1Conformal transformations
III Schramm--Loewner Evolutions
1.2 Brownian motion and harmonic functions
Recall that a function
f
=
u
+
iv
is holomorphic iff it satisfies the Cauchy–
Riemann equations
∂u
∂x
=
∂v
∂y
,
∂u
∂y
= −
∂v
∂x
.
This in particular implies that both u and v are harmonic.
Definition
(Harmonic function)
.
A function
f
:
R
k
→ R
is harmonic if it is
C
2
and
∆f =
∂
2
∂x
2
1
+ ··· +
∂
2
∂x
2
k
f = 0.
Indeed, we can calculate
∂
2
u
∂x
2
+
∂
2
u
∂y
2
=
∂
∂x
∂v
∂y
+
∂
∂y
−
∂v
∂x
= 0.
In Advanced Probability, we saw that Brownian motion is very closely related
to harmonic functions.
Definition
(Complex Brownian motion)
.
We say a process
B
=
B
1
+
iB
2
is a
complex Brownian motion if (B
1
, B
2
) is a standard Brownian motion in R
2
.
Recall some results from Advanced Probability:
Theorem.
Let
u
be a harmonic function on a bounded domain
D
which is
continuous on
¯
D
. For
z ∈ D
, let
P
z
be the law of a complex Brownian motion
starting from z, and let τ be the first hitting time of D. Then
u(z) = E
z
[u(B
τ
)].
Corollary
(Mean value property)
.
If
u
is a harmonic function, then, whenever
it makes sense, we have
u(z) =
1
2π
Z
2π
0
u(z + re
iθ
) dθ.
Corollary
(Maximum principle)
.
Let
u
be harmonic in a domain
D
. If
u
attains
its maximum at an interior point in D, then u is constant.
Corollary
(Maximum modulus principle)
.
Let
D
be a domain and let
f
:
D → C
be holomorphic. If
|f|
attains its maximum in the interior of
D
, then
f
is constant.
Proof.
Observe that if
f
is holomorphic, then
log |f|
is harmonic and attains its
maximum in the interior of
D
. It then follows that
|f|
is constant (if
|f|
vanishes
somewhere, then consider
log |f
+
M|
for some large
M
, and do some patching if
necessary). It is then a standard result that a holomorphic function of constant
modulus is constant.
The following lemma should be familiar from the old complex analysis days
as well:
Lemma
(Schwarz lemma)
.
Let
f
:
D → D
be a holomorphic map with
f
(0) = 0.
Then
|f
(
z
)
| ≤ |z|
for all
z ∈ D
. If
|f
(
z
)
|
=
|z|
for some non-zero
z ∈ D
, then
f(w) = λw for some λ ∈ C with |λ| = 1.
Proof. Consider the map
g(z) =
(
f(z)/z z 6= 0
f
0
(0) z = 0
.
Then one sees that
g
is holomorphic and
|g
(
z
)
| ≤
1 for all
z ∈ ∂D
, hence for all
z ∈ D
by the maximum modulus principle. If
|f
(
z
0
)
|
=
|z
0
|
for some
z
0
∈ D \{
0
}
,
then g must be constant, so f is linear.