Math 411: Honours Complex Variables - University of Alberta

102 CHAPTER 14. HARMONIC FUNCTIONS
for all r ∈ [0,R]. As the integrand is continuous, we conclude that g(z0 +re iθ ) = 0
for all r ∈ [0,R] and θ ∈ [0,2π], i.e. g ≡ 0 on BR[z0]. This means that, for z ∈ BR[z0],
we have
|h(z)|≤ |h(z0)|= Reh(z) ≤ |h(z)|,
so that Reh(z) = |h(z)|= |h(z0)| for z ∈ BR[z0]. That is, h(z) = |h(z)|= |h(z0)|=
|f(z0)|, so that f(z) = f(z0) for z ∈ BR[z0].
Corollary 14.4.1. Let D ⊂ C be open, let f : D → R be continuous and have
the mean value property, and suppose that f has a local maximum or minimum at
z0 ∈ D. Then f is constant on a neighbourhood of z0.
Proof. We only consider the case of a local maximum (for a local minimum, replace
f by −f).
Let R > 0 be such that BR[z0] ⊂ D and f(z) ≤ f(z0) for all z ∈ BR[z0]. Choose C
such that f(z)+C ≥ 0 for all z ∈ BR[z0]. It follows that |f+C| has a local maximum
at z0. Hence, f +C is constant on a neighbourhood of z0, as is f.
Corollary 14.4.2. Let D ⊂ C be open, connected, and bounded, and let f : D →
R be continuous such that f|D has the mean value property. Then f attains its
maximum and minimum on ∂D.
Proof. Without loss of generality, suppose that f is not constant. Let z0 ∈ D be such
that f(z0) is maximal. Set
V := {z ∈ D : f(z) < f(z0)}.
Then V is open and not empty. Let z ∈ D \ V, i.e. f(z) = f(z0). Then f has a
local maximum at z, so that, by Corollary 14.4.1, f(w) = f(z) = f(z0) for w in a
neighbourhood, say W ⊂ D, of z. Consequently, W ⊂ D \V holds, so that z is an
interior point of D \V. Since z ∈ D \V is arbitrary, this shows that D \V is open.
Since D is connected, and V �= ∅, we must have D \V = ∅, i.e. V = D.
The case of a minimum is treated analogously.
Corollary 14.4.3 (EquivalenceofHarmonicandMean-ValueProperties). LetD ⊂ C
be open, and let f: D → R be continuous. Then the following are equivalent:
(i) f is harmonic;
(ii) f has the mean value property.
Proof. Only (ii) =⇒ (i) needs proof (cf. Corollary 14.1.4).
Let z0 ∈ D, and let R > 0 be such that BR[z0] ⊂ D. By Theorem 14.3, there is a
continuous function g: BR[z0] → R such that g|∂BR[z0]= f|∂BR[z0] and g|Br(z0) is harmonic.
Consequently, (g−f)|BR(z0) has the mean value property. By Corollary 14.4.2,
this means that g − f attains its maximum and minimum over BR[z0] on ∂BR[z0],
so that g = f on BR[z0]. Hence, f|BR(z0) is harmonic, i.e. ∆f ≡ 0 on BR[z0]. Since
z0 ∈ D was arbitrary, this means that ∆f ≡ 0.