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.
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