Math 411: Honours Complex Variables - University of Alberta

so that
u(z) =
� 2π
u(re
0
iθ )Pr(re iθ ,z)dθ.
Suppose now that u is arbitrary. For t ∈ (0,1), define
ut: Br(0)
→ R, z ↦→ u(tz).
t
Then ut is harmonic, and by the foregoing we have
ut(z) =
� 2π
ut(re
0
iθ )Pr(re iθ ,z)dθ
for z ∈ Br(0). Letting t → 1 − (cf. Problem 5.2), we obtain for z ∈ Br(0) that
u(z) = limut(z)
= lim
t→1 t→1
� 2π
ut(re
0
iθ )Pr(re iθ � 2π
,z)dθ = u(re
0
iθ )Pr(re iθ ,z)dθ.
Theorem 14.3. Let r > 0, and let f: ∂Br(0) → R be continuous. Define
�
f(z), z ∈ ∂Br(0),
g: Br[0] → R, z ↦→ � 2π
0 f(reiθ )Pr(reiθ ,z)dθ, z ∈ Br(0).
Then g is harmonic on Br(0) and continuous on Br[0].
Proof. There is no loss of generality if we suppose that r = 1.
For z ∈ D and ζ ∈ ∂D, note that
Re
ζ +z
ζ −z = Re(ζ +z)(¯ ζ − ¯z)
|ζ −z| 2 =
As the real part of a holomorphic function,
99
1
|ζ −z| 2Re(|ζ|2−|z| 2 +z¯ ζ−ζ¯z) = 1−|z|2
= 2πP1(ζ,z).
|ζ −z| 2
D → R, z ↦→ P1(ζ,z)
is therefore harmonic for each ζ ∈ ∂D. We thus obtain for z = x+iy ∈ D:
(∆g)(z) = ∂2g ∂x2(z)+ ∂2 � 2π
g
∂y2(z) = f(e iθ � 2 ∂
)
∂x2P1(e iθ ,z)+ ∂2
∂y2P1(e iθ �
,z) dθ = 0.
0
Consequently, g is harmonic on Br(0).
What remains to be shown is that g is continuous at any point z0 ∈ ∂D.
Let z0 = e iθ0 , and suppose without loss of generality (if necessary considering
instead g(−z)) that θ0 ∈ (0,2π). Let ǫ > 0. We need to find δ > 0 such that
|g(z0)−g(z)|< ǫ for all z ∈ D with |z0 −z|< δ.
For δ0 > 0, let J := [θ0 −2δ0,θ0 +2δ0]. By making δ0 > 0 sufficiently small, we
for θ ∈ J. Set
can ensure that J ⊂ [0,2π] and |f(e iθ )−f(z0)|< ǫ
2
S := {se iθ : s ∈ [0,1), θ ∈ [θ0 −δ0,θ0 +δ0]},
and note that C := inf{|e iθ −z|: θ ∈ [0,2π]\J, z ∈ S} > 0.