given: A harmonic function $u$ on an open set $Ω$ , a point $a \in Ω$ , and a radius $r > 0$ with $\overline{D (a, r)} \subseteq Ω$ .

[L1]

Every open disc is star-shaped and therefore homologically simply connected, and every harmonic function on such a domain is the real part of a holomorphic function (Star-shaped plane domains are homologically simply connected, Harmonic conjugates exist on homologically simply connected plane domains).

[L2]

A holomorphic function equals its average on every smaller concentric circle (A holomorphic function equals its average on every circle inside a larger concentric holomorphy disc).

Proof

technique · direct

1.1

Because the closed disc $\overline{D (a, r)}$ lies in the open set $Ω$ , choose $R > r$ with $D (a, R) \subseteq Ω$ . The restriction of $u$ to this disc is harmonic, so [L1] gives a holomorphic function $F$ on $D (a, R)$ with $Re F = u$ there.

L1choose

2.1

For every $0 < t \leq r$ , applying [L2] to $F$ on the circle of radius $t$ and taking real parts gives $u (a) = \frac{1}{2 π} \int_{0}^{2 π} u (a + t e^{i θ}) d θ .$

step 1.1L2

3.1

The circle formula of the definition is step 2.1 at $t = r$ .

step 2.1

4.1

Multiplying the identity of step 2.1 by $2 t / r^{2}$ and integrating from $0$ to $r$ gives the disc formula of the definition, because $\frac{2}{r^{2}} \int_{0}^{r} t d t = 1$ .

step 2.1algebra∎

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Used by

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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)