A First Course in Complex Analysis · Problem 6.3.13

Power Series — Beck, Marchesi, Pixton & Sabalka

Mathematics — Complex Analysis · harmonic-functions

Problem

Recall from Exercise 4.5.31 the Poisson kernel

P_r(\phi) = \frac{1-r^2}{1 - 2r\cos(\phi) + r^2},

where

0 < r < 1

. In this exercise, we will prove the Poisson Integral Formula: if

u

is harmonic on an open set containing the closed unit disk

\overline{D}[0,1]

then for any

r < 1

u\!\left(r e^{i\phi}\right) = \frac{1}{2\pi}\int_0^{2\pi} u\!\left(e^{it}\right) P_r(\phi - t)\,dt. \tag{6.2}

Suppose

u

is harmonic on an open set containing

\overline{D}[0,1]

. By Exercise 6.3.14 we can find

R_0 > 1

so that

u

is harmonic in

D[0, R_0]

. (a) Recall the Möbius function

f_a(z) = \frac{z-a}{1-\bar{a}z},

for some fixed

a \in \mathbb{C}

with

|a| < 1

, from Exercise 3.6.9. Show that

u(f_{-a}(z))

is harmonic on an open disk

D[0, R_1]

containing

\overline{D}[0,1]

. (b) Apply Theorem 6.2.1 to the function

u(f_{-a}(z))

with

w = 0

to deduce

u(a) = \frac{1}{2\pi i}\int_{C[0,1]} \frac{u(f_{-a}(z))}{z}\,dz. \tag{6.3}

f_a(z)

maps the unit circle to itself, apply a change of variables to (6.3) to prove

u(a) = \frac{1}{2\pi}\int_0^{2\pi} u\!\left(e^{it}\right) \frac{1-|a|^2}{|e^{it}-a|^2}\,dt.

(d) Deduce (6.2) by setting

a = r e^{i\phi}