2Loewner's theorem

III Schramm--Loewner Evolutions



2.1 Key estimates
Before we prove Loewner’s theorem, we establish some key identities and esti-
mates. As before, a useful thing to do is to translate everything in terms of
Brownian motion.
Proposition. Let A Q and B be a complex Brownian motion. Set
τ = inf{t 0 : B
t
6∈ H \ A}.
Then
If x > Rad(A), then
g
A
(x) = lim
y→∞
πy
1
2
P
iy
[B
τ
[x, )]
.
If x < Rad(A), then
g
A
(x) = lim
y→∞
πy
P
iy
[B
τ
(−∞, x]]
1
2
.
Proof. First consider the case A = and, by symmetry, x > 0. Then
lim
y→∞
πy
1
2
P
iy
[B
τ
[x, )]
= lim
y→∞
πy P
iy
[B
τ
[0, x)]
= lim
y→∞
πy
Z
x
0
y
π(s
2
+ y
2
)
ds
= x,
where the first equality follows from the fact that Brownian motion exits through
the positive reals with probability
1
2
; the second equality follows from the
previously computed exit distribution; and the last follows from dominated
convergence.
Now suppose A 6= . We will use conformal invariance to reduce this to the
case above. We write g
A
= u
A
+ iv
A
. We let
σ = inf{t > 0 : B
t
6∈ H}.
Then we know
P
iy
[B
τ
[x, )] = P
g
A
(iy)
[B
σ
[g
A
(x), )]
= P
iv
A
(iy)
[B
σ
[g
A
(x) u
A
(iy), )].
Since
g
A
(
z
)
z
0 as
z
, it follows that
v
A
(iy)
y
1 and
u
A
(
iy
)
0 as
y . So we have
P
iv
A
(iy)
B
σ
[g
A
(x) u
A
(iy), )
P
iy
B
σ
[g
A
(x), )
= o(y
1
)
as y . Combining with the case A = , the the proposition follows.
Corollary. If A Q, Rad(A) 1, then
x g
A
(x) x +
1
x
if x > 1
x +
1
x
g
A
(x) x if x < 1.
Moreover, for all A Q, we have
|g
A
(z) z| 3 Rad(A).
Proof.
Exercise on the first example sheet. Note that
z 7→ z
+
1
z
sends
H \
¯
D
to
H.
This corollary gives us a “zeroth order” bound on
g
A
(
z
)
z
. We can obtain
a better first-order bound as follows:
Proposition.
There is a constant
c >
0 so that for every
A Q
and
|z| >
2 Rad(A), we have
g
a
(z)
z +
hcap(A)
z
c
Rad(A) · hcap(A)
|z|
2
.
Proof. Performing a scaling if necessary, we can assume that Rad(A) = 1. Let
h(z) = z +
hcap(A)
z
g
A
(z).
We aim to control the imaginary part of this, and then use the Cauchy–Riemann
equations to control the real part as well. We let
v(z) = im(h(z)) = im(z g
A
(z)) =
im(z)
|z|
2
hcap(A).
Let B be a complex Brownian motion, and let
σ = inf{t 0 : B
t
6∈ H \
¯
D}
τ = inf{t 0 : B
t
6∈ H}.
Let
p
(
z, e
) be the density with respect to the Lebesgue measure at
e
for
B
σ
.
Then by the strong Markov property at the time σ, we have
im(z g
A
(z)) =
Z
π
0
E
e
[im(B
τ
)]p(z, e
) dθ.
In the first example sheet Q3, we show that
p(z, e
) =
2
π
im(z)
|z|
2
sin θ
1 + O
1
|z|

. ()
We also have
hcap(A) =
2
π
Z
π
0
E
e
[im(B
τ
)] sin θ dθ.
So
|v(z)| =
im(z g
A
(z))
im(z)
|z|
2
hcap(A)
=
Z
π
0
E
e
[im(B
τ
)]p(z, e
) dθ
im(z)
|z|
2
·
2
π
Z
π
0
E
e
[im(B
τ
)] sin θ dθ
.
By applying (), we get
|v(z)| c
c hcap(A) im(z)
|z|
3
.
where c is a constant.
Recall that
v
is harmonic as it is the imaginary part of a holomorphic function.
By example sheet 1 Q9, we have
|
x
v(z)|
c hcap(A)
|z|
3
, |
y
v(z)|
c hcap(A)
|z|
3
.
By the Cauchy–Riemann equations,
re
(
h
(
z
)) satsifies the same bounds. So we
know that
|h
0
(z)|
c hcap(A)
|z|
3
.
Then
h(iy) =
Z
y
h
0
(is) ds,
since h(iy) 0 as y . Taking absolute values, we have
|h(iy) =
Z
y
|h
0
(is)| ds
c hcap(A)
Z
y
s
3
ds
c
0
hcap(A)y
z
.
To get the desired bound for a general
z
, integrate
h
0
along the boundary of the
circle of radius |z| to get
|h(z)| = |h(re
)|
c hcap(A)
|z|
2
+ h(iz).
The following is a very useful fact about Brownian motion:
Theorem
(Beurling estimate)
.
There exists a constant
c >
0 so that the
following holds. Let
B
be a complex Brownian motion, and
A
¯
D
be connected,
0 A, and A
¯
D 6= . Then for z D, we have
P
z
[B[0, τ] A = ] c|z|
1/2
,
where τ = inf{t 0 : B
t
6∈ D}.
We will not prove this, since it is quite tricky to prove. The worst case
behaviour is obtained when A = [i, 0].
To the theorem, we need one more estimate, which uses the Beurling estimate.
Proposition.
There exists a constant
c >
0 so that the following is true:
Suppose A,
˜
A Q with A
˜
A and
˜
A \ A is coonnected. Then
diam(g
A
(
˜
A \ A)) c
(
(dr)
1/2
d r
Rad(
˜
A) d > r
,
where
d = diam(
˜
A \ A), r = sup{im(z) : z
˜
A}.
Proof. By scaling, we can assume that r = 1.
If d 1, then the result follows since
|g
A
(z) z| 3Rad(A),
and so
diam(g
A
(
˜
A \ A)) diam(A) + 6Rad(A) 8 diam(
˜
A).
If
d <
1, fix
z H
so that
U
=
B
(
z, d
)
˜
A \ A
. It then suffices to bound
the size of g
A
(U) (or, to be precise, g
A
(U \ A)).
Let B be a complex Brownian motion starting from iy with y 2. Let
τ = inf{t 0 : B
t
6∈ H \ A}.
For B[0, τ ] to reach U, it must
(i)
Reach
B
(
z,
1) without leaving
H \ A
, which occurs with probability
at most c/y for some constant c, by example sheet.
(ii)
It must then hit
U
before leaving
H \ A
. By the Beurling estimate,
this occurs with probability cd
1/2
.
Combining the two, we see that
lim sup
y→∞
P
iy
[B[0, τ] U 6= ] cd
1/2
.
By the conformal invariance of Brownian motion, if
σ
=
inf{t
0 :
B
t
6∈
H}, this implies
lim sup
y→∞
y P
y
[B[0, σ] g
A
(
˜
A \ A) 6= ] cd
1/2
.
Since g
A
(
˜
A \ A) is connected, by Q10 of example sheet 1, we have
diam(g
A
(
˜
A \ A)) cd
1/2
.
We can finally get to the key content of Loewner’s theorem. An important
definition is the following:
Definition.
Suppose (
A
t
)
t0
is a family of compact
H
-hulls. We say that (
A
t
)
is
(i) non-decreasing if s t implies A
s
A
t
.
(ii)
locally growing if for all
T >
0 and
ε >
0, there exists
δ >
0 such that
whenever 0 s t s + δ T , we have
diam(g
A
s
(A
t
\ A
s
)) ε.
This is a continuity condition.
(iii) parametrized by half-plane capacity if hcap(A
t
) = 2t for all t 0.
We write
A
be the set of families of compact
H
-hulls which satisfy (i) to (iii).
We write A
T
for the set of such families defined on [0, T ].
Example.
If
γ
is is a simple curve in
H
starting from 0 and
A
t
=
γ
[0
, t
]. This
clearly satisfies (i), and the previous proposition tells us this satsifies (ii). On the
first example sheet, we show that we can reparametrize
γ
so that
hcap
(
A
t
) = 2
t
for all t 0. Upon doing so, we have A
t
A.
Theorem.
Suppose that (
A
t
)
A
. Let
g
t
=
g
A
t
. Then there exists a continuous
function U : [0, ) R so that
t
g
t
(z) =
2
g
t
(z) U
t
, g
0
(z) = z.
This is known as the chordal Loewner ODE, and
U
is called the driving
function.
Proof.
First note that since the hulls are locally growing, the intersection
T
st
g
t
(
A
s
) consists of exactly one point. Call this point
U
t
. Again by the
locally growing property, U
t
is in fact a continuous function in t.
Recall that if A Q, then
g
A
(z) = z +
hcap(A)
z
+ O
hcap(A)Rad(A)
|z|
2
.
If x R, then as g
A+x
(z) x = g
A
(z x), we have
g
A
(z) = g
A
(z + x) x = z +
hcap(z)
z + x
+ O
hcap(A)Rad(A + x)
|z + x|
2
. ()
Fix ε > 0. For 0 s t, let
g
s,t
= g
t
g
1
s
.
Note that
hcap(g
T
(A
t+ε
\ A
t
)) = 2ε.
Apply (
) with
A
=
g
t
(
A
t+ε
\ A
t
),
x
=
U
t
, and use that
Rad
(
A
+
x
) =
Rad(A U
t
) diam(A) to see that
g
A
(z) = g
t,t+ε
(z) = z +
2ε
z U
t
+ 2ε diam(g
t
(A
t+ε
\ A
t
))O
1
|z U
t
|
2
.
So
g
t+ε
(z) g
t
(z) = (g
t,t+ε
g
t,t
) g
t
(z)
=
2ε
g
t
(z) U
t
+ 2ε diam(g
t
(A
t+ε
\ A
t
))O
1
|g
t
(z) U
t
|
2
.
Dividing both sides by
ε
and taking the limit as
ε
0, the desired result follows
since diam(g
t
(A
t+ε
\ A
t
)) 0.
We can ask about the reverse procedure given a continuous function
U
t
,
can we find a collection
A
t
A
whose driving function is
U
t
? We can simply let
g
t
be the solution to this differential equation. We can then let
A
t
= H \ domain(g
t
).
On the example sheet, we show that this is indeed a family in
A
, and
g
t
are
indeed conformal transformations.