1.2. Boundary value problems 21

This matrix is often called the Poisson kernel. For a given set A, we

can solve the Dirichlet problem for any boundary function in terms

of the Poisson kernel.

♦

Analysts who are not comfortable with

probability1

think of the Poisson

kernel only as the matrix for the transformation which takes boundary data to

values on the interior. Probabilists also have the interpretation of HA(x, y) as

the probability that the random walk starting at x exits A at y.

What happens in Theorem 1.5 if we allow A to be an infinite

set? In this case it is not always true that the solution is unique.

Let us consider the one-dimensional example with A = {1, 2, 3,...}

and ∂A = {0}. Then for every c ∈ R, the function F (x) = cx is

harmonic in A with boundary value 0 at the origin. Where does

our proof break down? This depends on which proof we consider

(they all break down!), but let us consider the martingale version.

Suppose F is harmonic on A with F (0) = 0 and suppose Sn

is a

simple random walk starting at positive integer x. As before, we let

T = min{n ≥ 0 : Sn = 0} and Mn = F (Sn∧T ). The same argument

shows that Mn is a martingale and

F (x) = E[Mn] =

∞

y=0

F (y) P{Sn∧T = y}.

We have shown in a previous section that with probability one T ∞.

This implies that P{Sn∧T = 0} tends to 1, i.e.,

lim

n→∞

y0

P{Sn∧T = y} = 0.

However, if F is unbounded, we cannot conclude from this that

lim

n→∞

y0

F (y) P{Sn∧T = y} = 0.

However, we do see from this that there is only one bounded function

that is harmonic on A with a given boundary value at 0. We state

the theorem leaving the details as Exercise 1.7.

1The

politically correct term is stochastically challenged.