Math 411: Honours Complex Variables - University of Alberta

Set K := supζ∈∂D|f(ζ)|. For z ∈ S, we then have
�
|I2| ≤
(|f(e
[0,2π]\J
iθ )|+|f(z0)|)P1(e iθ ,z)dθ
= 1
�
2π
≤ K
�
π
≤ K
πC2 �
[0,2π]\J
[0,2π]\J
[0,2π]\J
≤ 2K
C 2 (1−|z|2 )
(|f(e iθ )|+|f(z0)|) 1−|z|2
|eiθ dθ
−z| 2
1−|z| 2
|eiθ dθ
−z| 2
(1−|z| 2 )dθ, because z ∈ S,
Choose δ ∈ (0,δ0) so small that |z0 −z|< δ for z ∈ D implies
1−|z| 2 < C2 ǫ
2K 2 .
For z ∈ D with |z0 − z|< δ, we then have z ∈ S and hence |I2|< ǫ.
On combining
2
these results, we see that |g(z0)−g(z)|< ǫ.
Definition. Let D ⊂ C be open, and let f : D → C be continuous. We say that f
has the mean value property if, for every z0 ∈ D, there exists R > 0 with BR[z0] ⊂ D
such that
for all r ∈ [0,R].
f(z0) = 1
� 2π
f(z0 +re
2π 0
iθ )dθ
Theorem 14.4. Let D ⊂ C be open, and let f : D → C have the mean value
property such that |f| attains a local maximum at z0 ∈ D. Then f is constant on a
neighbourhood of z0.
Proof. Choose R > 0 with BR[z0] ⊂ D such that |f(z0)|≥ |f(z)| for all z ∈ BR[z0]
and f(z0) = 1
� 2π
2π 0 f(z0 + reiθ )dθ for all r ∈ [0,R]. If f(z0) = 0 then the result is
trivial. Otherwise, let
h(z) = |f(z0)|
f(z0) f(z)
and set g := Reh−|h(z0)|. Then g has the mean value property and satisfies
for z ∈ BR[z0]. It follows that
g(z) ≤ |h(z)|−|h(z0)|≤ 0
0 = g(z0) =
� 2π
0
g(z0 +re iθ ) dθ
� �� �
≤0
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