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Poisson integrals are harmonic on the unit disc

Statement

For every continuous boundary datum $φ : \partial D \to R$ , the Poisson integral $P [φ]$ is harmonic on $D$ .

Facts & Assumptions

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

[L1]

For fixed $t$ , the function $H_{t} (z) : = \frac{e^{i t} + z}{e^{i t} - z} φ (e^{i t})$ is holomorphic on $D$ , and the family is jointly continuous in $(t, z)$ on $[0, 2 π] \times D$ ; therefore $F (z) : = \frac{1}{2 π} \int_{0}^{2 π} H_{t} (z) d t$ is holomorphic on $D$ (A jointly continuous finite-interval parameter integral of holomorphic functions is holomorphic).

[L2]

The real part of $H_{t} (z)$ is $P (z, e^{i t}) φ (e^{i t})$ , by the defining algebra of the Poisson kernel (The Poisson kernel on the unit disc).

[L3]

Holomorphic functions are smooth in their real coordinates, and the $C^{2}$ real part of a holomorphic function is harmonic (Holomorphic functions are real analytic and smooth in their two real coordinates, The $C^{2}$ real and imaginary parts of a holomorphic function satisfy Laplace's equation and form a harmonic-conjugate pair).

Proof

technique · direct

1.1L1

By [L1], the parameter integral $F (z) = \frac{1}{2 π} \int_{0}^{2 π} \frac{e^{i t} + z}{e^{i t} - z} φ (e^{i t}) d t$ is holomorphic on $D$ .

2.1step 1.1L2

Taking real parts under the integral and using [L2] gives $Re F (z) = \frac{1}{2 π} \int_{0}^{2 π} P (z, e^{i t}) φ (e^{i t}) d t = P [φ] (z) .$

3.1step 1.1step 2.1L3∎

By [L3], the real part of the holomorphic function $F$ is harmonic. Since step 2.1 identifies that real part with $P [φ]$ , the Poisson integral is harmonic on $D$ .

Depends on

Dependency tree · 2 levels

Sources

Sigurdur Helgason, MIT 18.112 Lecture 16: Harmonic Functions