proof of Riemann mapping theorem

proof of Riemann mapping theorem

This proof relies on the

existence

of a solution to the Dirichlet problem, and hence is applicable to any domain for which it can be proven that the Dirichlet problem has a solution. For simplicity, I will assume that the region

U

is

bounded

;

Since this proof uses real techniques to prove a

complex

result, a few simple conventions will make it easier to read. The letter

x

and

y

will always denote real quantities and

z

will always equal

x

+

i

⁢

y

.

Functions

of a complex variable will be written as functions of two real variables without further warning; thus,

f

⁢

(

z

)

and

f

⁢

(

x

,

y

)

will denote the same entity.

Consider the following boundary value problem:

∂

2

⁡

u

∂

⁡

x

2

+

∂

2

⁡

u

∂

⁡

y

2

=

0

u

⁢

(

x

,

y

)

=

log

⁡

|

x

+

i

⁢

y

-

a

|

when

x

+

i

⁢

y

lies on the

boundary

of

U

. Note that, since

a

lies in the

interior

of

U

,

log

⁡

|

x

+

i

⁢

y

-

a

|

bounded on the boundary of

U

.
Hence, by the

existence theorem

, there exists a unique function

u

satisfying this boundary value problem. Since the boundary values of

u

are bounded,

u

will be bounded on the interior of

U

as well. By the

regularity theorem

for the

Laplace equation

,

h

will be

differentiable

(in fact,

analytic

) on the interior of

U

.

Because

u

satisfies

the two-dimensional Laplace equation, the curl of the

vector field

(

-

∂

⁡

u

∂

⁡

y

,

∂

⁡

u

∂

⁡

x

)

vanishes

. Since

U

is simply connected, Poincare’s lemma implies that there must exist a function

v

such that this vector field is the

gradient

of

v

. Since

v

is only determind up to an

additive

constant

, we may impose the condition

v

⁢

(

a

)

=

0

. Written out explicitly, the condition that the gradient of

v

equals the vector field looks like

∂

⁡

v

∂

⁡

x

=

-

∂

⁡

u

∂

⁡

y

∂

⁡

v

∂

⁡

x

=

∂

⁡

u

∂

⁡

x

Note that these equations are exactly the

Cauchy-Riemann equations

for

u

and

v

and, hence,

u

+

i

⁢

v

is an

analytic function

.

Define

p

q

, and

f

as

p

⁢

(

z

)

=

u

⁢

(

z

)

-

log

⁡

|

z

-

a

|

q

⁢

(

z

)

=

v

⁢

(

z

)

-

arg

⁡

|

z

-

a

|

f

⁢

(

z

)

=

exp

⁡

(

-

p

⁢

(

z

)

-

i

⁢

q

⁢

(

z

)

)

=

(

z

-

a

)

⁢

exp

⁡

(

-

u

⁢

(

z

)

-

i

⁢

v

⁢

(

z

)

)

Because

u

and

v

satisfies Laplace’s equation,

p

and

q

will also satisfy Laplace’s equation in

U

∖

{

a

}

. Note that, whilst

p

is

single-valued

,

q

is multiply-valued with

branch point

at

a

. Upon circling

a

once, the value of

q

increases by

2

⁢

π

⁢

i

. On the other hand,

f

is single valued since the

exponentiation

operation

cancels out the multiple-valuedness of

q

. Because

p

and

q

satisfy the Cauchy-Riemann equations and the

exponential function

is analytic,

f

is analytic.

By construction,

p

⁢

(

z

)

=

0

when

z

lies on the boundary of

U

. It will now be shown that

p

⁢

(

z

)

≥

0

whenever

z

∈

U

∖

{

a

}

. Since

log

⁡

|

z

-

a

|

→

-

∞

as

z

→

a

and

h

⁢

(

a

)

is finite, it follows that there exists

ϵ

>

0

, such that

p

⁢

(

z

)

>

0

when

0

<

|

z

-

a

|

≤

ϵ

Consider the region

V

=

{

z

∈

U

:

|

z

|

>

ϵ

}

. For a point

z

to lie on the boundary of

V

, either

|

z

-

a

|

=

ϵ

or

z

must lie on the boundary of

U

. Either way,

p

⁢

(

z

)

≥

0

, so the

maximum principle

implies that

p

⁢

(

z

)

≥

0

whenever

z

∈

V

. We already saw that

p

⁢

(

z

)

≥

0

when

0

<

|

z

|

<

ϵ

, so

p

⁢

(

z

)

>

0

whenever

z

∈

U

∖

{

a

}

.

As a

consequence

,

|

f

⁢

(

z

)

|

≤

1

when

z

∈

U

because

|

f

⁢

(

z

)

|

=

exp

⁡

(

-

p

⁢

(

z

)

)

≤

1

Also, note that

f

⁢

(

a

)

=

0

and that

f

′

⁢

(

a

)

is real and

positive

because

f

′

⁢

(

a

)

=

exp

⁡

(

-

u

⁢

(

a

)

+

i

⁢

v

⁢

(

a

)

)

=

exp

⁡

(

-

u

⁢

(

a

)

)

∈

ℝ

since

v

was chosen so that

v

⁢

(

a

)

=

0

.

To show that

f

is

bijective

, we shall study the

level sets

of

p

. To simplify this study, we shall exclude those points at which the

derivative

of

f

vanishes. For every

real number

r

>

0

, define

A

⁢

(

r

)

=

{

z

∈

U

:

|

f

⁢

(

z

)

|

≥

e

-

r

&

f

′

⁢

(

z

)

=

0

}

We will now show that

A

⁢

(

r

)

is finite. The set

{

z

∈

U

:

|

f

⁢

(

z

)

|

≥

e

-

r

}

is

compact

. Hence, were

A

⁢

(

r

)

infinite

, it would have an

accumulation point

. Since

f

⁢

(

z

)

=

0

whenever

s

∈

A

⁢

(

r

)

and

f

is analytic, this would imply that

f

⁢

(

z

)

=

0

identically, which is not the case. Hence,

A

⁢

(

r

)

must be finite.

Choose

r

such that

f

′

⁢

(

z

)

≠

0

whenever

|

z

|

=

r

. Let

C

⁢

(

r

)

denote the level set

C

⁢

(

r

)

=

{

z

:

p

⁢

(

z

)

=

r

}

We shall now show that

C

is smooth and is

homeomorphic

to a circle. As the level set of a

continuous function

on a compact set,

C

is compact. Let

w

be an point of

C

⁢

(

r

)

. By

assumption

,

f

′

⁢

(

w

)

≠

0

. Hence, by the

inverse function theorem

, there exists a

neighborhood

N

of

w

on which

f

is

invertible

. Furthermore,

f

-

1

is an anlytic function. Since

z

∈

C

⁢

(

r

)

if and only if

|

f

⁢

(

z

)

|

=

e

-

r

, it follows that

C

⁢

(

r

)

∩

N

is the image of an arc the circle

{

z

:

|

z

|

=

e

-

r

}

under

f

-

1

. Hence,

C

⁢

(

r

)

∩

N

is diffeomorphic to a

line segment

.

Since this is true of every point

w

∈

C

⁢

(

r

)

,

C

⁢

(

r

)

is a compact one-dimensional manifold. A compact, one-dimensional manifold must be either a circle or a finite union of circles. Suppose that

C

is a union of more than one circle. By the

Jordan curve theorem

, each of these circles divides the compex plane into an interior and an

exterior

reigion. Hence, given two of the circles which would comprise

C

⁢

(

r

)

, one of these circles would have to lie inside the other and there would have to be an

open set

which has these two circles as boundary. However, it is assumed that

p

is constant on both circles and assumes the same value on both. By the maximum principle, this would imply that

p

is constant in the region between the circles which would, in turn, imply that

f

is constant on

U

, which is impossible. Hence,

C

consists of a single circle.

Next, note that

a

must lie inside

C

. If

a

did not lie in

C

, it would follow that

|

f

⁢

(

z

)

|

<

r

when

z

lies outside of

C

. But this is not possible because the boundary of

u

lies on the outsde of

C

as well and

|

f

|

was chosen to satisfy the boundary value

|

f

⁢

(

z

)

|

=

1

when

z

lies on the boundary of

U

.

Since

a

lies on the interior of the circle

C

, the

winding number

of

C

about

a

is 1. Hence, upon traversing

C

once, the phase of

q

will increase by

2

⁢

π

.

Since

f

-

1

is analytic,

C

is not only homeomorphic to a circle, it is also a

smooth curve

. Hence it makes sense to speak of the

tangent

to

C

and the normal to

C

. In terms of the normal and tangential derivatives, the Cauchy-Riemann equations my be written as

∂

⁡

p

∂

⁡

t

=

∂

⁡

q

∂

⁡

n

,

∂

⁡

p

∂

⁡

n

=

-

∂

⁡

q

∂

⁡

t

Since

p

⁢

(

x

)

=

r

when

x

lies on

C

⁢

(

r

)

and

p

⁢

(

x

)

≥

r

when

x

lies in the interior of

C

⁢

(

r

)

, it follows that

∂

⁡

p

/

∂

⁡

t

=

0

and

∂

⁡

p

/

∂

⁡

n

>

0

. By the Cauchy-Riemann eqations, this means that

∂

⁡

q

/

∂

⁡

t

≤

0

. It is not possible that

∂

⁡

q

/

∂

⁡

t

≤

0

because, if that were so, then all the derivatives of

p

and

q

would vanish, which would imply that

f

′

=

0

, contrary to

hypothesis

. Hence

q

is a monotonically decreasing function on

C

⁢

(

r

)

. We already saw that

q

decreases by

2

⁢

π

upon traversing

C

⁢

(

r

)

. These two facts together imply that

exp

⁡

(

-

i

⁢

q

)

is a bijection from

C

⁢

(

r

)

to the

unit circle

. Hence

f

is a bijection from

C

⁢

(

r

)

to the circle of radius

e

-

r

.

To finish the proof, we need to deal with the points

z

such that

f

′

⁢

(

z

)

=

0

. We shall use the

argument principle

to show that these points do not exist! As before, choose

r

such that

f

′

⁢

(

z

)

≠

0

whenever

|

z

|

=

r

. Then

C

⁢

(

r

)

is a smooth close curve and we may apply the argument principle to

f

′

along

C

⁢

(

r

)

. Differentiating the definition,

f

′

⁢

(

z

)

=

-

f

⁢

(

z

)

⁢

(

u

′

⁢

(

z

)

+

i

⁢

v

′

⁢

(

z

)

)

Since the

argument

of a

product

is the sum of the arguments of the factors, we may consider

f

and

u

′

+

i

⁢

v

′

separately. The argument principle states that, because

f

has only one simple zero (located at

a

) inside

C

⁢

(

r

)

, the argument of

f

will increase by

2

⁢

π

upon traversing

C

⁢

(

r

)

. To compute the argument of

u

′

+

i

⁢

v

′

, we may make use of the fact that the derivative of an analytic function is the same no matter what direction one chooses to compute the derivative. To compute the derivative of

u

+

i

⁢

v

at a point on

C

⁢

(

r

)

, choose the normal direction. From what we had seen earlier, it follows that

u

′

+

i

⁢

v

′

will point along the inward normal to

C

⁢

(

r

)

. Since the inward normal rotates by by

-

2

⁢

π

upon traversing the curve, the argument of

u

′

+

i

⁢

v

′

changes by

-

2

⁢

π

upon traversing

C

⁢

(

r

)

. Hence, the argument of

f

′

stays the same after traversing

C

⁢

(

r

)

. Since

f

′

is analytic inside

C

⁢

(

r

)

, the argument principle states that

f

′

can have no zeros inside of

C

⁢

(

r

)

.

Title

proof of Riemann mapping theorem

Canonical name

ProofOfRiemannMappingTheorem

Date of creation

2013-03-22 14:35:17

Last modified on

2013-03-22 14:35:17

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

22

Author

rspuzio (6075)

Entry type

Proof

Classification

msc 30A99

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