64 2 The Dirichlet Problem
continuous on Ω,and|u(x)| = |L(u : x, δ)|≤L(|u| : x, δ) whenever B
−
x,δ
⊂ Ω,
|u| is sub-mean-valued on Ω and is therefore subharmonic on Ω.
Example 2.3.11 If u is harmonic on the open set Ω,thenu
+
=max(u, 0)
is subharmonic on Ω and u
−
=max(−u, 0) is subharmonic on Ω. Finiteness
and continuity of u
+
and u
−
follow from that of u.Sinceu(x)=L(u : x, δ) ≤
L(u
+
: x, δ) whenever B
−
x,δ
⊂ Ω, u
+
(x) ≤ L(u
+
: x, δ) whenever B
−
x,δ
⊂ Ω.
Thus, u
+
is sub-mean-valued on Ω and therefore subharmonic on Ω.
The last example is a special case of a more general statement that a
convex function of a harmonic function is a subharmonic function. Recall
that a real-valued function φ on a finite or infinite interval (a, b)isconvex if
φ(λx +(1− λ)y) ≤ λφ(x)+(1− λ)φ(y)
whenever a<x<y<band 0 <λ<1. Such functions are continuous on
(a, b) and have right- and left-hand derivatives at each point of (a, b). See
Royden [54] for such facts. Let μ be a unit measure on the Borel subsets B
of the open set Ω,andletf be a real-valued function taking on values in the
domain (a, b) of the convex function φ. The inequality
φ
Ω
fdμ
≤
Ω
φ(f) dμ
is known as Jensen’s Inequality.Ifφ is convex on (−∞, +∞), if f is
integrable relative to μ,andiff takes on the values +∞ and −∞,then
Jensen’s Inequality holds if φ is arbitrarily defined at +∞ and −∞.This
can be seen by applying Jensen’s Inequality to
∼F
fdμ where F = {x ∈
Ω; |f (x)| =+∞}.
Theorem 2.3.12 If u is harmonic on the open set Ω and φ is a convex func-
tion on an open interval containing the range of u,thenφ(u) is subharmonic
on Ω.
Proof: φ(u) is finite-valued and continuous since both φ and u have these
properties. If B
−
x,δ
⊂ Ω,thenφ(u(x)) = φ(L(u : x, δ)) ≤ L(φ(u):x, δ)by
Jensen’s Inequality; that is, φ(u) is sub-mean-valued on Ω and is therefore
subharmonic on Ω.
Example 2.3.10 is a special case of this theorem with φ(x)=|x|,x∈ R.
Theorem 2.3.13 If u is subharmonic on the open set Ω and φ is an in-
creasing convex function on an open interval containing the range of u,then
φ(u) is subharmonic on Ω.
Proof: Since φ is real-valued, φ(u) is real-valued. Since a convex function
is continuous, φ(u) is u.s.c. By Jensen’s Inequality and the fact that φ is