FirstWorkshopinComplexAnalysisandOperatorTheoryM´alaga2016 ManuelD.Contreras Boundarybehavioroftheiteratesofaself-mapoftheunitdisc

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Introduction Automorphisms LFM Denjoy-Wolff Theorem Boundary behavior

Boundary behavior of the iterates of a self-map of the unit disc

Manuel D. Contreras

Departamento de Matem ´atica Aplicada II and IMUS Universidad de Sevilla

First Workshop in Complex Analysis and Operator Theory M ´alaga 2016

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Introduction.

Let M be a complex manifold (the unit disc D, the complex plane C, the Riemann sphere C∞, a half plane, an annulus, ...) and φ : M → M a holomorphic function.

Since φ(M) ⊆ M, we can define the iterates φn= φ^n-times◦...◦ φ.

One of the goals ofComplex Dynamicsis to analyze the behavior of the sequence of iterates {φn}.

Many classical authors worked in Complex Dynamics: Poincar é, Julia, Fatou, Denjoy, Wolff, Carath éodory, ... In the unit disc, we could add to the above list the names of Valir ón, Baker, Pommerenke, Cowen, ...

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Introduction.

Let M be a complex manifold (the unit disc D, the complex plane C, the Riemann sphere C∞, a half plane, an annulus, ...) and φ : M → M a holomorphic function.

Since φ(M) ⊆ M, we can define the iterates φn= φ^n-times◦...◦ φ.

One of the goals ofComplex Dynamicsis to analyze the behavior of the sequence of iterates {φn}.

Many classical authors worked in Complex Dynamics:

Poincar ´e, Julia, Fatou, Denjoy, Wolff, Carath ´eodory, ...

In the unit disc, we could add to the above list the names of Valir ´on, Baker, Pommerenke, Cowen, ...

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Introduction.

φ :M → M holomorphic.

Remark

Let g : M → N be a biholomorphism. Then the function ψ :=g ◦ φ ◦ g⁻¹:N −→ N

is holomorphic and

ψ_n =

n-times

(g ◦ φ ◦ g⁻¹) ◦ (g ◦ φ ◦ g⁻¹) ◦ ... ◦ (g ◦ φ ◦ g⁻¹)

=

n-times

g ◦ φ ◦ φ ◦ ... ◦ φ ◦ g⁻¹=g ◦ φn◦ g⁻¹.

Thus, the behavior of {φn} on M is similar to the behavior of {ψ_n} on N.

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Introduction.

Theorem (Uniformization Theorem of Poincar ´e and Koebe) Everysimply connectedRiemann surface is biholomorphic either to the unit disc D, or to the complex plane C, or to the Riemann sphere C∞.

So, in the setting ofsimply connectedRiemann surfaces we can reduce to:

The Riemann sphere C∞: iteration of rational functions.

The complex plane C: iteration of entire functions.

The unit disc D.

Theorem (Montel)

Let F be a family of homomorphic functions on a domain Ω. If there are two points that are omitted by every f ∈ F , then F is a normal family.

If φ : D −→ D, then the family {φn:n ∈ N} is normal.

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Introduction.

The unit disc D.

Theorem (Montel)

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Introduction.

The unit disc D.

Theorem (Montel)

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Introduction.

Iteration in the Riemann sphere and the in complex plane:

J. Milnor,Dynamics in One Complex Variable,2000.

L. Carleson and T.W. Gamelin,Complex Dynamics, 1993.

N. Fagella and X. Jarque,Iteraci ´on Compleja y Fractales, 2007.

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The easiest examples: automorphisms of the unit disc.

Examine theautomorphismsof the unit disc Aut(D), that is, Aut(D) := {T : D → D such that T is a biholomorphism}.

For a ∈ D, let Ta: D → C be the holomorphic map defined by Ta(z) := a − z

1 − az.

It is easy to see that Ta(D) = D and Ta⁻¹(z) = Ta(z), that is, Ta

is an automorphism of D. Note also that Ta(a) = 0, T_a(0) = a.

Proposition

Let T ∈ Aut(D). Then there exists θ ∈ R and a ∈ D such that T (z) = e^iθT_a(z).

In particular, every automorphism of D extends as a homeomorphism from D to D.

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The easiest examples: automorphisms of the unit disc.

Examine theautomorphismsof the unit disc Aut(D), that is, Aut(D) := {T : D → D such that T is a biholomorphism}.

For a ∈ D, let Ta: D → C be the holomorphic map defined by Ta(z) := a − z

1 − az.

It is easy to see that Ta(D) = D and Ta⁻¹(z) = Ta(z), that is, Ta

is an automorphism of D. Note also that Ta(a) = 0, T_a(0) = a.

Proposition

Let T ∈ Aut(D). Then there exists θ ∈ R and a ∈ D such that T (z) = e^iθT_a(z).

In particular, every automorphism of D extends as a homeomorphism from D to D.

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The easiest examples: automorphisms of the unit disc.

Lemma

Let T ∈ Aut(D) \ {idD}. Then T has at least one fixed point in D. Moreover, ifT has no fixed points in D, then it has two fixed points τ, σ ∈ ∂D, possibly τ = σ, such that T⁰(σ) ·T⁰(τ ) =1.

Proof.

T ∈ Aut(D) ⇒ T (z) = λ_1−az^a−z for a a ∈ D and λ ∈ C, |λ| = 1. Then

T (z) = z if and only if az²− (1 + λ)z + λa = 0.

If a = 0, since T 6= id_D, then the unique fixed point of T in C is z = 0. If a 6= 0, the above equation has two solutions z₁,z₂∈ C which satisfy z₁z₂= λ^a_a ∈ ∂D (in particular, |z1||z₂| = 1) and az₁+az₂=1 + λ. Moreover,

T⁰(z₁)T⁰(z₂) = λ² (1 − |a|²)²

((1 − az₁)(1 − az₂))² =1.

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The easiest examples: automorphisms of the unit disc.

Lemma

Proof.

T ∈ Aut(D) ⇒ T (z) = λ_1−az^a−z for a a ∈ D and λ ∈ C, |λ| = 1.

Then

If a = 0, since T 6= id_D, then the unique fixed point of T in C is z = 0.

If a 6= 0, the above equation has two solutions z1,z2∈ C which satisfy z₁z₂= λ^a_a ∈ ∂D (in particular, |z1||z₂| = 1) and az₁+az₂=1 + λ.

Moreover,

T⁰(z₁)T⁰(z₂) = λ² (1 − |a|²)²

((1 − az₁)(1 − az₂))² =1.

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The easiest examples: automorphisms of the unit disc.

Lemma

Proof.

T ∈ Aut(D) ⇒ T (z) = λ_1−az^a−z for a a ∈ D and λ ∈ C, |λ| = 1.

Then

If a = 0, since T 6= id_D, then the unique fixed point of T in C is z = 0.

If a 6= 0, the above equation has two solutions z1,z2∈ C which satisfy z₁z₂= λ^a_a ∈ ∂D (in particular, |z1||z₂| = 1) and az₁+az₂=1 + λ.

Moreover,

T⁰(z₁)T⁰(z₂) = λ² (1 − |a|²)²

((1 − az₁)(1 − az₂))² =1.

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The easiest examples: automorphisms of the unit disc.

Definition

Let T ∈ Aut(D) \ {idD}. Then we say that

1 T isellipticif it has a fixed point in D,

2 T isparabolicif it has a unique fixed point in ∂D,

3 T ishyperbolicif it has two different fixed points in ∂D.

We investigate now the dynamics of an automorphism of D according to the previous classification.

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The easiest examples: automorphisms of the unit disc.

The elliptic case:

Assume T is an elliptic automorphism with a fixed point τ ∈ D.

The map T_τ◦ T ◦ T_τ is an automorphism that fixes the origin.

So, there exists λ ∈ ∂D s. t. (Tτ ◦ T ◦ Tτ)(z) = λz, for all z ∈ D. The automorphism T is then holomorphically conjugated to a rotation.

Thus, given z ∈ D,

T₂(z) = T_τ(λT_τ(T_τ(λT_τ(z)))) = T_τ(λ²T_τ(z)), ... T_n(z) = T_τ(λT_τ(T_τ(λⁿ⁻¹T_τ(z)))) = T_τ(λⁿT_τ(z)). Remark: T⁰(τ ) = λ.

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The easiest examples: automorphisms of the unit disc.

The elliptic case:

So, there exists λ ∈ ∂D s. t. (Tτ ◦ T ◦ Tτ)(z) = λz, for all z ∈ D.

The automorphism T is then holomorphically conjugated to a rotation.

T₂(z) = T_τ(λT_τ(T_τ(λT_τ(z)))) = T_τ(λ²T_τ(z)), ... T_n(z) = T_τ(λT_τ(T_τ(λⁿ⁻¹T_τ(z)))) = T_τ(λⁿT_τ(z)). Remark: T⁰(τ ) = λ.

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The easiest examples: automorphisms of the unit disc.

The elliptic case:

So, there exists λ ∈ ∂D s. t. (Tτ ◦ T ◦ Tτ)(z) = λz, for all z ∈ D.

The automorphism T is then holomorphically conjugated to a rotation.

T₂(z) = T_τ(λT_τ(T_τ(λT_τ(z)))) = T_τ(λ²T_τ(z)), ...

T_n(z) = T_τ(λT_τ(T_τ(λⁿ⁻¹T_τ(z)))) = T_τ(λⁿT_τ(z)).

Remark: T⁰(τ ) = λ.

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The easiest examples: automorphisms of the unit disc.

The hyperbolic case:

Let τ, σ ∈ ∂D be its fixed points, τ 6= σ.

Since T⁰(τ )T⁰(σ) =1, we may assume that |T⁰(τ )| ≤1.

Consider C(z) = ^{τ +z}_{τ −z}, C : D → H, we can conjugate T to an automorphism of the right half plane Φ = C ◦ T ◦ C⁻¹that fixes the point ∞.

Thus Φ(w ) = aw + b, for all w ∈ C.

Since Φ(H) = H we conclude that: Re b = 0, Im a = 0, a > 0. Also, a 6= 1, for otherwise Φ would not have fixed points in C. Now, a direct computation shows that a = _T0¹(τ ). So

T⁰(τ ) ∈ (0, 1) and T⁰(σ) = _T0¹(τ ) ∈ (1, +∞).

Moreover, since Φn(w ) = aⁿw +^1−a_1−aⁿb for all w ∈ H, it follows that {Φ_n} converges to ∞ uniform on compacta of H. Hence, {Tn} = {C⁻¹◦ Φ_n◦ C} converges uniformly on compacta to τ .

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The easiest examples: automorphisms of the unit disc.

Since Φ(H) = H we conclude that: Re b = 0, Im a = 0, a > 0.

Also, a 6= 1, for otherwise Φ would not have fixed points in C. Now, a direct computation shows that a = _T0¹(τ ). So

T⁰(τ ) ∈ (0, 1) and T⁰(σ) = _T0¹(τ ) ∈ (1, +∞).

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The easiest examples: automorphisms of the unit disc.

Also, a 6= 1, for otherwise Φ would not have fixed points in C.

Now, a direct computation shows that a = _T0¹(τ ). So T⁰(τ ) ∈ (0, 1) and T⁰(σ) = _T0¹(τ ) ∈ (1, +∞).

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The easiest examples: automorphisms of the unit disc.

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The easiest examples: automorphisms of the unit disc.

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The easiest examples: automorphisms of the unit disc.

The parabolic case:

Assume that T is parabolic, with a unique fixed point τ ∈ ∂D.

Notice that T⁰(τ )²=1.

Arguing as in the hyperbolic case, the function

Φ(w ) = C ◦ T ◦ C⁻¹(w ) = aw + b is a M ¨obius transformation that has only one fixed point in C∞and Φ(H) = H.

Hence, Φ(w ) = w + b, for all w ∈ H, with a = _T⁰¹_{(τ )} >0 and Re b = 0. In particular, T⁰(τ ) =1.

Thus Φn(w ) = w + nb, which implies Φn(w ) → ∞ for all w ∈ H. Therefore, {Tn} converges to τ for all z ∈ D, and, even in this case, {T_n} converges uniformly on compacta to τ .

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The easiest examples: automorphisms of the unit disc.

The parabolic case:

Hence, Φ(w ) = w + b, for all w ∈ H, with a = _T0¹(τ ) >0 and Re b = 0. In particular, T⁰(τ ) =1.

Thus Φn(w ) = w + nb, which implies Φn(w ) → ∞ for all w ∈ H. Therefore, {Tn} converges to τ for all z ∈ D, and, even in this case, {T_n} converges uniformly on compacta to τ .

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The easiest examples: automorphisms of the unit disc.

The parabolic case:

Hence, Φ(w ) = w + b, for all w ∈ H, with a = _T0¹(τ ) >0 and Re b = 0. In particular, T⁰(τ ) =1.

Thus Φn(w ) = w + nb, which implies Φn(w ) → ∞ for all w ∈ H.

Therefore, {Tn} converges to τ for all z ∈ D, and, even in this case, {T_n} converges uniformly on compacta to τ .

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The easiest examples: automorphisms of the unit disc.

We have proved:

Conclusion I

Let T ∈ Aut(D) \ {idD}. If T is not elliptic, then there is τ ∈ ∂D such that {Tn} converges uniformly on compacta to the constant map D 3 z 7→ τ .

τ is the fixed point such that T⁰(τ ) ∈ (0, 1].

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Linear fractional maps.

Let φ : D → D be a linear fractional maps (LFM) given by φ(z) = az + b

cz + d with ad − bc 6= 0.

Then |c| < |d | and φ : D → D.

Theorem (Brouwer’s fixed-point theorem)

Every continuous function from a closed disk to itself has at least one fixed point.

Thus there is τ ∈ D such that φ(τ ) = τ .

We follow a similar argument to study the behavior of {φ_n}.

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Linear fractional maps.

Let φ : D → D be a linear fractional maps (LFM) given by φ(z) = ^az+b_cz+d with ad − bc 6= 0.

Assume that there isτ ∈ Dsuch thatφ(τ ) = τ andφ /∈ Aut(D).

Write again Tτ(z) = _{1−τ z}^{τ −z} ∈ Aut(D).

The map ψ = T_τ◦ φ ◦ T_τ : D → D is a LFM that fixes the origin, let us say, ψ(z) = _βz+γ^αz . We may assume that γ = 1.

Then

|ψ⁰(0)| = |α| < 1, ψ_n(z) = αⁿz

βw_nz + 1, where wn= 1 − αⁿ 1 − α. So {ψn} converges to 0 uniform on compacta of D. Hence, {φ_n} = {T_τ ◦ ψ_n◦ T_τ} converges uniformly on compacta to τ . Important: This is not satisfied if φ ∈ Aut(D).

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Linear fractional maps.

Then

|ψ⁰(0)| = |α| < 1, ψ_n(z) = αⁿz

βw_nz + 1, where wn= 1 − αⁿ 1 − α.

So {ψn} converges to 0 uniform on compacta of D. Hence, {φ_n} = {T_τ ◦ ψ_n◦ T_τ} converges uniformly on compacta to τ . Important: This is not satisfied if φ ∈ Aut(D).

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Linear fractional maps.

Then

|ψ⁰(0)| = |α| < 1, ψ_n(z) = αⁿz

βw_nz + 1, where wn= 1 − αⁿ 1 − α. So {ψn} converges to 0 uniform on compacta of D. Hence, {φ_n} = {T_τ ◦ ψ_n◦ T_τ} converges uniformly on compacta to τ . Important: This is not satisfied if φ ∈ Aut(D).

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Linear fractional maps.

Assume that φ hasno fixed point in D. Then τ ∈ ∂D.

Consider C(z) = ^{τ +z}_{τ −z}, C : D → H, we can conjugate φ to an automorphism of the right half plane Φ = C ◦ φ ◦ C⁻¹that fixes the point ∞.

Thus Φ(w ) = αw + β, for all w ∈ C.

, Re β ≥ 0, α > 0. Since Φ(H) ⊆ H we conclude that Re β = Re Φ(0) ≥ 0. Moreover

Re Φ(ic) = c Re (iα)+Re β = −c Im α+Re β ≥ 0, for all c ∈ R. So Im α = 0 and α ∈ R. On the other hand,

Re Φ(r ) = r α + Re β ≥ 0, for all r > 0. Thus α > 0.

If α = 1, then Φ_n(w ) = w + nβ, for all w ∈ H. If α 6= 1, then Φn(w ) = αⁿw +^1−α_1−αⁿβ for all w ∈ H.

In any case, it follows that {Φ_n} converges either to ∞ or to 0 uniformly on compacta of H.

Hence, {φn} converges uniformly on compacta of D to the constant map z 7→ σ where σ = τ or σ = C⁻¹(β/(1 − α)). Remark: φ⁰(τ ) ≤1.

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Linear fractional maps.

Thus Φ(w ) = αw + β, for all w ∈ C

.

, Re β ≥ 0, α > 0.

Since Φ(H) ⊆ H we conclude that Re β = Re Φ(0) ≥ 0.

Moreover

Re Φ(ic) = c Re (iα)+Re β = −c Im α+Re β ≥ 0, for all c ∈ R.

So Im α = 0 and α ∈ R. On the other hand,

Re Φ(r ) = r α + Re β ≥ 0, for all r > 0.

Thus α > 0.

If α = 1, then Φ_n(w ) = w + nβ, for all w ∈ H. If α 6= 1, then Φn(w ) = αⁿw +^1−α_1−αⁿβ for all w ∈ H.

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Linear fractional maps.

.

, Re β ≥ 0, α > 0.

Since Φ(H) ⊆ H we conclude that Re β = Re Φ(0) ≥ 0. Moreover

If α = 1, then Φ_n(w ) = w + nβ, for all w ∈ H.

If α 6= 1, then Φn(w ) = αⁿw +^1−α_1−αⁿβ for all w ∈ H.

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Linear fractional maps.

.

, Re β ≥ 0, α > 0.

Since Φ(H) ⊆ H we conclude that Re β = Re Φ(0) ≥ 0. Moreover

If α = 1, then Φ_n(w ) = w + nβ, for all w ∈ H.

If α 6= 1, then Φn(w ) = αⁿw +^1−α_1−αⁿβ for all w ∈ H.

Hence, {φn} converges uniformly on compacta of D to the constant map z 7→ σ where σ = τ or σ = C⁻¹(β/(1 − α)).

Remark: φ⁰(τ ) ≤1.

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Linear fractional maps.

We have proved:

Conclusion I

Let T ∈ Aut(D) \ {id_D}. If T is not elliptic, then there is τ ∈ ∂D such that {Tn} converges uniformly on compacta to the constant map D 3 z 7→ τ .

Conclusion II

Let φ : D → D be a LFM different from an elliptic automorphism.

Then there is τ ∈ D such that {φn} converges uniformly on compacta to the constant map D 3 z 7→ τ .

If τ ∈ ∂D, then φ⁰(τ ) ∈ (0, 1].

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Other examples

Example

(i) φ : D → D given by φ(z) = z².Then, φ_n(z) = z²ⁿ y φn(z) → 0 for all z ∈ D.

(ii) φ : D → D given by φ(z) = 1 − (1 − z)^α,where 0 < α < 1. Then, φ_n(z) = 1 − (1 − z)^αⁿ and φ_n(z) → 0 for all z ∈ D. In both cases, 0 is a fixed point of φ.

Example

(iii) φ : D → D given by φ(z) = 1 −_(1+(1−z)^1−z2)^1/2.Then φ_n(z) = 1 − _(1+n(1−z)^1−z ₂

)^1/2 and φ_n(z) → 1 for all z ∈ D. Notice that 1 is a “fixed point” of φ.

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Other examples

Example

(ii) φ : D → D given by φ(z) = 1 − (1 − z)^α,where 0 < α < 1.

Then, φ_n(z) = 1 − (1 − z)^αⁿ and φ_n(z) → 0 for all z ∈ D.

In both cases, 0 is a fixed point of φ. Example

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Other examples

Example

In both cases, 0 is a fixed point of φ.

Example

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Other examples

Example

(iii) φ : D → D given by φ(z) = 1 −_(1+(1−z)^1−z2)^1/2.Then φn(z) = 1 − _(1+n(1−z)^1−z ₂

)^1/2 and φ_n(z) → 1 for all z ∈ D.

Notice that 1 is a “fixed point” of φ.

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Other examples

Example

(iii) φ : D → D given by φ(z) = 1 −_(1+(1−z)^1−z2)^1/2.Then φn(z) = 1 − _(1+n(1−z)^1−z ₂

)^1/2 and φ_n(z) → 1 for all z ∈ D.

Notice that 1 is a “fixed point” of φ.

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Denjoy-Wolff Theorem

Theorem (Denjoy, Wolff, 1926)

Let φ : D → D be holomorphic, not an elliptic automorphism.

Then there exists a unique point τ ∈ D such that the sequence {φ_n} converges uniformly on compacta of the unit disc to the constant map D 3 z → τ .

Remark

τ ∈ D if and only if φ has a fixed point in D. In such case, φ(τ ) = τ .

Definition

The point τ is called theDenjoy-Wolff pointof φ.

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Denjoy-Wolff Theorem

Theorem (Denjoy, Wolff, 1926)

Let φ : D → D be holomorphic, not an elliptic automorphism.

Then there exists a unique point τ ∈ D such that the sequence {φ_n} converges uniformly on compacta of the unit disc to the constant map D 3 z → τ .

Remark

τ ∈ D if and only if φ has a fixed point in D.

In such case, φ(τ ) = τ . Definition

The point τ is called theDenjoy-Wolff pointof φ.

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Example in another domain: Newton’s method

In numerical analysis, Newton’s method is a method for finding successively better approximations to the roots (or zeroes) of a real-valued function.

The method starts with a function f defined over R and an initial guess x₀for a root of the equation f (x ) = 0. If the function satisfies a certain assumptions about f⁰ and the initial guess is close enough, then a better approximation x₁is given by

x₁=x₀− f (x₀) f⁰(x₀). The process is repeated as

x_n+1=x_n− f (x_n) f⁰(xn). until a sufficiently accurate value is reached.

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Example in another domain: Newton’s method

Let us see what happens with the roots of z²− 1 = 0. Applying Newton’s method we generate the sequence

z_n+1 =zn− f (zn) f⁰(zn) = 1

2

zn+ 1

zn

, where f (z) = z²− 1.

Define φ(z) = ¹₂ z + ¹_z .

To analyze the convergence of Newton’s method is equivalent to characterize the values z such that {φn(z)} converges. Problem:We have to find a domain Ω, Ω 6= C, s. t. φ : Ω → Ω. Since φ : H −→ H and φ(1) = 1, applying Denjoy-Wolff

Theorem to the simply connected domain H (biholomorphic to D) and the function φ that fixes a point in H, we obtain

if z ∈ H, then {φn(z)} converges to 1.

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Example in another domain: Newton’s method

z_n+1 =zn− f (zn) f⁰(zn) = 1

2

zn+ 1

zn

, where f (z) = z²− 1.

Define φ(z) = ¹₂ z + ¹_z .

To analyze the convergence of Newton’s method is equivalent to characterize the values z such that {φn(z)} converges.

Problem:We have to find a domain Ω, Ω 6= C, s. t. φ : Ω → Ω. Since φ : H −→ H and φ(1) = 1, applying Denjoy-Wolff

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Example in another domain: Newton’s method

z_n+1 =zn− f (zn) f⁰(zn) = 1

2

zn+ 1

zn

, where f (z) = z²− 1.

Define φ(z) = ¹₂ z + ¹_z .

Problem:We have to find a domain Ω, Ω 6= C, s. t. φ : Ω → Ω.

Since φ : H −→ H and φ(1) = 1, applying Denjoy-Wolff

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Example in another domain: Newton’s method

z_n+1 =zn− f (zn) f⁰(zn) = 1

2

zn+ 1

zn

, where f (z) = z²− 1.

Define φ(z) = ¹₂ z + ¹_z .

Problem:We have to find a domain Ω, Ω 6= C, s. t. φ : Ω → Ω.

Since φ : H −→ H and φ(1) = 1, applying Denjoy-Wolff

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Example in another domain: Newton’s method

z_n+1 =z_n− f (z_n) f⁰(zn) = 1

2

z_n+ 1

zn

, where f (z) = z²− 1.

Define φ(z) = ¹₂ z + ¹_z .

If Re z > 0, then {φn(z)} converges to 1.

If Re z < 0, then {φ_n(z)} converges to −1.

What happens with the sequence {φ_n(z)} if Re z = 0?

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Example in another domain: Newton’s method

z_n+1 =z_n− f (z_n) f⁰(zn) = 1

2

z_n+ 1

zn

, where f (z) = z²− 1.

Define φ(z) = ¹₂ z + ¹_z .

If Re z > 0, then {φn(z)} converges to 1.

If Re z < 0, then {φ_n(z)} converges to −1.

What happens with the sequence {φ_n(z)} if Re z = 0?

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Boundary behavior of the iterates.

What happens with the sequence {φn(ξ)}when ξ ∈ ∂D?

Does the sequence {φ_n(ξ)}make sense when ξ ∈ ∂D?

Theorem (Fatou)

Let φ : D → D be a holomorphic function. Then the redial limit φ(ξ) := lim

r →1φ(r ξ) ∈ D exists a.e. ξ ∈ ∂D. Moreover, the function

φ : ∂D −→ D

ξ 7−→ φ(ξ) is measurable.

φis calledinnerif φ(ξ) ∈ ∂D a.e. ξ ∈ ∂D.

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Boundary behavior of the iterates.

What happens with the sequence {φn(ξ)}when ξ ∈ ∂D?

Does the sequence {φ_n(ξ)}make sense when ξ ∈ ∂D?

Theorem (Fatou)

Let φ : D → D be a holomorphic function. Then the redial limit φ(ξ) := lim

r →1φ(r ξ) ∈ D exists a.e. ξ ∈ ∂D. Moreover, the function

φ : ∂D −→ D

ξ 7−→ φ(ξ) is measurable.

φis calledinnerif φ(ξ) ∈ ∂D a.e. ξ ∈ ∂D.

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Boundary behavior of the iterates.

So, there exists a set A of measure 0 such that φn(ξ) := lim

r →1φn(r ξ) exists for all ξ ∈ ∂D \ A and for all n ∈ N.

Therefore, we can reformulate our question:

Problem

Let φ : D → D be a holomorphic function with Denjoy-Wolff point τ ∈ D.

Is it satisfied that {φ_n(ξ)}converges to τ a.e. ξ ∈ ∂D?

NO.

Example: φ(z) = z². Its Denjoy-Wolff point is 0. If ξ ∈ ∂D, then |φn(ξ)| =1 (φis inner).

So if ξ ∈ ∂D, then {φn(ξ)}does not converge to 0.

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Boundary behavior of the iterates.

So, there exists a set A of measure 0 such that φn(ξ) := lim

r →1φn(r ξ) exists for all ξ ∈ ∂D \ A and for all n ∈ N.

Therefore, we can reformulate our question:

Problem

Let φ : D → D be a holomorphic function with Denjoy-Wolff point τ ∈ D.

Is it satisfied that {φ_n(ξ)}converges to τ a.e. ξ ∈ ∂D?

NO.

Example: φ(z) = z². Its Denjoy-Wolff point is 0.

If ξ ∈ ∂D, then |φn(ξ)| =1 (φis inner).

So if ξ ∈ ∂D, then {φn(ξ)}does not converge to 0.

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Boundary behavior of the iterates.

Theorem (Bourdon-Matache-Shapiro, Poggi-Corradini (2005)) Let φ : D → D be holomorphic with Denjoy-Wolff point τ ∈ D.

Then (φ_n(ξ))converges to τ a.e. ξ ∈ ∂D if and only if φ is not an inner function.

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Hyperbolic distance.

Thehyperbolic distanceis defined by ρ(z, w ) := 1

2log 1 +

z−w 1−zw

1 −

z−w 1−zw

for all z, w ∈ D.

Theorem (Schwarz’s Lemma)

Let φ : D → D be holomorphic. Assume that φ(0) = 0. Then

|φ(z)| ≤ |z| for all z ∈ D and |φ⁰(0)| ≤ 1.

Theorem (Schwarz-Pick’s Lemma)

Let φ : D → D be holomorphic. Then, for all z, w ∈ D,

φ(z) − φ(w ) 1 − φ(z)φ(w )

≤

z − w 1 − zw

.

(56)

Hyperbolic distance.

Thehyperbolic distanceis defined by

ρ(z, w ) := 1 2log

1 +

z−w 1−zw

1 −

z−w 1−zw

for all z, w ∈ D.

Theorem (Schwarz-Pick’s Lemma)

Let φ : D → D be holomorphic. Then, for all z, w ∈ D,

φ(z) − φ(w ) 1 − φ(z)φ(w )

≤

z − w 1 − zw

.

Since the map x 7→ ¹₂log^1+x_1−x is non-decreasing we obtain ρ(φ(z), φ(w )) ≤ ρ(z, w ) for all z, w ∈ D.

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Hyperbolic distance.

Therefore Proposition

Let φ ∈ Hol(D, D) be holomorphic

ρ(φ(z), φ(w )) ≤ ρ(z, w ) for all z, w ∈ D.

Equality holds for some z 6= w if and only if φ ∈ Aut(D).

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The hyperbolic step.

Let φ : D → D be holomorphic, z0∈ D y zn= φn(z₀):

ρ(zn,z_n+1) = ρ(φ(z_n−1), φ(zn)) ≤ ρ(z_n−1,zn).

Thus, the sequence {ρ(z_n,z_n+1)}is decreasing. So, there exists limnρ(zn,z_n+1) ∈ [0, +∞). Some examples: τ =Denjoy-Wolff point.

- If τ ∈ D and φ is not an elliptic automorphism, then lim_nρ(z_n,z_n+1) = ρ(τ, τ ) =0. - If φ ∈ Aut(D), then ρ(zn,z_n+1) = ρ(z_n−1,zn)for all n. So, if z₀6= τ then limnρ(z_n,z_n+1) >0.

- If φ is a LFM with τ ∈ ∂D and φ⁰(τ ) <1, then limnρ(zn,z_n+1) >0. - If φ is a LFM with τ ∈ ∂D and φ⁰(τ ) =1, then

lim_nρ(z_n,z_n+1) >0 if and only if φ ∈ Aut(D).

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The hyperbolic step.

Thus, the sequence {ρ(z_n,z_n+1)}is decreasing. So, there exists limnρ(zn,z_n+1) ∈ [0, +∞).

Some examples: τ =Denjoy-Wolff point.

- If τ ∈ D and φ is not an elliptic automorphism, then lim_nρ(z_n,z_n+1) = ρ(τ, τ ) =0. - If φ ∈ Aut(D), then ρ(zn,z_n+1) = ρ(z_n−1,zn)for all n. So, if z₀6= τ then limnρ(z_n,z_n+1) >0.

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The hyperbolic step.

- If τ ∈ D and φ is not an elliptic automorphism, then lim_nρ(z_n,z_n+1) = ρ(τ, τ ) =0.

- If φ ∈ Aut(D), then ρ(zn,z_n+1) = ρ(z_n−1,z_n)for all n.

So, if z₀6= τ then limnρ(z_n,z_n+1) >0.

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The hyperbolic step.

- If τ ∈ D and φ is not an elliptic automorphism, then lim_nρ(z_n,z_n+1) = ρ(τ, τ ) =0.

- If φ ∈ Aut(D), then ρ(zn,z_n+1) = ρ(z_n−1,z_n)for all n.

So, if z₀6= τ then limnρ(z_n,z_n+1) >0.

- If φ is a LFM with τ ∈ ∂D and φ⁰(τ ) <1, then limnρ(zn,z_n+1) >0.

- If φ is a LFM with τ ∈ ∂D and φ⁰(τ ) =1, then