1.5. Space of harmonic functions 39

Theorem 1.16. If A is a proper cofinite subset of Zd (d ≥ 3), then

the only bounded functions on Zd that vanish on Zd \ A and are har-

monic on A are of the form

(1.30) F (x) = r P{TA = ∞ | S0 = x}, r ∈ R.

We will first consider the case A =

Zd

\ {0} and assume that

F :

Zd

→ [−M, M] is a function satisfying F (0) = 0 and LF (x) = 0

for x = 0. Let α = LF (0) and let

f(x) = F (x) + α G(x, 0).

Then f is a bounded harmonic function and hence must be equal to

a constant. Since G(x, 0) → 0 as x → ∞, the constant must be r and

hence

F (x) = r − α G(x, 0)

= r P{τ0 = ∞ | S0 = x} + P{τ0 ∞ | S0 = x}[r − αG(0, 0)].

Since F (0) = 0 and P{τ0 = ∞ | S0 = 0} = 0, we know that r −

αG(0, 0) = 0 and hence F is of the form (1.30).

For other cofinite A, assume F is such a function with |F | ≤ 1.

Then F satisfies

LF (x) = −g(x), x ∈ A

for some function g that vanishes on A. In particular,

f(x) = F (x) +

y∈Zd\A

G(x, y) g(x),

is a bounded harmonic function (why is it bounded?) and hence

constant. This tells us that there is an r such that

F (x) = r −

y∈Zd\A

G(x, y) g(x),

which implies, in particular, that F (x) → r as r → ∞. Also, if

x ∈ Zd \ A, F (x) = 0 which implies

y∈Zd\A

G(x, y) g(x) = r.

If we show that G(x, y) is invertible on

Zd

\ A, then we know there

is a unique solution to this equation, which would determine g, and

hence F .