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.