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|

.