THE CHINESE UNIVERSITY OF HONG KONG
Department of Mathematics
2018-2019 semester 1 MATH4060
week 5 tutorial
Underlined contents were not included in the tutorial because of time constraint, but
included here for completeness.
Below is a brief introduction to properties of harmonic functions. Removable singularity
theorem and Liouville’s theorem for harmonic functions are proven by maximum prin-
ciple and Poisson integral formula. The main reference is Chapter 2 of Gilbarg and
Trudinger’s Elliptic Partial Differential Equations of Second Order. Below, Ω always
denotes a nonempty connected open set in R2 = C.
1 Properties of Harmonic Functions
A C 2 function u : Ω → R is harmonic iff ∆u = uxx + uyy = 0.
Harmonic functions and holomorphic functions are intimately related.
1. f is holomorphic iff ∂z̄ f = 0, whereas u is harmonic iff ∂z ∂z̄ u = 0.
2. If f is holomorphic, then <f , =f and log |f | are harmonic whenever finitelyR defined.
If Ω is simply connected and u is harmonic, then f = u + iv, where v = (ux dy −
uy dx), is holomorphic, and log |ef | = u.
3. (Cauchy integral formul and mean-value property) If f is holmorphic, then
Z
1 f (w)
f (z) = dw.
2πi ∂B(z,r) w − z
If u is harmonic, then
Z Z
1 1 u(w)
u(z) = u(w)dw = dw. (1)
2πr ∂B(z,r) 2π ∂B(z,r) |w − z|
4. (strong maximum (modulus) principle) If a holomorphic f attains the maximum
modulus in the interior, then it is constant. If a harmonic u attains the maximum
in the interior, then it is constant.
5. (weak maximum (modulus) principle) The maximum modulus of a holomorphic
function or a harmonic function on a bounded domain is attained on the boundary.
Mean-value property for harmonic function is more rigid than that for holomorphic func-
tion because the domain of integration in (1) cannot be any ∂B(w, r) containing z. Indeed,
the offset mean-value property is given by the more involved Poisson integral formula.
2
Proposition 1 (Poisson integral formula). Suppose u is harmonic on a neighbourhood
of B(0, R). Let ϕ = u|∂B(0, R). Then for x ∈ B(0, R)
Z
u(x) = ϕ(y)P2 (x, y)dy, (2)
∂B(0,R)
n
1 R2 −|x|2 R
where Pn (x, y) = |∂B(0,R)| R2 |x−y|
.
Conversely, if ϕ is a continuous function on ∂B(0, R), then (2) defines a harmonic function
on B(0, R) whose continuous extension to ∂B(0, R) exists and agrees with ϕ.
Corollary 2. Harmonic functions are smooth.
Below, we prove removable singularity theorem and Liouville’s theorem for harmonic
functions.
Proposition 3 (Removable singularity theorem). Suppose u is harmonic on B(0, r)\{0}.
If u(z) = o(log |z|) as z → 0, then u extends to a harmonic function on B(0, r).
Proof. It suffices to show u agrees to ũ defined by Poisson integral formula, which is a
harmonic function on B(0, r). Let w = ũ − u. Then w(z) = o(log |z|) = o(log |z| − log r).
Note that both w and log |z| − log r vanish on ∂B(0, r). By maximum principle, for ε > 0,
since ±w(z) + ε log |z| → −∞, supB(0,r)\{0} ±w + ε(log |z| − log r) ≤ 0. The result follows
by letting ε → 0.
To prove Liouville’s property, it is handy to have an estimate on the gradient.
Proposition 4 (gradient estimate). Suppose u is harmonic on a neighbourhood of B(0, R).
Then
n
|∂i u(0)| ≤ kukL∞ (B(0,R)) .
R
Remark. Repeated application of the gradient estimate shows harmonic functions are in
fact analytic.
Proof. Apply differentiation under the integral sign on Poisson integral formula.
Proposition 5 (Liouville’s theorem). If a harmonic function on R2 is bounded, then it
is constant.
Proof. Let R → ∞ in the gradient estimate.
Exercise 6. Complete the following alternative proof of Liouville’s theorem:
By Poisson integral formula, we have the following Harnack inequality for nonnegative
harmonic u on R2
1 R − |x| 1 R + |x|
u(0) ≤ u(x) ≤ u(0).
(R + |x|) R (R − |x|) R
Liouville’s theorem for nonnegative functions then follows by letting R → ∞ on the far
right. The general case follows by translation.
