Complex analysis review 5

Maximum modulus principle and Schwarz lemma

Average Vaule Properties

f(z0)=12πi∫∂D(z0,r)f(ξ)ξ−z0dξ=12πi∫2π0f(z0+reit)ireitreitdt=12π∫2π0f(z0+reit)dt.

Shows that f(z0) is equal to the integration average on the circle ∂D(z0,r).

Maximum modulus principle

Suppose that f(z) is analytic on U⊂C, and there is a z0∈U such that |f(z0)|≥|f(z)|,∀z∈U, then f(z) must be a constant function on U.

Multiple by a constant with modulus 1, such that M=f(z0)≥0, let

S={z∈U|f(z)=f(z0}.

Then S≠∅. Since f is a continuous function on U, so S is closed set. Now we want to prove that S is also an open set, then since U is a single connected domain, we concluded that S=U.

If w∈S, choose r, such that D(w,r)⊂U, and let 0<r′<r, then

M=f(w)=|12π∫2π0f(w+r′eit)dt|≤M.

So

f(w+r′eit)=|f(w+r′eit)|=M

for any t and 0<r′<r, which means

D(w,r′)⊂S.

From the maximum modulus principle, we have that if f is analytic on U⊂C, U is a bounded domain, and f is continuous on U¯, then if f is not identical to a constant function, |f(z)| can only attains its maximum on the boundary ∂U.

Schwarz Lemma

Suppose that f is an analytic function which maps D=D(0,1) into D, and f(0)=0, then

|f(z)|≤|z|,|f′(0)|≤1.

Moreover |f(z)|=|z|,z≠0 or |f′(0)|=1 holds if and only if f(z)=eiτz.τ∈R.

Let

G(z)=⎧⎩⎨f(z)zf′(0)ifz≠0ifz=0

Then G(z) is analytic on D. Consider {z||z|≤1−ϵ}, by maximum modulus principle, we have

|G(z)|≤max|z|=1−ϵ|f(z)|1−ϵ<11−ϵ.

Let ϵ→0, we have |G(z)|≤1 on D. Therefore, when z≠0, |f(z)|≤z and when z=0, |G(0)|=|f′(0)|≤1. The case when equality holds are easy.

Aut(D)

Let a∈D,

ϕa(z)=−z+a1−a¯z∈Aut(D)

And ϕ−1a=ϕa. Then mapping above is called the Mobius mapping.

Let τ∈R, and define the rotation mapping

ξ=ρτ(z)=eiτz

Theorem

If f∈Aut(D), then there is a∈D,τ∈R, such that

f(z)=ϕa∘ρτ(z).

Which means that the element of Aut(D) is the component of Mobius tranforamtion and rotation transformation.

Let b=f(0) then let

G=ϕb∘f

G is also in Aut(D), and G(0)=ϕb∘f(0)=ϕb(b)=0, by Schwarz lemma, we have |G′(0)|≤1. And G is invertible with G−1∈Aut(D),G−1(0)=0. By Schwarz lemma again,

|1G′(0)|=|(G−1)′(0)|≤1.

Then |G′(0)|=1. Then

G(z)=eiτz=ρτ(z).

Which shows that

f=ϕ−b∘ρτ.

Schwarz-Pick lemma

Suppose that f is an analytic function which maps D=D(0,1) into D, and z1,z2∈D, w1=f(z1),w2=f(z2), then

|w1−w21−w1w¯2|≤|z1−z21−z1z¯2|,

and

|dw|1−|w|2≤|dz|1−|z|2.

Construct

ϕ(z)=z+z11+z¯1z,ψ(z)=z−w11−w¯1z

Then ϕ,ψ∈Aut(D).

And consider ψ∘f∘ϕ, use Schwarz lemma, then the remaining are easy (let z=ϕ−1(z2)).

In the above theorem, we actually define a measure called Poincare measure. Then if that f is an analytic function which maps D=D(0,1) into D, the Poincare is nonincreasing.