5Brownian motion

III Advanced Probability



5.2 Harmonic functions and Brownian motion
Recall that a Markov chain plus a harmonic function gave us a martingale. We
shall derive similar results here.
Definition (Domain). A domain is an open connected set D R
d
.
Definition (Harmonic function). A function u : D R is called harmonic if
f =
d
X
i=1
2
f
x
2
i
= 0.
There is also an alternative characterization of harmonic functions that
involves integrals instead of derivatives.
Lemma.
Let
u
:
D R
be measurable and locally bounded. Then the following
are equivalent:
(i) u is twice continuously differentiable and u = 0.
(ii) For any x D and r > 0 such that B(x, r) D, we have
u(x) =
1
Vol(B(x, r))
Z
B(x,r)
u(y) dy
(iii) For any x D and r > 0 such that B(x, r) D, we have
u(x) =
1
Area(B(x, r))
Z
B(x,r)
u(y) dy.
The latter two properties are known as the mean value property.
Proof. IA Vector Calculus.
Theorem.
Let (
B
t
)
t0
be a standard Brownian motion in
R
d
, and
u
:
R
d
R
be harmonic such that
E|u(x + B
t
)| <
for any
x R
d
and
t
0. Then the process (
u
(
B
t
))
t0
is a martingale with
respect to (F
+
t
)
t0
.
To prove this, we need to prove a side lemma:
Lemma.
If
X
and
Y
are independent random variables in
R
d
, and
X
is
G
-
measurable. If f : R
d
× R
d
R is such that f(X, Y ) is integrable, then
E(f(X, Y ) | G) = Ef (z, Y )|
z=X
.
Proof. Use Fubini and the fact that µ
(X,Y )
= µ
X
µ
Y
.
Observe that if
µ
is a probability measure in
R
d
such that the density of
µ
with respect to the Lebesgue measure depends only on
|x|
, then if
u
is harmonic,
the mean value property implies
u(x) =
Z
R
d
u(x + y) dµ(y).
Proof of theorem. Let t s. Then
E(u(B
t
) | F
+
s
) = E(u(B
s
+ (B
t
B
s
)) | F
+
s
)
= E(u(z + B
t
B
s
))|
Z=B
s
= u(z)|
z=B
s
= u(B
s
).
In fact, the following more general result is true:
Theorem.
Let
f
:
R
d
R
be twice continuously differentiable with bounded
derivatives. Then, the processes (X
t
)
t0
defined by
X
t
= f(B
t
)
1
2
Z
t
0
f(B
s
) ds
is a martingale with respect to (F
+
t
)
t0
.
We shall not prove this, but we can justify this as follows: suppose we have
a sequence of independent random variables {X
1
, X
2
, . . .}, with
P(X
i
= ±1) =
1
2
.
Let S
n
= X
1
+ ··· + X
n
. Then
E(f(S
n+1
) | S
1
, . . . , S
n
)f(S
n
) =
1
2
(f(S
n
1)+f(S
n
+1)2f(s
n
))
1
2
˜
f(S
n
),
and we see that this is the discretized second derivative. So
f(S
n
)
1
2
n1
X
i=0
˜
f(S
i
)
is a martingale.
Now the mean value property of a harmonic function
u
says if we draw a
sphere
B
centered at
x
, then
u
(
x
) is the average value of
u
on
B
. More generally,
if we have a surface
S
containing
x
, is it true that
u
(
x
) is the average value of
u
on S in some sense?
Remarkably, the answer is yes, and the precise result is given by Brownian
motion. Let (
X
t
)
t0
be a Brownian motion started at
x
, and let
T
be the first
hitting time of S. Then, under certain technical conditions, we have
u(x) = E
x
u(X
T
).
In fact, given some function ϕ defined on the boundary of D, we can set
u(x) = E
x
ϕ(X
T
),
and this gives us the (unique) solution to Laplace’s equation with the boundary
condition given by ϕ.
It is in fact not hard to show that the resulting
u
is harmonic in
D
, since
it is almost immediate by construction that
u
(
x
) is the average of
u
on a small
sphere around
x
. The hard part is to show that
u
is in fact continuous at the
boundary, so that it is a genuine solution to the boundary value problem. This
is where the technical condition comes in.
First, we quickly establish that solutions to Laplace’s equation are unique.
Definition
(Maximum principle)
.
Let
u
:
¯
D R
be continuous and harmonic.
Then
(i) If u attains its maximum inside D, then u is constant.
(ii) If D is bounded, then the maximum of u in
¯
D is attained at D.
Thus, harmonic functions do not have interior maxima unless it is constant.
Proof. Follows from the mean value property of harmonic functions.
Corollary.
If
u
and
u
0
solve
u
=
u
0
= 0, and
u
and
u
0
agree on
D
, then
u = u
0
.
Proof. u u
0
is also harmonic, and so attains the maximum at the boundary,
where it is 0. Similarly, the minimum is attained at the boundary.
The technical condition we impose on D is the following:
Definition
(Poincar´e cone condition)
.
We say a domain
D
satisfies the Poincar´e
cone condition if for any
x D
, there is an open cone
C
based at
X
such that
C D B(x, δ) =
for some δ 0.
Example.
If
D
=
R
2
\
(
{
0
} × R
0
), then
D
does not satisfy the Poincar´e cone
condition.
And the technical lemma is as follows:
Lemma.
Let
C
be an open cone in
R
d
based at 0. Then there exists 0
a <
1
such that if |x|
1
2
k
, then
P
x
(T
B(0,1)
< T
C
) a
k
.
Proof. Pick
a = sup
|x|≤
1
2
P
x
(T
B(0,1)
< T
C
) < 1.
We then apply the strong Markov property, and the fact that Brownian motion is
scale invariant. We reason as follows if we start with
|x|
1
2
k
, we may or may
not hit
B
(2
k+1
) before hitting
C
. If we don’t, then we are happy. If we are not,
then we have reached
B
(2
k+1
). This happens with probability at most
a
. Now
that we are at
B
(2
k+1
), the probability of hitting
B
(2
k+2
) before hitting
the cone is at most
a
again. If we hit
B
(2
k+3
), we again have a probability of
a
of hitting
B
(2
k+4
), and keep going on. Then by induction, we find that
the probability of hitting B(0, 1) before hitting the cone is a
k
.
The ultimate theorem is then
Theorem.
Let
D
be a bounded domain satisfying the Poincar´e cone condition,
and let ϕ : D R be continuous. Let
T
D
= inf{t 0 : B
t
D}.
This is a bounded stopping time. Then the function u :
¯
D R defined by
u(x) = E
x
(ϕ(B
T
D
)),
where
E
x
is the expectation if we start at
x
, is the unique continuous function
such that u(x) = ϕ(x) for x D, and u = 0 for x D.
Proof.
Let
τ
=
T
B(x,δ)
for
δ
small. Then by the strong Markov property and
the tower law, we have
u(x) = E
x
(u(x
τ
)),
and
x
τ
is uniformly distributed over
B
(
x, δ
). So we know
u
is harmonic in the
interior of
D
, and in particular is continuous in the interior. It is also clear that
u|
D
= ϕ. So it remains to show that u is continuous up to
¯
D.
So let
x D
. Since
ϕ
is continuous, for every
ε >
0, there is
δ >
0 such
that if y D and |y x| < δ, then |ϕ(y) ϕ(x)| ε.
Take
z
¯
D
such that
|z x|
δ
2
. Suppose we start our Brownian motion at
z
. If we hit the boundary before we leave the ball, then we are in good shape. If
not, then we are sad. But if the second case has small probability, then since
ϕ
is bounded, we can still be fine.
Pick a cone
C
as in the definition of the Poincar´e cone condition, and assume
we picked δ small enough that C B(x, δ) D = . Then we have
|u(z) ϕ(x)| = |E
z
(ϕ(B
T
D
)) ϕ(x)|
E
z
|ϕ(B
T
D
ϕ(x))|
εP
z
(T
B(x,δ)
> T
D
) + 2 sup kϕkP
z
(T
D
> T
B(x,δ)
)
ε + 2kϕk
P
z
(T
B(x,δ)
T
C
),
and we know the second term 0 as z x.