1.5. Space of harmonic functions 37
The argument uses the explicit formula that we derive for the rec-
tangle. Although we cannot get such a nice formula in general, we
can derive two important facts. Suppose A is a finite subset of
Z2
containing AN . Then for x AN , z ∂A,
HA(x, z) =
y∈∂AN
HAN (x, y) HA(y, z).
Using this and (1.27) we get for x, ˜ x
ˆ
A
N
,
HA(x, z)
c2
HA(˜ x, z),
|HA(x, z) HA(˜ x, z)| c1
|x ˜| x
N
HA(x, z).
We can extend this to harmonic functions.
Theorem 1.14 (Difference estimates). There is a c such that
if A is a finite subset of
Zd
and F : A [−M, M] is harmonic on A,
then if x, z A with |z x| = 1,
(1.29) |F (z) F (x)|
c M
dist(x, ∂A)
.
Theorem 1.15 (Harnack principle). Suppose K is a compact subset
of Rd and U is an open set containing K. There is a c = c(K, U)
such that the following holds. Suppose N is a positive integer, A is
a finite subset of Zd contained in NU = {z Rd : z/N U}, and
ˆ
A is a subset of A contained in NK. Suppose F : A [0, ∞) is a
harmonic function. Then for all x, z
ˆ,
A
F (x) c F (z).
As an application of this, let us show that: the only bounded
functions on Zd that are harmonic everywhere are constants. For d =
1, this is immediate from the fact that the only harmonic functions
are the linear functions. For d 2, we suppose that F is a harmonic
function on Zd with |F (z)| M for all z. If x Zd and AR is a
bounded subset of Zd containing all the points within distance R of
the origin, then (1.29) shows that
|F (x) F (0)| c M
|x|
R |x|
.
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