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 .
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