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Derivation of the Poisson Kernel
From the Use of the Argument Function
We are going to derive the Poisson kernel by using the argument function
and plane Euclidean geometry. We start out with the basic building block of
the Poisson kernel. It is the harmonic measure. Given two points e
iα and e
iβ
on the unit circle, we have the harmonic measure associated to the arc which
goes from e
iα
to e
iβ
in the counterclockwise sense. We denote it by ω α,β
. The
harmonic measure ω α,β
is a function on the unit disk which is characterized
by the following properties.
(i) ω α,β
is harmonic on the open unit disk,
(ii) ωα,β assumes the value 1 on the arc in the counterclockwise sense be-
tween e
iα and e
iβ , and
(iii) ω α,β
assumes the value 0 on the arc in the counterclockwise sense be-
tween e
iβ
and e
iα
.
The Poisson kernel is actually the harmonic measure for the case where α = θ
and β = θ + dθ or more precisely
lim
β→α
ω α,β
β − α
α=θ
One way to get the harmonic measure ω α,β
is to use the argument func-
tion. The idea is to use the function
arg
z − e
iβ
z − e
iα
This is a branch of a holomorphic function and is defined as the difference
arg
z − e
iβ
− arg (z − e
iα
) of the two branch functions arg
z − e
iβ
and
arg (z − e
iα ).
The branch arg
z − e
iβ
is defined by removing the cut
re
iβ
r ≥ 1
from C. We have to specify the range of the numerical value of arg
z − e
iβ
Let A be the point e
iα
and B be the point e
iβ
and O be the origin. We draw a
tangent line to the unit circle {|z| = 1} at the point B and put a point T B
on
it so that the direction from B to T B
represents the direction of the velocity
vector at B for a point moving in counterclockwise sense along the unit circle
{|z| = 1}. We use the interval (0, π) as the range for arg
z − e
iβ
− arg
BT
B
for a point in {|z| = 1} different from B (that is, measuring from the direction
BT
B
). We can choose the value
π
2
+β for arg
BT
B
, because we can choose the
value β for arg
OB and the vector direction arg
BT
B
is obtained by turning
the vector direction
OB one right angle in the counterclockwise sense. So
we can use the range β +
π
2
≤ arg
z − e
iβ
≤ β +
3 π
2
to define the branch
arg
z − e
iβ
, because β +
3 π
2
β +
π
2
Similarly, The branch arg (z − e
iα ) is defined by removing from C the cut
{
re
iα
r ≥ 1
and use the range α +
π
2
≤ arg (z − e
iα ) ≤ α +
3 π
2
. Strictly
speaking, we should also specify how to choose the value β for e
iβ and how to
choose the value α for e
iα
. It matters when we are allowed to let α and β go
through a range whose length is at least 2π. However, we are only interested
in finding the Poisson integral kernel which, as we saw earlier, involves only
the case of α = θ and β = θ + dθ. So the length of the range for the values
of α and β under consideration is limited to an extremely small number and
certainly will be less than 2π. It is immaterial how we choose the value β for
e
iβ and how we choose the value α for e
iα .
When z = e
iθ
is on the arc between e
iβ
and e
iα
in the counterclockwise
sense so that β < θ < 2 π + α, we have
arg
z − e
iβ
z − e
iα
(β − α)
for the following reason. Take e
iθ very close to e
iβ but is just after e
iβ
in the counterclockwise sense along the unit circle {|z| = 1}. The value
arg
e
iθ − e
iβ
is very close to that of arg
BT
B
. On the other hand, the
value arg
e
iθ − e
iα
is very close to that of arg
AB. Let
A be a point on the
line AB which is not on the same side of B as A so that when we go from A
to B to
A we are along the direction of the vector
AB. Then the constant
value of
arg
z − e
iβ
z − e
iα
should be equal to the angle ∠
ABT
B
which can be computed as follows. The
angle ∠AOB is equal to β − α. The triangle ∆AOB is an isosceles triangle
and the sum of its three angle ∠AOB, ∠OBA, ∠OAB is π. Since the two
angles ∠OBA and ∠OAB are equal, each one is equal to
π
2
1
2
(β − α). Let
be the intersection of the line joining B and P with the unit circle. We are
going to express ω α,β
in terms of the arc-length
_
A
′ B
′ from A
′ to B
′ around the
unit circle. Since the exterior angle of a triangle equals the sum of the two
non-adjacent interior angles, it follows that (when we consider the triangle
∆B
′
P A
′
and the exterior angle ∠B
′
P A
′
at the vertex P )
arg
z − e
iβ
z − e
iα
= ∠B
′ P A
′ = ∠P B
′ A + ∠P AB
′
_
AB +
_
A
′ B
′
Thus
ω α,β
π
arg
z − e
iβ
z − e
iα
(β − α)
π
_
AB +
_
A
′
B
′
−
(β − α)
_
A
′ B
′
2 π
because
_
AB= β − α. Dividing this by β − α and taking limit, we get
() lim
β→α
ω α,β
β − α
2 π
lim
β→α
_
A
′
B
′
β − α
We now derive the formula for the computation of
_
A
′
B
′
which would give us
the Poisson integral formula. We denote the point A on the unit circle by ζ.
Let P be a point z in the open unit disk and let w be the point A
′ which is
the intersection of the line joining A and P with the unit circle. Since w, z,
and ζ are collinear, there exists λ ∈ R such that w − z = λ (ζ − z). We have
w = λ (ζ − z) + z. Since |w|
2
= 1, it follows that
(λ (ζ − z) + z) (λ (ζ − z) + z) = 1,
which can be expanded into
λ
2
|ζ − z|
2
ζ − ¯z
2
= 1
or
(∗) λ
2 |ζ − z|
2
ζ z¯ + z
ζ − 2 r
2
2 = 1.
From
|ζ − z|
2
= (ζ − z) (ζ − z) = |ζ|
2
− ζ z¯ −
ζz + |z|
2
it follows that
ζ ¯z +
ζz = − |ζ − z|
2
2
2
= 1 + r
2
− |ζ − z|
2
,
which we put into (∗) and get
(†) λ
2
|ζ − z|
2
1 − r
2
− |ζ − z|
2
2
= 1.
Rewrite (†) as
λ
2
1 − r
2
|ζ − z|
2
1 − r
2
|ζ − z|
2
which we can factor into
(
λ +
1 − r
2
|ζ − z|
2
(λ − 1) = 0.
We have two solutions
λ = −
1 − r
2
|ζ − z|
2
and λ = 1. The solution λ = 1 just gives us back the intersection point A
between the unit circle and the straight line joining A and P. Thus
|w − z|
|ζ − z|
= |λ| =
1 − r
2
|ζ − z|
2
From the similar triangles ∆A
′
P B
′
and ∆BP A, we conclude that
A
′ B
′
BA
A
′ P
BP
and
lim
B→A
_
A
′ B
′
_
AB
= lim
B→A
A
′ B
′
BA
= lim
B→A
A
′ P
BP
which gives (due to () and
_
AB= β − α) the formula
lim
β→α
ω α,β
β − α
2 π
lim
β→α
_
A
′ B
′
β − α
2 π
lim
B→A
_
A
′ B
′
_
AB
2 π
lim
B→A
A
′ P
BP
2 π
A
′ P
AP
2 π
1 − r
2
|ζ − z|
2
This finishes the derivation of the Poisson integral formula
u(z) =
2 π
|ζ|=
1 − |z|
2
|ζ − z|
2
u(ζ) d (arg ζ).
