· Intro duction ry Histo First example of a andering w domain The rst example of a andering w domain is due to er. Bak function considered as w f (z) = C · 2 Y∞ j = 1 1 + z r

On boundaries of multiply onne ted wandering domainsMarkus BaumgartnerChristian-Albre hts-Universität zu KielBar elona, 11 June 2013M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 1 / 22

Outline1 Introdu tion

M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 2 / 22

Outline1 Introdu tion2 Results


Outline1 Introdu tion2 Results3 Proof


Outline1 Introdu tion2 Results3 Proof4 Examples


Introdu tion HistoryIntrodu tionDe�nition (Wandering domain)Let f be a rational or entire fun tion. A Fatou omponent U is alledwandering domain if f n(U) ∩ f m(U) = ∅ for all m < n.


Introdu tion HistoryIntrodu tionDe�nition (Wandering domain)Let f be a rational or entire fun tion. A Fatou omponent U is alledwandering domain if f n(U) ∩ f m(U) = ∅ for all m < n.Theorem (Sullivan 1982)There are no wandering domains for rational fun tions.


Introdu tion HistoryFirst example of a wandering domainThe �rst example of a wandering domain is due to Baker. The fun tion onsidered was f (z) = C · z2 ∞∏j=1(1+ zrj) ,


Introdu tion HistoryFirst example of a wandering domainThe �rst example of a wandering domain is due to Baker. The fun tion onsidered was f (z) = C · z2 ∞∏j=1(1+ zrj) ,where C > 0 is a small onstant, r1 is large and (rn)n∈N is a sequen e ofpositive real numbers that satis�es the re urren e relationrn+1 = C · r2n n∏j=1(1+ rnrj ) .


Introdu tion HistoryFirst example of a wandering domainThe �rst example of a wandering domain is due to Baker. The fun tion onsidered was f (z) = C · z2 ∞∏j=1(1+ zrj) ,where C > 0 is a small onstant, r1 is large and (rn)n∈N is a sequen e ofpositive real numbers that satis�es the re urren e relationrn+1 = C · r2n n∏j=1(1+ rnrj ) .In 1963 Baker showed that f has multiply onne ted Fatou omponents Unwith f (Un) ⊂ Un+1, but the question whether the Un are all di�erentremained open. Those were the �rst known multiply onne ted Fatou omponents.M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 4 / 22

Introdu tion HistoryFirst example of a wandering domainThe �rst example of a wandering domain is due to Baker. The fun tion onsidered was f (z) = C · z2 ∞∏j=1(1+ zrj) ,where C > 0 is a small onstant, r1 is large and (rn)n∈N is a sequen e ofpositive real numbers that satis�es the re urren e relationrn+1 = C · r2n n∏j=1(1+ rnrj ) .In 1963 Baker showed that f has multiply onne ted Fatou omponents Unwith f (Un) ⊂ Un+1, but the question whether the Un are all di�erentremained open. Those were the �rst known multiply onne ted Fatou omponents.In 1976 Baker was able to show that the Un are all di�erent and thereforewandering domains.M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 4 / 22

Introdu tion Historyrn rn+1 rn+2f f


Introdu tion HistoryBn−1 Bn Bn+1 Bn+2rn rn+1 rn+2An An+1 An+2


Introdu tion HistoryBn−1 Bn Bn+1 Bn+2rn rn+1 rn+2f f ff (Bn)An An+1 An+2f (Bn) ⊂ Bn+1 (and therefore An+1 ⊂ f (An))


Introdu tion HistoryBn−1 Bn Bn+1 Bn+2An An+1 An+2Un−1 Un Un+1 Un+2f (Bn) ⊂ Bn+1 (and therefore An+1 ⊂ f (An))This implies that Bn belongs to a multiply onne ted Fatou omponent Un.


Introdu tion HistoryBn−1 Bn Bn+1 Bn+2An An+1 An+2Un−1 Un Un+1 Un+2f (Bn) ⊂ Bn+1 (and therefore An+1 ⊂ f (An))This implies that Bn belongs to a multiply onne ted Fatou omponent Un.Assume that Un = Um for n 6= m, then this implies that Un = Um forall n,m.M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 5 / 22

Introdu tion HistoryBn−1 Bn Bn+1 Bn+2An An+1 An+2Un−1 Un Un+1 Un+2f (Bn) ⊂ Bn+1 (and therefore An+1 ⊂ f (An))This implies that Bn belongs to a multiply onne ted Fatou omponent Un.Assume that Un = Um for n 6= m, then this implies that Un = Um forall n,m.M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 5 / 22

Introdu tion HistoryBn−1 Bn Bn+1 Bn+2An An+1 An+2Un−1 Un Un+1 Un+2f (Bn) ⊂ Bn+1 (and therefore An+1 ⊂ f (An))This implies that Bn belongs to a multiply onne ted Fatou omponent Un.Assume that Un = Um for n 6= m, then this implies that Un = Um forall n,m.Baker showed that there are no unbounded multiply onne ted Fatou omponents.M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 5 / 22

Introdu tion MotivationTheorem (Baker and Dominguez 2000)Let f be an entire fun tion. If J (f ) is not onne ted, then it is not lo ally onne ted at any point of J (f ).


Introdu tion MotivationTheorem (Baker and Dominguez 2000)Let f be an entire fun tion. If J (f ) is not onne ted, then it is not lo ally onne ted at any point of J (f ).This implies that J (f ) an not be lo ally onne ted at any point for anentire fun tion f that has a multiply onne ted wandering domain.M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 6 / 22

Introdu tion MotivationTheorem (Baker and Dominguez 2000)Let f be an entire fun tion. If J (f ) is not onne ted, then it is not lo ally onne ted at any point of J (f ).This implies that J (f ) an not be lo ally onne ted at any point for anentire fun tion f that has a multiply onne ted wandering domain.QuestionAre at least the di�erent omponents of J (f ) lo ally onne ted?M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 6 / 22

Introdu tion MotivationTheorem (Baker and Dominguez 2000)Let f be an entire fun tion. If J (f ) is not onne ted, then it is not lo ally onne ted at any point of J (f ).This implies that J (f ) an not be lo ally onne ted at any point for anentire fun tion f that has a multiply onne ted wandering domain.QuestionAre at least the di�erent omponents of J (f ) lo ally onne ted?We want to show that under suitable onditions every boundary omponentof a multiply onne ted wandering domain is a urve or even a Jordan urve and therefore lo ally onne ted.M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 6 / 22

Results Preparations for the resultsResultsDe�nition (Inner and outer boundary)Let U ⊂ C be a domain and let a ∈ C \ U. We denote by C (a,U) the omponent of C \ U that ontains a.U 0a


Results Preparations for the resultsResultsDe�nition (Inner and outer boundary)Let U ⊂ C be a domain and let a ∈ C \ U. We denote by C (a,U) the omponent of C \ U that ontains a.U 0C (a,U)

aM. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 7 / 22

Results Preparations for the resultsResultsDe�nition (Inner and outer boundary)Let U ⊂ C be a domain and let a ∈ C \ U. We denote by C (a,U) the omponent of C \ U that ontains a.For 0 /∈ U we de�ne ∂0U = ∂C (0,U) as inner boundary omponent and∂∞U = ∂C (∞,U) as outer boundary omponent. We all ∂0U and ∂∞Ubig boundary omponents. U 0C (a,U)


Results Preparations for the resultsResultsDe�nition (Inner and outer boundary)Let U ⊂ C be a domain and let a ∈ C \ U. We denote by C (a,U) the omponent of C \ U that ontains a.For 0 /∈ U we de�ne ∂0U = ∂C (0,U) as inner boundary omponent and∂∞U = ∂C (∞,U) as outer boundary omponent. We all ∂0U and ∂∞Ubig boundary omponents. U 0

∂0UC (a,U)


Results Preparations for the resultsResultsDe�nition (Inner and outer boundary)Let U ⊂ C be a domain and let a ∈ C \ U. We denote by C (a,U) the omponent of C \ U that ontains a.For 0 /∈ U we de�ne ∂0U = ∂C (0,U) as inner boundary omponent and∂∞U = ∂C (∞,U) as outer boundary omponent. We all ∂0U and ∂∞Ubig boundary omponents. U 0

∂∞U∂0UC (a,U)


Results Preparations for the resultsDe�nition (Conne tivity)Let U ⊂ C be a domain. By (U) we denote the onne tivity of U, that isthe number of onne ted omponents of C \ U.


Results Preparations for the resultsDe�nition (Conne tivity)Let U ⊂ C be a domain. By (U) we denote the onne tivity of U, that isthe number of onne ted omponents of C \ U.U


Results Preparations for the resultsDe�nition (Conne tivity)Let U ⊂ C be a domain. By (U) we denote the onne tivity of U, that isthe number of onne ted omponents of C \ U.U (U) = 6


Results Preparations for the resultsDe�nition (Conne tivity)Let U ⊂ C be a domain. By (U) we denote the onne tivity of U, that isthe number of onne ted omponents of C \ U.For a sequen e of domains Un we all the eventual onne tivity of Un if (Un) = for all large n. U (U) = 6M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 8 / 22

Results Preparations for the resultsDe�nition (Conne tivity)Let U ⊂ C be a domain. By (U) we denote the onne tivity of U, that isthe number of onne ted omponents of C \ U.For a sequen e of domains Un we all the eventual onne tivity of Un if (Un) = for all large n. U (U) = 6Kisaka and Shishikura showed that the eventual onne tivity of a multiply onne ted wandering domain is either 2 or ∞.M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 8 / 22

Results Preparations for the resultsTheorem (Bergweiler, Rippon, Stallard 2013)Let f be an entire fun tion with a multiply onne ted wandering domainU = U0. Denote Un = f n(U).


Results Preparations for the resultsTheorem (Bergweiler, Rippon, Stallard 2013)Let f be an entire fun tion with a multiply onne ted wandering domainU = U0. Denote Un = f n(U).Then every Un ontains an annulus Bn su h that every ompa t subsetC ⊂ Un is mapped inside Bn+m for all large m ∈ N.


Results Preparations for the resultsTheorem (Bergweiler, Rippon, Stallard 2013)Let f be an entire fun tion with a multiply onne ted wandering domainU = U0. Denote Un = f n(U).Then every Un ontains an annulus Bn su h that every ompa t subsetC ⊂ Un is mapped inside Bn+m for all large m ∈ N.Un Un+mM. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 9 / 22

Results Preparations for the resultsTheorem (Bergweiler, Rippon, Stallard 2013)Let f be an entire fun tion with a multiply onne ted wandering domainU = U0. Denote Un = f n(U).Then every Un ontains an annulus Bn su h that every ompa t subsetC ⊂ Un is mapped inside Bn+m for all large m ∈ N.Un Un+mBn Bn+mM. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 9 / 22

Results Preparations for the resultsTheorem (Bergweiler, Rippon, Stallard 2013)Let f be an entire fun tion with a multiply onne ted wandering domainU = U0. Denote Un = f n(U).Then every Un ontains an annulus Bn su h that every ompa t subsetC ⊂ Un is mapped inside Bn+m for all large m ∈ N.Un Un+mBn Bn+mC f m(C )


Results Preparations for the resultsTheorem (Bergweiler, Rippon, Stallard 2013)Let f be an entire fun tion with a multiply onne ted wandering domainU = U0. Denote Un = f n(U).Then every Un ontains an annulus Bn su h that every ompa t subsetC ⊂ Un is mapped inside Bn+m for all large m ∈ N.Un Un+mBn Bn+mC f m(C )

De�nition (Inner onne tivity)We all (Un ∩ C (0,Bn)) the inner onne tivity of Un and de�ne theeventual inner onne tivity respe tively.M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 9 / 22

Results Main resultsTheorem 1Let f be an entire fun tion with a multiply onne ted wandering domainU = U0. Denote Un = f n(U).


Results Main resultsTheorem 1Let f be an entire fun tion with a multiply onne ted wandering domainU = U0. Denote Un = f n(U).Suppose that there exists a sequen e of positive real numbers (rn)n∈N aswell as α, β > 0 su h that for a sequen e of annuliCn := {z ∈ C : αrn ≤ |z | ≤ βrn} the following onditions hold:


Results Main resultsTheorem 1Let f be an entire fun tion with a multiply onne ted wandering domainU = U0. Denote Un = f n(U).Suppose that there exists a sequen e of positive real numbers (rn)n∈N aswell as α, β > 0 su h that for a sequen e of annuliCn := {z ∈ C : αrn ≤ |z | ≤ βrn} the following onditions hold:∂0Cn ⊂ Un−1, ∂∞Cn ⊂ Un


Results Main resultsTheorem 1Let f be an entire fun tion with a multiply onne ted wandering domainU = U0. Denote Un = f n(U).Suppose that there exists a sequen e of positive real numbers (rn)n∈N aswell as α, β > 0 su h that for a sequen e of annuliCn := {z ∈ C : αrn ≤ |z | ≤ βrn} the following onditions hold:∂0Cn ⊂ Un−1, ∂∞Cn ⊂ UnCn+1 ⊂ f (Cn)


Results Main resultsTheorem 1Let f be an entire fun tion with a multiply onne ted wandering domainU = U0. Denote Un = f n(U).Suppose that there exists a sequen e of positive real numbers (rn)n∈N aswell as α, β > 0 su h that for a sequen e of annuliCn := {z ∈ C : αrn ≤ |z | ≤ βrn} the following onditions hold:∂0Cn ⊂ Un−1, ∂∞Cn ⊂ UnCn+1 ⊂ f (Cn)There exists m > α

βsu h that for all z ∈ f −1(Cn+1) ∩ Cn

∣

∣

∣

∣

z · f ′(z)f (z) ∣

∣

∣

∣

≥ m and Re(z · f ′(z)f (z) )

> 0.M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 10 / 22

Results Main resultsTheorem 1Let f be an entire fun tion with a multiply onne ted wandering domainU = U0. Denote Un = f n(U).Suppose that there exists a sequen e of positive real numbers (rn)n∈N aswell as α, β > 0 su h that for a sequen e of annuliCn := {z ∈ C : αrn ≤ |z | ≤ βrn} the following onditions hold:∂0Cn ⊂ Un−1, ∂∞Cn ⊂ UnCn+1 ⊂ f (Cn)There exists m > α

βsu h that for all z ∈ f −1(Cn+1) ∩ Cn

∣

∣

∣

∣

z · f ′(z)f (z) ∣

∣

∣

∣

≥ m and Re(z · f ′(z)f (z) )

> 0.Then all big boundary omponents are Jordan urves and ∂∞Un−1 = ∂0Un.M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 10 / 22

Results Main resultsTheorem 2Suppose we have additionally to the onditions of theorem 1 that theeventual inner onne tivity of Un is 2.


Results Main resultsTheorem 2Suppose we have additionally to the onditions of theorem 1 that theeventual inner onne tivity of Un is 2.Then all wandering domains, whi h belong to the orbit of Un, are boundedby a ountable number of losed urves.


Results Main resultsTheorem 2Suppose we have additionally to the onditions of theorem 1 that theeventual inner onne tivity of Un is 2.Then all wandering domains, whi h belong to the orbit of Un, are boundedby a ountable number of losed urves.With the onditions of theorem 2 we an obtain the following orollary:M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 11 / 22

Results Main resultsTheorem 2Suppose we have additionally to the onditions of theorem 1 that theeventual inner onne tivity of Un is 2.Then all wandering domains, whi h belong to the orbit of Un, are boundedby a ountable number of losed urves.With the onditions of theorem 2 we an obtain the following orollary:CorollaryIf Z is a boundary omponent of a wandering domain, whi h belongs to theorbit of Un, su h that f j(Z ) does not ontain any riti al points for allj ∈ N0, then Z is a Jordan urve.M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 11 / 22

Results Main resultsTheorem 2Suppose we have additionally to the onditions of theorem 1 that theeventual inner onne tivity of Un is 2.Then all wandering domains, whi h belong to the orbit of Un, are boundedby a ountable number of losed urves.With the onditions of theorem 2 we an obtain the following orollary:CorollaryIf Z is a boundary omponent of a wandering domain, whi h belongs to theorbit of Un, su h that f j(Z ) does not ontain any riti al points for allj ∈ N0, then Z is a Jordan urve.Both theorems work for Baker's �rst example of a wandering domain.M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 11 / 22

Proof Idea of the proofProofUnderstanding the setting of the theorem 1:


Proof Idea of the proofProofUnderstanding the setting of the theorem 1:Cn Cn+1 Cn+2rn βrnαrn

Cn := {z ∈ C : αrn ≤ |z | ≤ βrn}M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 12 / 22

Proof Idea of the proofProofUnderstanding the setting of the theorem 1:Cn Cn+1 Cn+2rn βrnαrnUn−1 Un Un+1Cn := {z ∈ C : αrn ≤ |z | ≤ βrn}∂0Cn ⊂ Un−1, ∂∞Cn ⊂ UnM. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 12 / 22

Proof Idea of the proofProofUnderstanding the setting of the theorem 1:Cn Cn+1 Cn+2rn βrnαrnUn−1 Un Un+1fCn := {z ∈ C : αrn ≤ |z | ≤ βrn}∂0Cn ⊂ Un−1, ∂∞Cn ⊂ UnCn+1 ⊂ f (Cn)M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 12 / 22

Proof Idea of the proofWe want to show that ∂∞Un−1 and ∂0Un are both urves that oin ide.


Proof Idea of the proofWe want to show that ∂∞Un−1 and ∂0Un are both urves that oin ide.De�ne for all k ∈ N

Γk := {z ∈ Cn : f j(z) ∈ Cn+j for all j=1,. . . ,k}.



Γk := {z ∈ Cn : f j(z) ∈ Cn+j for all j=1,. . . ,k}.Cn Cn+1 Cn+2 Cn+3



Γk := {z ∈ Cn : f j(z) ∈ Cn+j for all j=1,. . . ,k}.Γ1Cn Cn+1 Cn+2 Cn+3f



Γk := {z ∈ Cn : f j(z) ∈ Cn+j for all j=1,. . . ,k}.Γ1

Γ2Cn Cn+1 Cn+2 Cn+3ff 2M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 13 / 22


Γk := {z ∈ Cn : f j(z) ∈ Cn+j for all j=1,. . . ,k}.Γ1

Γ3 Γ2Cn Cn+1 Cn+2 Cn+3ff 2f 3M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 13 / 22

Proof Idea of the proofThe inequality ∣∣∣

z ·f ′(z)f (z) ∣∣∣ ≥ m implies that there are no riti al points insidethe Γk . So all Γk are topologi al annuli by the Riemann-Hurwitz-formulathat are bounded by Jordan urves.



z ·f ′(z)f (z) ∣∣∣ ≥ m implies that there are no riti al points insidethe Γk . So all Γk are topologi al annuli by the Riemann-Hurwitz-formulathat are bounded by Jordan urves.The inequality also implies the following lemma:



z ·f ′(z)f (z) ∣∣∣ ≥ m implies that there are no riti al points insidethe Γk . So all Γk are topologi al annuli by the Riemann-Hurwitz-formulathat are bounded by Jordan urves.The inequality also implies the following lemma:LemmaThere exists ̺ > 1 su h that for all k ∈ N and z ∈ Γk|(f k)′(z)| ≥ ̺k · rn+krn



z ·f ′(z)f (z) ∣∣∣ ≥ m implies that there are no riti al points insidethe Γk . So all Γk are topologi al annuli by the Riemann-Hurwitz-formulathat are bounded by Jordan urves.The inequality also implies the following lemma:LemmaThere exists ̺ > 1 su h that for all k ∈ N and z ∈ Γk|(f k)′(z)| ≥ ̺k · rn+krnTherefore f k is expanding inside Γk and this implies that f −k : Cn+k → Γkis ontra ting.



z ·f ′(z)f (z) ∣∣∣ ≥ m implies that there are no riti al points insidethe Γk . So all Γk are topologi al annuli by the Riemann-Hurwitz-formulathat are bounded by Jordan urves.The inequality also implies the following lemma:LemmaThere exists ̺ > 1 su h that for all k ∈ N and z ∈ Γk|(f k)′(z)| ≥ ̺k · rn+krnTherefore f k is expanding inside Γk and this implies that f −k : Cn+k → Γkis ontra ting.We parametrise now ∂0Γk and ∂∞Γk as urves by γ0k and γ∞k respe tively.Thereby one has to he k that the parametrisations are ompatible withea h other. Here Re( z ·f ′(z)f (z) ) > 0 is used. It ensures that the urves are notdistorted too mu h under iteration.M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 14 / 22

Proof Idea of the proof∂0Un∂∞Un−1

∂0Cn ∂∞Cn



∂0Cn ∂∞Cnγ01 γ∞1



∂0Cn ∂∞Cnγ02γ01 γ∞1γ∞2



∂0Cn ∂∞Cnγ03γ02γ01 γ∞1γ∞3 γ∞2


Proof Idea of the proofγ

∂0Un∂∞Un−1∂0Cn ∂∞Cn

γ03γ02γ01 γ∞1γ∞3 γ∞2

Then we use that f −k is ontra ting to show that the urves γ0k and γ∞k onverge uniformly to the same urve γ withtra e(γ) = ⋂k∈N Γk .M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 15 / 22

Proof Idea of the proofBy positioning of Cn to Un−1 and Un we have∂∞Un−1 = tra e(γ) = ∂0Un.


Proof Idea of the proofBy positioning of Cn to Un−1 and Un we have∂∞Un−1 = tra e(γ) = ∂0Un.Now we have that all big boundary omponents are urves, so it remains toshow that they are Jordan urves.


Proof Idea of the proofBy positioning of Cn to Un−1 and Un we have∂∞Un−1 = tra e(γ) = ∂0Un.Now we have that all big boundary omponents are urves, so it remains toshow that they are Jordan urves.Sin e ∂∞Un−1 and ∂0Un are urves and therefore lo ally onne ted everypoint on tra e(γ) is a essible in Un−1 and in Un.


Proof Idea of the proofBy positioning of Cn to Un−1 and Un we have∂∞Un−1 = tra e(γ) = ∂0Un.Now we have that all big boundary omponents are urves, so it remains toshow that they are Jordan urves.Sin e ∂∞Un−1 and ∂0Un are urves and therefore lo ally onne ted everypoint on tra e(γ) is a essible in Un−1 and in Un.Thus a theorem of S hön�ies yields that γ is in fa t a Jordan urve.


Proof Idea of the proofBy positioning of Cn to Un−1 and Un we have∂∞Un−1 = tra e(γ) = ∂0Un.Now we have that all big boundary omponents are urves, so it remains toshow that they are Jordan urves.Sin e ∂∞Un−1 and ∂0Un are urves and therefore lo ally onne ted everypoint on tra e(γ) is a essible in Un−1 and in Un.Thus a theorem of S hön�ies yields that γ is in fa t a Jordan urve.We have proven theorem 1, so it remains to prove theorem 2.


Proof Idea of the proofLemmaLet f be an entire fun tion with a multiply onne ted wandering domain Uwith eventual inner onne tivity 2. Let Z be a boundary omponent of U.


Proof Idea of the proofLemmaLet f be an entire fun tion with a multiply onne ted wandering domain Uwith eventual inner onne tivity 2. Let Z be a boundary omponent of U.If Z 6= ∂∞U there exists q ∈ N0 su h that f j(Z ) = ∂0Uj for all j ≥ q.If Z = ∂∞U we have f j(Z ) = ∂∞Uj for all j ∈ N0.


Proof Idea of the proofLemmaLet f be an entire fun tion with a multiply onne ted wandering domain Uwith eventual inner onne tivity 2. Let Z be a boundary omponent of U.If Z 6= ∂∞U there exists q ∈ N0 su h that f j(Z ) = ∂0Uj for all j ≥ q.If Z = ∂∞U we have f j(Z ) = ∂∞Uj for all j ∈ N0.Un Un+1 Un+2 Un+3


Proof Idea of the proofLemmaLet f be an entire fun tion with a multiply onne ted wandering domain Uwith eventual inner onne tivity 2. Let Z be a boundary omponent of U.If Z 6= ∂∞U there exists q ∈ N0 su h that f j(Z ) = ∂0Uj for all j ≥ q.If Z = ∂∞U we have f j(Z ) = ∂∞Uj for all j ∈ N0.Un Un+1 Un+2 Un+3ZM. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 17 / 22

Proof Idea of the proofLemmaLet f be an entire fun tion with a multiply onne ted wandering domain Uwith eventual inner onne tivity 2. Let Z be a boundary omponent of U.If Z 6= ∂∞U there exists q ∈ N0 su h that f j(Z ) = ∂0Uj for all j ≥ q.If Z = ∂∞U we have f j(Z ) = ∂∞Uj for all j ∈ N0.Un Un+1 Un+2 Un+3Z f (Z )fM. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 17 / 22

Proof Idea of the proofLemmaLet f be an entire fun tion with a multiply onne ted wandering domain Uwith eventual inner onne tivity 2. Let Z be a boundary omponent of U.If Z 6= ∂∞U there exists q ∈ N0 su h that f j(Z ) = ∂0Uj for all j ≥ q.If Z = ∂∞U we have f j(Z ) = ∂∞Uj for all j ∈ N0.Un Un+1 Un+2 Un+3Z f (Z )f f f 2(Z )M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 17 / 22

Proof Idea of the proofLemmaLet f be an entire fun tion with a multiply onne ted wandering domain Uwith eventual inner onne tivity 2. Let Z be a boundary omponent of U.If Z 6= ∂∞U there exists q ∈ N0 su h that f j(Z ) = ∂0Uj for all j ≥ q.If Z = ∂∞U we have f j(Z ) = ∂∞Uj for all j ∈ N0.Un Un+1 Un+2 Un+3Z f (Z )f f f 2(Z ) f ∂0Un+3M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 17 / 22

Proof Idea of the proofLemmaLet f be an entire fun tion with a multiply onne ted wandering domain Uwith eventual inner onne tivity 2. Let Z be a boundary omponent of U.If Z 6= ∂∞U there exists q ∈ N0 su h that f j(Z ) = ∂0Uj for all j ≥ q.If Z = ∂∞U we have f j(Z ) = ∂∞Uj for all j ∈ N0.Un Un+1 Un+2 Un+3Z f (Z )f f f 2(Z ) ff∂∞Un∂0Un+3∂∞Un+1


Proof Idea of the proofLemmaLet f be an entire fun tion with a multiply onne ted wandering domain Uwith eventual inner onne tivity 2. Let Z be a boundary omponent of U.If Z 6= ∂∞U there exists q ∈ N0 su h that f j(Z ) = ∂0Uj for all j ≥ q.If Z = ∂∞U we have f j(Z ) = ∂∞Uj for all j ∈ N0.Un Un+1 Un+2 Un+3For this reason it makes sense to group the boundary omponents in ertain 'levels' indi ated by the iterations they need to be mapped onto abig boundary omponent.M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 17 / 22

Proof Idea of the proofLemmaLet f be an entire fun tion with a multiply onne ted wandering domain Uwith eventual inner onne tivity 2. Let Z be a boundary omponent of U.If Z 6= ∂∞U there exists q ∈ N0 su h that f j(Z ) = ∂0Uj for all j ≥ q.If Z = ∂∞U we have f j(Z ) = ∂∞Uj for all j ∈ N0.1Un Un+1 Un+2 Un+3

For this reason it makes sense to group the boundary omponents in ertain 'levels' indi ated by the iterations they need to be mapped onto abig boundary omponent.M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 17 / 22

Proof Idea of the proofLemmaLet f be an entire fun tion with a multiply onne ted wandering domain Uwith eventual inner onne tivity 2. Let Z be a boundary omponent of U.If Z 6= ∂∞U there exists q ∈ N0 su h that f j(Z ) = ∂0Uj for all j ≥ q.If Z = ∂∞U we have f j(Z ) = ∂∞Uj for all j ∈ N0.1 2Un Un+1 Un+2 Un+3


Proof Idea of the proofLemmaLet f be an entire fun tion with a multiply onne ted wandering domain Uwith eventual inner onne tivity 2. Let Z be a boundary omponent of U.If Z 6= ∂∞U there exists q ∈ N0 su h that f j(Z ) = ∂0Uj for all j ≥ q.If Z = ∂∞U we have f j(Z ) = ∂∞Uj for all j ∈ N0.1 2 3Un Un+1 Un+2 Un+3


Proof Idea of the proofBy the maximum modulus prin iple it is lear that only outer boundary omponents are mapped onto outer boundary omponents.


Proof Idea of the proofBy the maximum modulus prin iple it is lear that only outer boundary omponents are mapped onto outer boundary omponents.Suppose Z is a boundary omponent of U and Z 6= ∂∞U.


Proof Idea of the proofBy the maximum modulus prin iple it is lear that only outer boundary omponents are mapped onto outer boundary omponents.Suppose Z is a boundary omponent of U and Z 6= ∂∞U.U UnZ


Proof Idea of the proofBy the maximum modulus prin iple it is lear that only outer boundary omponents are mapped onto outer boundary omponents.Suppose Z is a boundary omponent of U and Z 6= ∂∞U.U UnZγ


Proof Idea of the proofBy the maximum modulus prin iple it is lear that only outer boundary omponents are mapped onto outer boundary omponents.Suppose Z is a boundary omponent of U and Z 6= ∂∞U.BnU UnZγ


Proof Idea of the proofBy the maximum modulus prin iple it is lear that only outer boundary omponents are mapped onto outer boundary omponents.Suppose Z is a boundary omponent of U and Z 6= ∂∞U.BnU UnZγ f n(γ)


Proof Idea of the proofBy the maximum modulus prin iple it is lear that only outer boundary omponents are mapped onto outer boundary omponents.Suppose Z is a boundary omponent of U and Z 6= ∂∞U.BnU UnZ f n(Z )γ f n(γ)


Proof Idea of the proofBy the maximum modulus prin iple it is lear that only outer boundary omponents are mapped onto outer boundary omponents.Suppose Z is a boundary omponent of U and Z 6= ∂∞U.BnU UnZ f n(Z )γ f n(γ)This implies that every boundary omponent of U will be eventuallymapped onto a big boundary omponent.


Proof Idea of the proofBy the maximum modulus prin iple it is lear that only outer boundary omponents are mapped onto outer boundary omponents.Suppose Z is a boundary omponent of U and Z 6= ∂∞U.BnU UnZ f n(Z )γ f n(γ)This implies that every boundary omponent of U will be eventuallymapped onto a big boundary omponent.Under the onditions of theorem 1 those big boundary omponents areJordan urves, so every boundary omponent of U is either a urve or evena Jordan urve if there are no riti al points in its forward orbit.M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 18 / 22

Examples Bergweiler's and Zheng's exampleExamplesIn the following we are looking at three di�erent examples of entirefun tions with multiply onne ted wandering domains to whi h we anapply the theorems.


Examples Bergweiler's and Zheng's exampleExamplesIn the following we are looking at three di�erent examples of entirefun tions with multiply onne ted wandering domains to whi h we anapply the theorems.Bergweiler's and Zheng's examplef (z) = C · zk ∞∏j=1(1− zaj) ,where C > 0, k ∈ N and (aj )j∈N is a omplex sequen e with |aj | = rj and

(rj)j∈N is a fast growing sequen e of positive real numbers.M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 19 / 22

Examples Bergweiler's and Zheng's exampleExamplesIn the following we are looking at three di�erent examples of entirefun tions with multiply onne ted wandering domains to whi h we anapply the theorems.Bergweiler's and Zheng's examplef (z) = C · zk ∞∏j=1(1− zaj) ,where C > 0, k ∈ N and (aj )j∈N is a omplex sequen e with |aj | = rj and

(rj)j∈N is a fast growing sequen e of positive real numbers.This example in ludes the �rst example of Baker.M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 19 / 22

Examples Baker's in�nite onne tivity exampleBaker's in�nite onne tivity examplef (z) = C ·

∞∏j=1(1− zrj)k

,where C > 0, k ∈ N and (rj)j∈N is a fast growing sequen e of positive realnumbers.M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 20 / 22


∞∏j=1(1− zrj)k

,where C > 0, k ∈ N and (rj)j∈N is a fast growing sequen e of positive realnumbers.This example in ludes the �rst example of Baker (1984) with a wanderingdomain with in�nite onne tivity.M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 20 / 22


∞∏j=1(1− zrj)k

,where C > 0, k ∈ N and (rj)j∈N is a fast growing sequen e of positive realnumbers.This example in ludes the �rst example of Baker (1984) with a wanderingdomain with in�nite onne tivity.Bergweiler and Zheng showed that Baker's �rst example of a wanderingdomain has also in�nite onne tivity.M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 20 / 22

Examples Bishop's exampleThe following example is due to Bishop (2011). It was the initial examplefor my work.


Examples Bishop's exampleThe following example is due to Bishop (2011). It was the initial examplefor my work.Bishop's example f (z) = F0(z) · ∞∏j=1(1− 12 ( zrj)kj)

,where F0(z) is a ertain polynomial and (rj)j∈N and (kj)j∈N are fastgrowing sequen es of positive real numbers.M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 21 / 22


,where F0(z) is a ertain polynomial and (rj)j∈N and (kj)j∈N are fastgrowing sequen es of positive real numbers.Bishop showed that the Julia set of this fun tion has Hausdor� dimension 1and that the Fatou set onsists of multiply onne ted wandering domainswhi h are bounded by C 1 − urves.M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 21 / 22


,where F0(z) is a ertain polynomial and (rj)j∈N and (kj)j∈N are fastgrowing sequen es of positive real numbers.Bishop showed that the Julia set of this fun tion has Hausdor� dimension 1and that the Fatou set onsists of multiply onne ted wandering domainswhi h are bounded by C 1 − urves.Our proofs use some of Bishop's ideas. But the arguments to show that theboundaries are C 1 − urves do not work for the other examples.M. Baumgartner (University of Kiel) Boundaries of wandering domains Bar elona, 11 June 2013 21 / 22

The EndThank you for your attention.


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· Intro duction ry Histo First example of a andering w domain The rst example of a andering w domain is due to er. Bak function considered as w f (z) = C · 2 Y∞ j = 1 1 + z r