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The Poisson integral gives the unique continuous harmonic extension on the closed unit disc

Statement

Let $φ : \partial D \to R$ be continuous. Then its Poisson integral $P [φ]$ is harmonic on $D$ , extends continuously to $\overline{D}$ , agrees with $φ$ on $\partial D$ , and is the unique function with those properties.

Facts & Assumptions

Given: A continuous boundary datum $φ : \partial D \to R$ .

[L1]

The Poisson integral is harmonic on $D$ (Poisson integrals are harmonic on the unit disc).

[L2]

The Poisson integral converges to the boundary data uniformly as $r \to 1^{-}$ (The Poisson kernel is a boundary approximate identity).

[L3]

A bounded-domain continuous harmonic extension of fixed boundary data is unique (The bounded plane Dirichlet problem has at most one continuous harmonic solution).

Proof

technique · direct

1.1L1

By [L1], the function $P [φ]$ is harmonic on $D$ .

1.2L2

For $z = r e^{i α}$ with $0 \leq r < 1$ , [L2] gives $P [φ] (r e^{i α}) \to φ (e^{i α})$ uniformly in $α$ as $r \to 1^{-}$ . Therefore defining the boundary values of $P [φ]$ by $φ$ produces a continuous extension to $\overline{D}$ .

2.1step 1.1step 1.2L3∎

If $u$ is any other continuous harmonic function on $\overline{D}$ with $u = φ$ on $\partial D$ , then [L3] applied to $u$ and the extended Poisson integral forces $u = P [φ]$ on $\overline{D}$ .

Depends on

Dependency tree · 2 levels

Sources

Sigurdur Helgason, MIT 18.112 Lecture 16: Harmonic Functions