3.2 The Perron Method 63
(iii) First v ≤ ¯v,sincev is subharmonic. Let Ω
⊂⊂ Ω, u harmonic on Ω
,
v ≤ u on ∂Ω
.Sincev ≤ v,alsov ≤ u on ∂Ω
, and thus, since v is
subharmonic, v ≤ u on Ω
and thus v ≤ u on Ω
\
˚
B(y, R). Therefore,
also
v ≤ u on Ω
∩ ∂B(y, R). Since v is harmonic, hence subharmonic
on Ω
∩ B(y, R), we get v ≤ u on Ω
∩ B(y, R). Altogether, we obtain
v ≤ u on Ω
. This shows that v is subharmonic.
For the sequel, let ϕ be a bounded function on Ω (not necessarily contin-
uous).
Definition 3.2.2: A subharmonic function u ∈ C
0
(
¯
Ω) is called a subfunc-
tion with respect to ϕ if
u ≤ ϕ for all x ∈ ∂Ω.
Let S
ϕ
be the set of all subfunctions with respect to ϕ. (Analogously, a su-
perharmonic function u ∈ C
0
(
¯
Ω) is called superfunction with respect to ϕ if
u ≥ ϕ on ∂Ω.)
The key point of the Perron method is contained in the following theorem:
Theorem 3.2.1: Let
u(x):= sup
v∈S
ϕ
v(x). (3.2.1)
Then u is harmonic.
Remark: If w ∈ C
2
(Ω) ∩ C
0
(
¯
Ω)isharmoniconΩ,andifw = ϕ on ∂Ω,the
maximum principle implies that for all subfunctions v ∈ S
ϕ
,wehavev ≤ w
in Ω and hence
w(x)= sup
v∈S
ϕ
v(x).
Thus, w satisfies an extremal property. The idea of the Perron method (and
the content of Theorem 3.2.1) is that, conversely, each supremum in S
ϕ
yields
a harmonic function.
Proof of Theorem 3.2.1: First of all, u is well-defined, since by the maximum
principle v ≤ sup
∂Ω
ϕ<∞ for all v ∈ S
ϕ
.Nowlety ∈ Ω be arbitrary.
By (3.2.1) there exists a sequence {v
n
}⊂S
ϕ
with lim
n→∞
v
n
(y)=u(y).
Replacing v
n
by max(v
1
,...,v
n
, inf
∂Ω
ϕ), we may assume without loss of
generality that (v
n
)
n∈N
is a monotonically increasing, bounded sequence. We
now choose R with B(y, R) ⊂⊂ Ω and consider the harmonic replacements
¯v
n
for B(y,R). The maximum principle implies that (¯v
n
)
n∈N
likewise is a
monotonically increasing sequence of subharmonic functions that are even