Skew products
Jan Pel
18-07-2014
Bachelor thesis
Supervisor: dr. Han Peters
v0 v1 v2 u0 u1 u2 v0 v1 v2 u0 u1 u2 Hv1
Korteweg-de Vries Instituut voor Wiskunde
Abstract
In this thesis we discuss Skew products and there dynamical properties. First we define the Fatou and Julia set and introduce the reader to complex analysis in sev-eral variables. After that, we state and prove the Hadamard-Perron theorem. We will then discuss skew products and prove that parabolic Fatou components bulge. Finally we will explain the last chapter how the bulging of Fatou components was used by Krastio Lilov in his thesis to show that there are no non wandering of Fatou components in a neighborhood of super attracting fixed fibre, but that his prove cannot be used for the attracting case since Han Peters and Liz Raquel Vivas have shown that the theorem he uses is not valid in the attracting case.
Titel: Skew products Auteur: Jan Pel,
Supervisor: dr. Han Peters Einddatum: 18-07-2014
Korteweg-de Vries Instituut voor Wiskunde Universiteit van Amsterdam
Science Park 904, 1098 XH Amsterdam http://www.science.uva.nl/math
Contents
Introduction 4
1 Complex dynamics 6
1.1 Basics . . . 6
1.2 Riemann sphere . . . 8
1.2.1 Parabolic fixed points . . . 9
1.3 Holomorphic functions in several variables . . . 14
2 Hadamard-Perron Theorem 18 2.1 Linear maps . . . 18
2.2 Hadamard-Perron Theorem . . . 19
2.2.1 Step 1 of the proof . . . 21
2.2.2 Step 2 of the proof . . . 26
2.2.3 Extension to nearby points . . . 35
3 Skew products in C2 37 3.1 Definitions . . . 37
3.2 Bulging of Fatou components . . . 38
3.2.1 Attraction to (0,0) through parabolic fatou component . . . . 48
4 Non wandering Fatou components in C2 50
Reflection 53
Populaire samenvatting 55
Introduction
In the 1920 century, Pierre Fatou and Gaston Julie were early pioneers in the subject of complex dynamics and they proofed some big theorems. The following 60 years not much happened, some small theorems were proofed, but the subject did not really pick up steam until computers started becoming widespread in the 1980, and with the help of computers people got the intuition with which they were able to proof a lot of theorems.
Nowadays the subject of one-dimensional complex dynamics is well understood. In more dimensions a lot is still unknown. Skew products are two-dimensional holo-morphic functions
F : C2 → C2 : (t, z) → (p(t), ψ(t, z)).
They are two-dimensional functions that look a bit like one-dimensional functions. Suppose now that b is a fixed point of p with |p0(b)| < 1. Let A be the open set of points which converge to b. The behavior for points on the fibre Cb := {(b, z) ∈ C2}
is well understood. Since F|Cb is just the one-dimensional function Fb : C → C : z →
ψ(b, z).
The main goal of this thesis is to find out if the long time behavior of F on A ⊕ C looks eventually like the long term behavior of Fb.
To be able to talk less vague and more mathematical we will introduce in section one of chapter one the basic of complex dynamics. We introduce the Fatou and Julia set and prove some theorems about them. In section two we explain why for rational functions the Fatou set is well understood. Since we are working with two dimensional fuctions we introduce the reader in section three to complex analysis of several variables.
To state and proof the Hadamard-Perron theorem is the main goal of chapter two, with this theorem we can proof theorem 3.4 easier and more elegantly, the theorem is also needed to state and proof theorem 3.8.
Chapter three is the main part of our thesis. In this chapter we give two proofs of theorem 3.4, which was not yet proven in the literature.
Then we will discuss in chapter four the possibility of wandering Fatou components in a neighborhood of the point b, about this Krastio Lilov proved a lot already in
2003, and about which Han peters and Liz Vivas wrote a paper recently. We will show where our theorem fits in.
Acknowledgments
I am most grateful to Han Peters for supervising my bachelor thesis, he put a lot of effort into it, and was willing to have weekly appointments where he really took his time to explain things to me. Furthermore i am most grateful to dhr. prof. dr. Ben van Linden van den Heuvell for willing to be my second grader.
1 Complex dynamics
1.1 Basics
Complex dynamics tries to characterize the long term behavior of holomorphic self maps on complex manifolds. So if M is a n-dimensional complex manifold with
F : M → M.
We want to say in which points p ∈ M , (M, F ) is stable, so that for point q close to p the long term behavior is similar. And for which points u ∈ M , (M, F ) is unstable. Meaning there are points q arbitrarily close such that the long term behavior of q and u looks different.
Different and similar are just word in our natural language, so to be able to speak unambiguously there needs to be defined when points are stable and when they are unstable. The Julia and Fatou set are such definitions. Basicly the Fatou set is the set of stable points and the Julia set consists of the unstable points.
To define them both properly we need to introduce the notion of a normal family. Definition 1.1. For complex manifolds M and S a sequence of holomorphic functions {fn}n∈N , fn : M → S is said to converge locally uniformly if there is a holomorphic
function g : M → S such that that fn converges uniformly on compact subsets of S
to g.
Definition 1.2. A sequence of functions {fn}n∈N, fn : S → T is set to be diverge
locally uniformly from T, if for every pairs of compact set L ⊂ S and K ⊂ T there is an N ∈ N such that for n > N , fn(L) ∩ K = ∅
Definition 1.3. A collection of maps H from a complex manifold S to a complex manifold T will be called normal if every sequence of maps from H contains either a subsequence which converges locally uniformly or a subsequence which diverges locally uniformly from T .
Definition 1.4. The Fatou set F is for a Dynamical system (M, F ) the set of points p ∈ M such that there is a neighborhood U around p. such that {F|U◦n}n∈N is normal.
Definition 1.5. The Julia set J consists of the points that are not in the Fatou set Definition 1.6. A Fatou component is a connected component of the Fatou set. Definition 1.7. For a point x ∈ M we call the orbit of x the sequence of points {F◦k(x)}
k∈N which we write very often as {xn}.
Definition 1.8. Let Fc be the set of points p ∈ M such that p ∈ F and there
is a compact subset K ⊂ T and an increasing function n : N → N such that {f◦ni(p)}
i∈N ⊂ K
Let Fd be the set of points p ∈ F such that p 6∈ Fc
Lemma 1.9. Both Fc and Fd are open subsets of S.
Proof. Suppose that p ∈ Fc and that δ > 0 such that {f|B◦nδ} is a normal family. the
sequence (fni
Bδ)i∈Ncan never have a subsequence which diverges locally uniformly from
T . So it should have a subsequence which converges locally uniformly, which implies that all points q Bδ have a function m : N → N such that {fmi(q)} is contained in a
compact set. (since a manifold is locally compact)
Suppose now that p ∈ Fd and let δ > 0 such that again {f|B◦nδ} is a normal family.
Now suppose that there is a sequence (f◦ni
|Bδ) such that it converges locally uniformly.
It follows now using again the local compactness of manifolds that that {fni(p)}
is contained in a compact subset K. But then p ∈ Fc. Since {f|B◦nδ} is a normal
family, every sequence is thus locally divergent (fni
|Bδ). Since {q} is a compact set
⇒ for every compact subset K ⊂ T there is an N ∈ N such that for n > N, fn(q) ∩ K = ∅. So there is no function n : N → N and no compact subset K ⊂ T
such that {f◦ni(q)} ⊂ K. And thus q ∈ F
d so that Fd is open.
Corollary 1.10. If H is a Fatou component then H ∩ Fd = ∅ or H ∩ Fc = ∅
Lemma 1.11. The Fatou set is open and the Julia set is closed.
Lemma 1.12. The Fatou and Julia set are invariant under F in the following sense: F (F) ⊂ F,
F−1(F) = F, F (J) ⊂ J, F−1(J) = J.
Lemma 1.13. For any k > 0, the Fatou set F(F◦k) of the k-fold iterates coincides with the Fatou set F(F )
Lemma 1.14. Every open subset of a manifold can be exhausted by compact subsets. If U is an open subset of M then there is a sequence {Ki}i∈N of compact subsets of
U, such that
[
i∈N
Ki = U
for j > i, Ki ⊂ Kj.
Furthermore for every point x ∈ U there is a i ∈ N such that Ki contains a
neigbor-hood around x.
Lemma 1.15. Let M be a manifold, then the sequence {F|F◦n
c}n∈N has a subsequence
{F◦ni
|FC}i∈N and a holomorphic function G : Fc → M such that {F
◦ni
|Fc }i∈N converges
uniformly on compact subsets of Fc to G.
Proof. We will proof this with a diagonal argument. Let {Ki}i∈N be a set of compact
subsets of Fc that exhausts Fc. Then for every Ki and every function k : N → N
there is a subsequence of {F|F◦k(n)
c }n∈N. So now we can inductively (using the axiom
of choice) find a sequence of sequences, {{F◦ki(n)
|Fc }n∈N}i∈N. Such that for j > i, kj is
a subsequence of ki.
h(i) := ki(i)
Then {F|F◦h(n)
c }n∈N is a subsequence of {F
◦(n)
|Fc }n∈N that converges uniformly on every
Ki and thus on every compact subset of Fc
From this lemma and the fact that for a Fatou H we have: H ⊂ Fd or H ⊂ Fc,
it follows that on H the sequence {F|H◦n}n∈Ndiverges locally uniformly from H, or has a subsequence which converges locally uniformly to a holomorphic function G : H → M . So the dynamical behavior is very similar on Fatou components, and that is the reason we are interested in them.
1.2 Riemann sphere
If M is the Riemann sphere( the complex plane with {∞}). Every holomorphic map f : ˆC →C is a rational map.ˆ
f (z) = p(z) q(z).
Where p and q are polynomials. For those maps the Fatou set and components are very well understood. Because for a Fatou component that is mapped onto itself by a rational map f with deg(f ) ≥ 2, the behavior is perfectly classified by the following theorem:
Theorem 1.16. If f maps the Fatou component U onto itself then either: (i) U is the immediate basin for an attracting fixed point
(ii) U is the immediate basin for one petal of a parabolic fixed point. (iii) U is a Siegel disk
(iv) U is a Herman ring.
From the following theorem it follows that for such a map every fatou component is eventually periodic.
Theorem 1.17. Sullivan’s non wandering theorem For every Fatou component U ⊂ ˆ
C there are n, m ∈ N with n < m such that f◦n(U ) = f◦m(U )
The Fatou component fn(U ) is mapped onto itself by the rational map fm−n and
so the behavior is understood by the classification given in the first theorem.
1.2.1 Parabolic fixed points
Since the theorems in chapter four are about the bulging of parabolic Fatou com-ponents, we will explain more about parabolic fixed points and parabolic Fatou components.
Definition 1.18. Let M be a one dimensional complex manifold and f : M → M
a holomorphic function. If p is a fixed point of f such that relative to a chart (U, h), (h ◦ f ◦ h−1)0(h(p)) = 1 (this is independent of the chosen chart), we call p a parabolic fixed point.
We can now assume that h(p) = 0 such that:
h ◦ f ◦ h−1 : U → C. h ◦ f ◦ h−1(0) = 0 (h ◦ f ◦ h−1)0(0) = 1
v0 v1 v2 u0 u1 u2
Figure 1.1: Attracting and repelling vectors in a case with m = 3
where U is a neighborhood of 0.
We will now write the map h ◦ f ◦ h−1 as a power series around 0. So for z small enough the following holds.
h ◦ f ◦ h−1(z) = z + a2z2+ a3z3....
If an = 0 for n > 1, then f is the identity function around the point p, and if the
manifold is connected, the function f is the identity on the whole manifold M . So in that case the behavior of f is well understood. Suppose now that m is the smallest natural number such that am+1 6= 0.
Definition 1.19. The attraction vectors are the m complex numbers v0, .., vm−1 such
that mam+1vjm = −1
Definition 1.20. The repulsion vectors are the m complex numbers u0, .., um−1 such
that mam+1umj = 1
From now on we’ll write f for h ◦ f ◦ h−1.
Definition 1.21. We say that the orbit of a point x converges to 0 from the direction vj if lim k→∞ m √ kxk = vj
Definition 1.22. An attracting petal for the vector vj is a non empty open set
P ⊂ U such that:
1. f maps P into itself.
2. the orbit of a point x converges to 0 from the direction vj iff there is an k ∈ N
such that f◦k(x) ∈ P .
Theorem 1.23. For every attracting vector vj there is an attracting petal, and for
every point x, whose orbit {xm} converges to 0, the orbit of x converges to 0 from
the direction vj for a j ∈ {0, .., m − 1}
Since the derivative of f is 1 in the point 0, f is locally invertible. For the inverse function, the derivative is again 1 in the point 0. And the attracting vectors for the inverse function, are precisely the repulsion vector for f . Since if f−1(z) = z +b2z2+..
for z small enough. So:
(z + b2z2+ ..) + am+1(z + b2z2..)m+1 = z
⇒ bj = 0 for 1 < j < m + 1
And bm+1 = −am+1
Definition 1.24. A repelling petal Rj is an attracting petal for the function f−1 for
the attracting vector uj(for this inverse function).
Theorem 1.25. Parabolic Flower Theorem
There are attracting petals Pj, and repelling petals such that there is a neighborhood
V of 0 such that
V ⊂ {0} ∪ P0∪ .. ∪ Pm−1∪ R0.. ∪ Rm−1
Proof. We will proof the Fatou flower theorem by giving the petals, thereby also proving theorem 1.21. Consider the following map:
ϕ : C∗ → C∗ : z → −1 mam+1zm
. If we consider the following sets:
Hvj = {z ∈ C
∗
; −π/m < arg(z vj
) < π/m}. then the map
ϕvj : Hvj → C
∗
v0 v1 v2 u0 u1 u2 v0 v1 v2 u0 u1 u2 Hv1
Figure 1.2: a schematic representation of the function ϕv1
is biholomorphic, just as the map
ϕuj : Huj → C
∗
\R>0
Now we take a look at the maps:
Fj := ϕ ◦ f ◦ ϕ−1vj , Gj := ϕ ◦ f−1ϕ−1uj. Fj(w) = w 1 + a −1 mamw + o(1 w) −m = w + 1 + o(1) as |w| → ∞. (1.1) Gj(w) = w − 1 + o(1) as |w| → ∞. (1.2)
So we can find an N such that for all j the following holds:
|w| > N ⇒ Re(Fj(w)) > Re(w) + 1/2 and Re(Gj(w)) < Re(w) − 1/2.
Consider then the following sets:
A1 := {w ∈ C∗; Re(w) > 2N − |Im(w)|},
A1
Figure 1.3: The set A1
Then for w ∈ A1 we have:
Re(Fj(w)) > Re(w) + 1/2,
|Im(w)| − 1/2 < |Im(Fj(w))| < |Im(w)| + 1/2.
So
Fj(A1) ⊂ A1, (1.3)
and by an analogous argument
Gj(A2) ⊂ A2. (1.4)
Now we define the following sets:
Pvj := ϕ −1 vj (A1), Puj := ϕ −1 uj(A2).
Claim 1.26. Pvj is an attracting petals for the vector vj, and Puj is a repelling vector
for the vector uj
Proof. That every vector v ∈ Pvj converges to 0 from the direction vj follows at once
orbit of x converges to 0 from the direction vj, then there is an M > 0 such that for n > M : |xn| < 1 4N, |m√ nxn− vj| < 1 2m.
Then Re(ϕ(xn)) > 2N so ϕ(xn) ∈ A1, and since xn∈ Hvj it follows that xn∈ Pvj
Claim 1.27. {0} ∪ Pv0 ∪ .. ∪ Pvm−1∪ Pu0 ∪ ... ∪ Pum−1 is an open neighborhood of 0
Proof. This follows at once from the fact that A1∪ A2 is a neighborhood of ∞
By proving this, we proved the Fatou flower theorem.
Definition 1.28. Let f : M → M And p is a parabolic fixed point, then Bp,vj is
the set of elements whose orbit converges to p from the direction vj
The basin of attraction is an open subset of the Fatou set, since there is a petal of attraction for vj, which is an open set. and clearly
∞
[
n=1
f−n(PvJ) = Bp,vj
1.3 Holomorphic functions in several variables
Since we are working with holomorphic functions of several variables in this project, we will give a short overview about what they are, and name a few of the basic theorems. The proofs can be found in the first chapter of [2].
If B ⊂ Cn and
F : B → C,
then we call F complex differentiable a point x ∈ B if there is a complex linear function A : Cn → C such that:
lim
y→x,y6=x
F (y) − A(y − x) − F (x)
ky − xk = 0.
In one variable a holomorphic function can be written locally as a power series. That can also be done in more dimensions. To do this we introduce the following: For v = (v1, .., vn) ∈ Nn: kvk = n X i=1 vi and zv = z1v1..z vn n .
To discuss power series we need to talk bout convergence of the following sum: X
v∈Nn
av,
but since there is no natural ordering on Nn this is not necessarily well defined. Definition 1.29. The series P
v∈Nnav is called convergent is there is a bijective
function g : N → Nn such that P∞
i=0|vg(i)| is convergent. Then:
X v∈Nn av := ∞ X i=0 ag(i)
This definition is clearly independent of g, since for an absolute convergent series the order of summation does not matter.
Definition 1.30. Pn(z
0, r) = {z ∈ C; |zi− zi0| < ri for i = 1, .., n}
A powerseries is the following formal expression: X
v∈Nn
av(z − z0)v
Claim 1.31. Suppose thatP
v∈Nnav(z−z0)v converges for a point z with |zi−zi0| = ri
for i = 1, .., n, then the series converges for all points in Pn(z0, r)
So the sets Pn(z
0, r) are the natural domains of convergence for power series in
more variables. Just as disks are the natural domains of convergence for 1 dimen-sional holomorphic functions.
Definition 1.32. We will call a function F : B → C
analytic if for every point x ∈ B there is a neighborhood U such that F|U can be
written as a power series. Definition 1.33. A function
F : B → C is called weakly holomorphic in a point x, if dzdF
i(x) exists for i = 1, .., n.
Theorem 1.34. If F : B → C is a function continuous function then the following are equivalent.
(i) F is analytic
(ii) F is holomorphic at every point x ∈ B
(iii) F is weakly holomorphic at every point x ∈ B
Since we will discuss real differentiable as well as complex differentiable functions it is convenient to introduce the following convention. Which we will use from now on without mentioning explicitly. We can identify Ck with R2k, by the following
map:
g : R2k → Ck : (x1, .., xk, y1, .., yk) → (x1 + iy1, .., xk+ iyk).
So a map f : B → Ck can be identified with a map f0 : B ⊂ R2n → R2m.
f0 = g ◦ f ◦ g−1
Definition 1.35. We call a function F = (F1, .., Fm) : B → Cm holomorphic iff Fi
is holomorphic for i = 1, .., n.
Definition 1.36. The function F : B → Cm is real differentiable if the function F0 is differentiable.
A function H : D ⊂ Rm → Rn is differentiable in the point x iff there is a real
linear function C : Rm → Rn such that
lim
y→x,y6=x
H(y) − C(y − x) − H(x)
ky − xk = 0
Claim 1.37. A function F : B → Cn is complex differentiable in a point x ∈ B iff
F0 is differentiable in g−1(x) with the real linear function C, and if g−1◦ C ◦ g is a complex linear function.
Just for the sake completness we will now list 3 theorems one learns when learning complex analysis in 1 variable, that are still true for functions in more dimensions. Theorem 1.38. Maximum principle
Let B be a connected open subset of Cm, if F : B → C is a holomorphic function and there is a point x such that |F | has a local maximum in x, Then F is a constant function.
Theorem 1.39. Identity theorem
Let B be a connected open subset of Cm, if F
1, F2 are 2 holomorphic functions Fi :
Theorem 1.40. Inverse mapping theorem
Consider a point z0 ∈ B and its image w0 ∈ F (B). The following are equivalent:
(i) There are open neighborhoods U = U (z0) ⊂ B and V = V (w0) ⊂ F (B) such
that F : U → V is a biholomorphic function. i.e. there is an holomorphic function G : V → U such that F ◦ G = Id = G ◦ F .
2 Hadamard-Perron Theorem
In this chapter we state and proof the Hadamard-Perron theorem, which is used in in the second proof of theorem 3.6 and in the formulation and proof of theorem 3.10. This whole chapter is paraphrased from chapter 3 of [5]
2.1 Linear maps
Definition 2.1. A Linear Map T : V → V admits a (λ, µ)-splitting. If there are subspaces: E+ and E−, such that: V = E+⊕ E−. T (E+) = E+, T (E−) ⊂ E−. ∀x ∈ E+, kT (x)k ≥ µkxk. ∀x ∈ E−, kT (x)k ≤ λkxk.
Definition 2.2. A sequence of linear maps {Tm}m∈Z, Tm : V → V admits a linear
splitting of unstable dimension k if there are sequences of linear subspaces {Em+} and {E− m} such that: Em+ has dimension k T (Em+) = Em+1+ , T (Em−) ⊂ Em+1− . ∀x ∈ Em+, kT (x)k ≥ µkxk. ∀x ∈ Em−, kT (x)k ≤ λkxk.
Theorem 2.3. Let U ⊂ Cn be a neighborhood of the origin, and F : U → Cn a Cr function with F (0) = 0. Then for every > 0 there is a δ > 0 and a Cr function F0
such that
F (x) = F0(x) for kxk < δ 2. kF0− dF0k1,Cn < .
Proof. Since F is continuous differentiable there is an 0 < η < 1 such that kF − DF0k1,B(0,η) < 8. Let g : Cn → Cn be the bump function that is 1 on B(0,12η) and
g(x) = 0 for kxk > η. And with kgk1,Cn ≤ 4
η. Now we define:
F0 = gF + (1 − g)dF0
The map is Cr, since it is the product of Cr and C∞ functions. We clearly have: kF0 − DF0k0,Cn = kF
0
− DF0k0,B(0,η)≤ kF − DF0k0,B(0,η) ≤
8, Since F0(x) = DF0(x) for kxk > η. Furthermore
kd(F0− DF0)k0,Cn = kd(F 0 − DF0)k0,B(0,η), and d(F0 − dF0) = d(g(F − dF0)) = dg ◦ (F − dF0) + gd(F − dF0). So kd(F0 − dF0)k0,Cn ≤ kgk1,Cn ◦ k(F − dF0)k0,Cn+ kgk0,CnkF − dF0k1,Cn ≤ 4 η η 8 + 8 < . From the fact that
kd(F0− dF0)k0,Cn = kF 0
− dF0k1,Cn
the theorem now follows.
2.2 Hadamard-Perron Theorem
Theorem 2.4. Hadamard-PerronLet 0 < λ < µ be given. Assume that we have a sequence fm : Cn→ Cn of C∞ maps
fixing the origin and a sequence Tm : Cn → Cn of complex linear maps admitting a
(λ, µ) splitting Cn = E+ m ⊕ E
−
m of unstable dimension k ≥ 0. Assume furthermore
that there is a δ > 0 such that for all m ∈ Z, fm is holomorphic on Bδ and the
following equations hold:
0 < γ < min{1,pµ/λ − 1}. (2.1) 0 < < 1 2min{ µ − λ γ + 2 + γ−1, µ − (1 + γ)2λ (1 + γ)(γ2+ 2γ + 2)}. (2.2) kfm− Tmk1,Rn < for all m ∈ Z. (2.3) λ0 = (1 + γ)(λ + (1 + γ)), µ0 = µ 1 + γ − . (2.4)
Then there is a unique family {W+
m} of k-dimensional C1 manifolds given as graphs
of C1 functions ϕ : Em+ → E−
m and a unique family {Wm−} of (n-k) dimensional C1
manifolds given as graphs of C1 functions ϕ− m : E
−
m → Em+ with supmkdφ ±
mk < γ And
such that the following properties hold: (i) 0 ∈ Wm+∩ W−
m;
(ii) the sequence {d(fm)0}admitsa(λ
0
, µ0) splitting Rn = E+(d(f
m)0) ⊕ E−(d(fm)0)
of unstable dimension k, and Wm± is tangent to E±(d(fm)0) at the origin;
(iii) fm(Wm−) ⊂ W − m+1, fm(Wm+) = W + m+1 and fm|w+ m is invertible; (iv) kfm(z)k ≤ λ 0
kzk for all z ∈ Wm−, and kfm(z)k ≥ µ
0
kzk for all z ∈ W+ m;
(v) given m0 ∈ Z, we have z0 ∈ Wm−0 iff there are λ 0
< ν < µ0 and C > 0 such that kfm0+m−1◦ ... ◦ fm0(z0)k ≤ Cν
mkz
0k for all m ≥ 1
(vi) given m0 ∈ Z, we have z0 ∈ Wm+0 iff there are λ 0
< ν < µ0, C > 0 and a m0-history ˆz for z0 such that kz−mk ≤ Cν−mkz0k for all m ≥ 1;
(vii) the manifolds Wm± are of class C∞; (viii) fm = f for all m ∈ Z then Wm+ and W
−
m do not depend on m;
(ix) W+
m0 is a complex manifold at all points p ∈ Bδ ∩ W
+
m0 such that there is an
m0-history ˆp of p so that pm ∈ Bδ for all m < m0
(x) Wm−
0 is a complex manifold at all points p ∈ Bδ ∩ W
−
m0 such that there is an
m0-history ˆp of p so that pm ∈ Bδ for all m > m0
Proof. The proof of the Hadamard-Perron Theorem is quite long, to make sure the reader does not get lost, we give a quick overview.
1. First we will construct using the contraction theorem a family of Lipschitz con-tinuous functions ϕ±m : Em± → E∓
m such that Fm( graph(ϕ+m)) = graph(ϕ + m+1)
and FM(graph(ϕ−m)) ⊂ graph(ϕ − m+1)
The graphs of the functions created in this way will be the desired manifolds. So we have the following:
2. We will show that the constructed Lipschitz continuous functions ϕ±m are C1.
Furthermore we will show that ϕ+m is complex differentiable in the points p ∈ Bδ∩Wm+0 such that there is an m0-history ˆp of p so that pm ∈ Bδfor all m < m0,
and φ−m is complex differentiable in the points p ∈ Bδ∩ Wm−0 such that there is
an m0-history ˆp of p so that pm ∈ Bδ for all m > m0
The observant reader will find that proofs of the fact that the functions ϕ±m are C∞ as well of (v) are missing. That is because we decided to omit them out of esthetic reasons, the proofs can be found on the pages 60 - 68 of [5].
2.2.1 Step 1 of the proof
For k ≤ n, let Cγ0(Ck) be the set of Lipschitz-continuous functions ψ : Ck → Cn−k such that:
kψ(x) − ψ(y)k ≤ γkx − yk, ψ(0) = 0. On this set we can define a metric:
d(ψ, φ) := sup
x∈Ck\{0}
kψ(x) − φ(x)k
kxk .
The proof that this is a metric we leave to the reader. Claim 2.5. C0
γ(Ck) is with this metric a complete metric space.
Proof. Suppose we have a cauchy-sequence (ψn)n∈N. For every x ∈ Ck, (ψn(x))n∈Nis
also a cauchy-sequence. So there is a unique y ∈ Cn−k such that limn→∞ψn(x) = y.
Now we define the function
ϕ : Ck → Cn−k : x → lim
n→∞ψn(x).
Since ψm(0) = 0 for all m ∈ N, ϕ(0) = 0.
Let x, y ∈ Ck, then:
So ϕ ∈ C0 γ(Ck).
We need to proof that limn→∞ψn= ϕ.
d(ϕ, ψm) = sup x∈Ck\{0} kψm(x) − ϕ(x)k kxk ≤ sup x∈Ck\{0} kψm(x) − ψn(x)k + kψn(x) − ϕ(x)k kxk for all n ∈ N.
Since (ψn)n∈N is a cauchy sequence, for every > 0 there is a N > 0 such that
n, m > N ⇒ d(ψn, ψm) < 2. If m > N we can choose now any n > N such that
d(ψn(x), ϕ(x)) < 2kxk and this implies that:
kψm(x) − ψn(x)k + kψn(x) − ϕ(x)k kxk ≤ 2 + 2 = Since this can be done for all x ∈ Ck\{0}
sup
x∈Ck\{0}
kψm(x) − ψn(x)k + kψn(x) − ϕ(x)k
kxk ≤
Claim 2.6. Suppose we have a linear map T : Cn→ Cn admitting a (λ, µ) splitting
with Cn = Ck⊕ Cn−k
, and a map F : Cn→ Cn fixing the origin. let the following be
given:
0 < γ and 0 < < µ − λ
γ + 2 + γ−1, (2.5)
kF − T k1.Cn < . (2.6)
If ϕ ∈ C0
γ(Ck), then there is a unique ψ such that
F (graph(ϕ)) = graph(ψ). Proof. We can write:
T = (A(x, y), B(x, y)) F = (A(x, y) + α(x, y), B(x, y) + β(x, u)). Then:
To show that this is again the graph of a function, we need to show that the function Gϕ : Ck→ Ck : x → A(x) + α(x, ϕ(x)) is a bijection. If x0 ∈ Ck then Gϕ(y) = x0 ⇐⇒ A−1(x0) = x + A−1(α(x, ϕ(x))). If ∀(x0)Fx0 : x → A −1(x
0) + A−1(α(x, ϕ(x))) is a contraction, then Gϕ is a bijection.
x1, x2 ∈ Ck (2.7)
kFx0(x1) − Fx0(x2)k = kA
−1
(α(x1, ϕ(x1)) − α(x2, ϕ(x2)))k (2.8)
≤ µ−1(1 + γ)k(x1− x2)k < kx1− x2k (2.9)
We define the following function:
ψ : Ck → Cn−k : x → B(ϕ(G−1 ϕ (x)) + β(G −1 ϕ (x), ϕ(G −1 ϕ (x)). For x, y ∈ Ck
kψ(x) − ψ(y)k ≤ λγkG−1ϕ (x) − G−1ϕ (y)k + (1 + γ)kG−1ϕ (x) − G−1ϕ (y)k ≤ (λ + (1 + γ))kG−1ϕ (x) − G−1ϕ (y)k and
kx − yk ≥ (µ − (1 + γ))kG−1ϕ (x) − G−1ϕ (y)k ⇒ kψ(x) − ψ(y)k ≤ λ + (1 + γ)
λ + (1 + γ)kx − yk ≤ γkx − yk.
In this way we can define the unstable graph transform:
F∗ : Cγ0(Ck) → Cγ0(Ck) : ϕ → the unique ψ such that F (graph(ϕ)) = graph(ψ).
(2.10) Claim 2.7. The function F∗ is a Contraction.
Proof. Let ϕ, ψ ∈ Cγ0(Ck) then d(ψ, ϕ) = supx∈Ck\{0}
kψ(x)−ϕ(x)k kxk ,for x ∈ C K kF∗(ϕ)(Gϕ(x)) − F∗(ψ)(Gϕ(x))k ≤ (λ + (1 + γ))kϕ(x) − ψ(x)k kGϕ(x)k ≥ (µ − (1 + γ))kxk ⇒ (since Gϕis a bijection) d(F∗(ϕ, F∗ψ) ≤ λ + (1 + γ) µ − (1 + γ)d(ϕ, ψ).
Now we’ll use the same type of argument to find a lipschitz function ψ : Cn−k → Ck
such that F (ψ) ⊂ ψ Claim 2.8. If ψ ∈ C0
γ(Cn−k) then there is a unique ϕ ∈ Cγ0(Cn−k) such that
F (graph(ϕ)) ⊂ graph(ψ)
Proof. The proof is similar to the previous proof, and can be found on the pages 48-51 of [5].
This gives us the stable graph transform:
F∗ : Cγ0(Cn−k) → Cγ0(Cn−k) : ψ → the unique ϕ such that F (graphϕ) ⊂ graphψ Claim 2.9. The stable graph transform F∗ is a contraction. By the following in-equality:
d(F∗(ψ), F∗(ϕ)) ≤ (λ + (1 + γ)
µ − (1 + γ)d(ψ, ϕ) Proof. See again page 52 of [5].
Let {Fm}m∈Z and {Tm}m∈Z be respectively a family of linear maps on Cn and a
family of maps on Cn. Suppose the following is given:
Cn = Ck⊕ Cn−k is a (λ, µ)-splitting of Tm
0 < γ , 0 < < µ − λ
γ + 2 + max(γ−1, γ)
kFm− Tmk1,Cn <
Definition 2.10. Consider the set: (C0
γ(Ck))Z. On this set we define the following
metric:
d(ϕ, ψ) = sup
m∈Z
d(ϕm, ψm) (2.11)
This is clearly again a complete metric space. On this space we can again define the unstable graph transform:
F∗ : (Cγ0(C
k))Z→ (C0 γ(C
k))Z, (2.12)
(F∗(ϕ))m = (Fm)∗(ϕm). (2.13)
We can alos define the stable graph transform: F∗ : (Cγ0(Cn−k))Z → (C0
γ(C n−k
))Z, (2.14)
(F∗(ϕ))m = (Fm)∗(ϕm+1). (2.15)
Lemma 2.11. For the family of linear maps {Tm} and maps {Fm} from the hadamard
perron theorem there is a unique family of lipschitz-continuous functions {ϕ+m}m∈Z and a unique family of lipzschitz-continuous functions {ϕ−m}m∈Z such that:
(i) ϕ±m : Em± → E∓ m
(ii) ϕ±m(0) = 0
(iii) Fm( graph(ϕ+m)) = graph(ϕ + m+1) , F (graph(ϕ − m)) ⊂ graphϕm+1 (iv) Fm|graph(ϕ+ m) is invertible (v) For x ∈ graph(ϕ−m) , kFm(x)k ≤ λ 0
kxk and for x ∈ graph(ϕ+
m) , kFm(x)k ≥
µ0(kx)k Where µ0 and λ0 are as in the Hadamard-Perron theorem. (vi) If Fm = F for all m ∈ Z , graph(ϕ±m) is independent of m.
Proof. We can find a family {Lm}m∈Z of orthogonal linear Isomorphisms such that:
Lm(Ck) = Em+
Lm(Cn−k) = Em−.
So then:
L−1m+1◦ Tm◦ Lm has Cn= Ck⊕ Cn−k as a (λ, µ)-splitting
So we can now assume that E+
m = Ck and that E −
m = Cn−k, and thus by using the
preceding paragraph. We can find a family {ϕ±m} of γ lipschitz-continuous functions that satisfy (i) , (ii), (iii) and (iv) immediatily. By the uniqueness of a fixed point of a contraction it also satisfies (vi). We need to proof (v). Let x = (x, ϕ+
m(x)) ∈ graph(ϕ+ m), then: kFm(x)k ≥ (µ − (1 + γ))kxk ≥ µ − (1 + γ) 1 + γ kxk = µ 0 kxk If x = (ϕ−m(x), x) ∈ graph(ϕ+ m), then: kFm(x)k ≤ (1 + γ)(λ + (1 + γ))kxk ≤ λ 0 kxk
2.2.2 Step 2 of the proof
We will show now that the functions ϕ±m are differentiable and to do that we use cones.
Definition 2.12. We define the Horizontal cone Hpγ and the vertical cone Vpγ as follows:
Hpγ := {(u, v) ∈ TpCn; kvk ≤ γkuk}, (2.16)
Vpγ := {(u, v) ∈ TpCn; kuk ≤ γkvk}. (2.17)
Claim 2.13. If Fm and Tm are maps satisfying the hypotheses of the
Hadamard-Perron theorem, such that Tm splits Cn as Ck⊕ Cn−k then:
(i) D(Fm)p(Hpγ) ⊂ Int(H γ F (p)) ∪ {0}, (ii) D(Fm)−1p (V γ F (p)) ⊂ Int(V γ p ) ∪ {0}.
Proof. Suppose (u, v) ∈ Hγ
p\{0} then kvk ≤ γkuk.
DFp(u, v) = (A(u) + α(u, v), B(v) + β(u, v)).
So
kA(u) + α(u, v)k ≥ µkuk − (1 + γ)kuk ⇒ kA(u) + α(u, v)k ≥ (µ − (1 + γ))kuk, and kB(v) + β(u, v)k ≤ λkvk + k(u, v)k ≤ λγkuk + k(u, v)k ⇒ kB(v) + β(u, v)k ≤ (λγ + (1 + γ))kuk. (i) follows from the fact that
λ + (1 + γ) µ − (1 + γ) < γ. The proof for (ii) is identical.
Claim 2.14. Vectors in horizontal cones are expanding and vectors in vertical cones are contracting, in the following sense:
For (u, v) ∈ Hγ p: k(DFm)p(u, v)k ≥ ( µ 1 + γ − )k(u, v)k. For (u, v) ∈ Vγ p : k(DFm)p(u, v)k ≤ (1 + γ)(λ + )k(u, v)k.
Proof. The proof is straightforward, and is therefore left out.
Now we’ll use these cones to create a family of invariant subspaces.
Claim 2.15. Suppose we have a family of (real) linear maps {Tm}m∈Z, Tm : Ck⊕
Cn−k → Ck⊕ Cn−k, such that for two fixed R+-sequences {γm}m∈Z and {γ
0 m}m∈Z and fixed λ0 < µ0: 1. Tm(Hγm) ⊂ Int(Hγm) ∪ {0}. 2. Tm−1(Vγm+1 ⊂ Int(Vγm) ∪ {0}. 3. (u, v) ∈ Hγm ⇒ kT m(u, v)k ≥ µ 0 k(u, v)k. 4. (u, v) ∈ L−1m−1(Vγm) ⇒ kT m−1(u, v)k ≤ λ 0 k(u, v)k. If we define: Em+ = ∞ \ n=1 (Tm−1◦ .. ◦ Tm−n)(Hγm−n), and Em−= ∞ \ n=1 (Tm−1◦ .. ◦ T−1 m+n)(V γm+n+1),
then the following holds: (i) {E+
m} is the unique family of 2k-dimensional (Real) spaces such that Tm(Em+) =
Em+.
(ii) {Em−} is the unique family of 2(n − k) dimensional (Real) spaces such that Tm−1(Em+1− ) = Em.
(iii) Cn= Em+⊕ E− m.
(iv) If {Sm} is a family of sets such that Sm ⊂ Hγm and Sm+1 ⊂ Tm(Sm) then
Sm ⊂ Em+.
(v) If {Km} is a family of sets such that Km ⊂ Vγm and Km ⊂ Tm−1(Bm+1) then
Km ⊂ Em−.
(vi) If the maps Tn are als complex linear for n ≤ m then Em+ is a complex subspaces
(vii) If the maps Tn are als complex linear for n ≥ m then Em− is a complex subspaces
of Cn.
Proof. Lm(Em+) ⊂ E +
m+1, since x ∈ Em+, x ∈ (Tm−1◦..◦Tm−n)(Hγm−n) for all n ∈ N ⇒
Tm(x) ∈ (Tm◦ .. ◦ Tm−n)(Hγm−n) for all n ∈ N ⇒ Tm(x) ∈ Em+1+ . In the same way
L−1m (Em+1− ) = Em−. We will proof (iii) first, therefore we first define the following: Aj = (Tm−1◦ .. ◦ Tm−j)(Ck⊕ {0}) and
Bj = (Tm−1◦ .. ◦ Tm−j)(Hγm−j),
then since Tm−1(Vγm+1) ⊂ Vγm, KernT
m ⊂ Vγm, so Aj is a 2k-dimensional real vector
space. For every Aj we can find a orthonormal basis {v1j, ..., v j
2k}. Because the
unit ball is a compact set, we can find a subsequence {aj} and a orthonormal basis
{v1, .., v2k} such that for every i limj→∞v aj
i = vi. Since Bj+1 ⊂ Bj and Bj is closed,
{v1, .., v2k} ⊂ Bj. If x ∈ Sp{v1, .., v2k}, then x = α1v1+ .. + α2kv2k = lim j→∞a1v aj 1 + ... + a2kv aj 2k
So x ∈ Bj for every j. And thus Sp{v1, .., v2k} ⊂ Em+, now we need to proof that
Sp{v1, .., v2k} = Em+. Suppose that x ∈ Em+ then since Sp{v1, .., v2k} is transverse to
{0} ⊕ Cn−k we have the following decomposition
x = (c, d) = (c0, d0) + (0, d00). Since (c, d), (c1, d1) ∈ Em+ there are cj, dj, c
0 j, d 0 j: Tm−1◦ .. ◦ Tm−j(cj, dj) = (c, d), Tm−1 ◦ .. ◦ Tm−j(c 0 j, d 0 j) = (c 0 , d0). From linearity it follows that:
Tm−1◦ .. ◦ Tm−j(cj − c 0 j, dj− d 0 j) = (0, d 00 ). Since (0, d00) ∈ Vγm k(0, d00)k ≤ (λ0)jk(cj− c 0 j, dj − d 0 j)k ≤ (λ 0 )j(k(cj, dj)k + k(c 0 j, d 0 j)k.
We also have that
k(c, d)k = kTm−1◦ .. ◦ Tm−j(cj, dj)k ≥ (µ 0 )jk(cj, dj)k and k(c0, d0)k = kTm−1◦ .. ◦ Tm−j(c 0 j, d 0 j)k ≥ (µ 0 )jk(c0j, d0j)k
so k(0, d00)k ≤ λ 0 µ0 j (k(cj, dj)k + k(c 0 j, d 0 j)k.
Since λ0 > µ0 this implies that k(0, d00)k = 0 and thus d00 = 0, so E+
m = Sp{v1, .., v2k}.
If for n ≤ m, Tm is complex linear, then every Aj is a complex linear subspace and
thus looking at the proof so is E+ m.
With a similar argument it can be shown that Em− is a real 2(n − k)-dimensional subspace, and a complex subspace if for n ≥ m, Tm is complex linear, thus:
Cn= Em+⊕ E − m. (Tm)|Em+ is injective, so Tm(E + m) = E +
m+1. Let {Sm} be a sequence of sets as in (iv).
then:
Sm ⊂ Tm−1(Sm−1) ⊂ Tm−1(Hγm−1).
Thus Sm ⊂ Em+. A similar argument proofs (v) and the uniqueness of (i) and (ii) is
now also implied by this.
Claim 2.16. We have under the assumptions of the previous claim that (u, v) ∈ Em−0 iff for all j ∈ N
kTm+j ◦ .. ◦ Tm(u, v)k ≤ (λ
0
)j+1k(u, v)k And (u, v) ∈ E+
m0 iff there is an history {(u−j, v−j)} with (u0, v0) = (u, v) and
Tm−j(uj, vj) = (uj−1, vj−1) such that for all j ∈ N
k(uj, vj)k ≤ (µ
0
)j+1k(u, v)k Proof. Again left to the reader.
Now using the cones we will finally show the differentiability of the lipschitz curves ϕ±m. First define for x 6= y
∆yϕ(x) =
(y, ϕ(y)) − (x, ϕ(x)) k(y, ϕ(y)) − (x, ϕ(x)). We also define the following set:
txϕ = {v ∈ T(x,ϕ(x))Cn; there is a sequence {yn} converging to x so that lim
n→∞∆ynϕ(x) → v}.
The following set is called the tangent cone of ϕ at x Txϕ =
[
v∈txϕ
and
T ϕ = G
x∈Ck⊕{0}
Txϕ
is called the tangent cone of ϕ. Claim 2.17. Let ϕ ∈ C0
γ(Ck), then:
(i) If π :→ Cn → Ck is the projection associated to the splitting Cn = Ck⊕ Cn−k
then π(Txϕ) = Ck for all x ∈ Ck.
(ii) We have Txϕ ⊂ H(x,ϕ(x))γ for all x ∈ Ck.
(iii) Txϕ is a 2k-dimensional real subspace iff ϕ is real differentiable in the point x.
(iv) Txϕ is a k-dimensional complex subspace iff ϕ is complex differentiable in the
point x.
(v) ϕ is C1 iff T ϕ is a continuous subbundle of T R2n of rank 2k.
(vi) If f : Cn→ Cn is a C1 map such that f (graphϕ) ⊂ graphψ then df (T ϕ) ⊂ T ψ.
If f (graphψ) =graphψ and Ker(df ) ∩ T ϕ = {0} then df (T ϕ) = T ψ. Proof. Let u ∈ Ck⊕{0} now consider the sequence {x+1
nu}, then since ∆x+n1uϕ(x) ∈
B(0, 1) which is compact, there is a subsequence {∆x+1 niuϕ(x)}i∈N such that: lim n→∞∆x+ 1 niuϕ(x) = v.
Since π is a continuous function and π(∆x+1
niuϕ(x)) ∈ Rv, π(v) ∈ Rv and from
lipschitz continuity it follows that kπ(v)k 6= 0. There is thus a c such that π(cv) = u. (ii) is clear from the lipschitz continuity of γ.
If A : Ck → Cn−k is a real linear function then we can write for a point x
0 ∈ Ck:
So:
ϕ(x) = ϕ(x0) + A(x − x0) + LA(x),
∆yϕ(x0) =
(y − x0, A(y − x0) + LA(y))
k(y − x0, ϕ(x0) + A(y − x0) + LA(y))k
= ( y−x0 ky−x0k, A( y−x0 ky−x0k) + LA(y) ky−x0k) k( y−x0 ky−x0k, A( y−x0 ky−x0k) + LA(y) ky−x0k)k
If {yn} is a sequence such that limn→∞∆ynϕ(x) = (u, v) then:
lim n→∞ y−x0 ky−x0k k( y−x0 ky−x0k, A( y−x0 ky−x0k) + LA(y) ky−x0k)k = u ⇒ lim n→∞k( y − x0 ky − x0k , A( y − x0 ky − x0k ) + LA(y) ky − x0k ))k = 1 kuk and thus lim n→∞ yn− x0 kyn− x0k = u kuk.
Here we use that ϕ is lipschitz continuous with constant γ and that thus
(1 + γ)−1 ≤ kuk ≤ 1 and thus kuk 6= 0 and the fact that for two converging sequences {an} , {bn} with limn→∞bn 6= 0 and bn6= 0
lim n→∞ an bn = limn→∞an limn→∞bn .
Suppose now that ϕ is real differentiable in the point x, then there is a linear function A : Ck→ Cn−k and a continuous function L
A: Ck→ Cn−k such that: ϕ(x + h) = ϕ(x) + A(h) + LA(h), lim h→0 LA(x + h) khk = 0.
Suppose now that (u, v) ∈ Txϕ then there is a sequence {xn} such that limn→∞∆xnϕ(x) =
c(u, v) and from the previous derived equations it now follows that: lim
Combined with (i) this implies that Txϕ is a 2k-dimensional real subspace. If Txϕ
is a 2k-dimensional real subspace then every (u, v) ∈ Txϕ is of the form (u, A(u)).
Now suppose that:
lim n→∞∆xnϕ(x) = limn→∞ ( xn−x kxn−xk, A( xn−x kxn−xk) + LA(xn) kxn−xk) k( xn−x kxn−xk, A( xn−x kxn−xk) + LA(xn) kxn−xk)k = (u, A(u)) ⇒ lim n→∞ LA(xn) kxn− xk = 0.
Suppose now towards contradiction that there is a sequence yn that converges to x
such that LA(yn)
kyn−xk > > 0 then there is a subsequence such that:
lim
n→∞∆ynϕ(x) = (b, A(b)).
but that is a contradiction. Thus lim
y→x
LA(y)
ky − xk = 0 So ϕ is real differentiable in the point x.
If ϕ x is complex differentiable in the point x, then A is a complex linear function and since Txϕ = {(y, A(y)); y ∈ CK ⊕ {0}} which is a k-dimensional complex linear
subspace. And if Txϕ is a complex linear subspace of Cn then it follows from that
A is a complex linear function. Thus in that case ϕ is complex differentiable at the point x.
(v) is obvious, so only (vi) is left to proof. Let F be a C1 map such that
F graph(ϕ) ⊂ graph(ψ) suppose that v ∈ Txϕ and v 6= 0 since if v = 0 then
dfx(0) = 0 ∈ TF (x)ψ. There is a sequence xn and a c > 0 which converges to x
such that limn→∞∆xnϕ(x) = cv. If pn = (xn, ϕ(xn)) then x 0
= π(F (pn)) is again a
sequence which converges (by continuity ofϕ) to π(F (p)). So there is a subsequence of x0 such that ∆x0
nψ(x
0
) converges. So we assume now that xn is so, such that
∆x0
nψ(x
0
converges. by the definition of the differential:
lim n→∞ f (xn, ϕ(xn)) − f (x, ϕ(x) k(xn, ϕ(xn)) − (x, ϕ(x))k = df(x,ϕ(x))(v) and: f (xn, ϕ(xn)) − f (x, ϕ(x) k(xn, ϕ(xn)) − (x, ϕ(x))k = kf (xn, ϕ(xn)) − f (x, ϕ(x)k k(xn, ϕ(xn)) − (x, ϕ(x))k ∆x0ψ(x 0 )
So if dfx,ϕ(x)(v) ∈ Tx0ψ and if the hyptheses of (vi) are satisfied then :
kf (xn, ϕ(xn)) − f (x, ϕ(x)k
k(xn, ϕ(xn)) − (x, ϕ(x))k
= kdf(x,ϕ(x))(v)k 6= 0.
If x0n is a sequence that converges to x0, then since F (graphϕ) = graphψ there is a sequence xn such that π(F (pn)) = x
0 n. If v = limn→∞∆xnϕ(x) dfx(v) = kf (xn, ϕ(xn)) − f (x, ϕ(x)k k(xn, ϕ(xn)) − (x, ϕ(x))k ∆x0ψ(x 0 )
so df(x,ϕ(x))(Txϕ) = TF (x,ϕ(x))ψ , a analogous result holds for maps ϕ ∈ Cγ0(Cn−k) only
then Txϕ ⊂ V(ϕ(x),x)γ
Now for the last part of the theorem:
Claim 2.18. Suppose we have a sequence of maps Fm : Cn → Cn, and a sequence
of linear maps Tm : Cn → Cn admitting a (λ, µ) splitting such that Em+ = Ck⊕ {0}
and Em− = {0} ⊕ Cn−k. And such that the following holds:
(a) 0 < γ <pµ/λ − 1. (b) 0 < < min{ , }. (c) kFm− Tmk1,Cn < .
(d) λ0 = (1 + γ)(λ + (1 + γ)). (e) µ0 = 1+γµ − .
Then we have the following: (i) the maps ϕ± are C1.
(ii) the sequence {d(Fm)0} admits a (λ
0
, µ0)-splitting with: Cn = E+(d(Fm)0) ⊕ E−(d(Fm)0).
Proof. We will proof it for ϕ+
m, the proof for ϕ −
m is similar. Let x ∈ Ck⊕ {0}, then
we know that p = (x, ϕ(x)) has a Z history ˆp such that pm−j ∈ graph(ϕm−j). Then
dFi(pi) is a sequence of linear maps satisfying the hypotheses of claim 2.15. There
is thus a sequence of 2k-dimensional subspaces E+ m. Tpiϕ
+
m which are a sequence of
can thus conclude by Claim 2.15(iv), Tpiϕ
+
m ⊂ Em+. Since Tpiϕ
+
m contains a
2k-dimensional real subspace
Tpiϕ
+ m = E
+ m.
Thus by Claim 2.17(iii), ϕ+m is differentiable in every point x ∈ Ck. We need to show that ϕ+
m is C1, this follows from a diagonal argument.
Suppose that {xl} is a sequence of points in Ck such that xl converges to x,
then pl = (xl, ϕ+m(xl)) is a sequence of points in graph(ϕ+m) that converges to p = (x, ϕ+
m(x)). For every point pl there is a unique history ˆpl with plm ∈ ϕ+m and since
(Fm)|ϕ+m is invertible, the function p → ˆp is continuous. We want to proof that
(Em+)p = lim(Em+)pl. Let vl ∈ (Em+)pl be sequence of points. From Claim 2.16 we
know that vl∈ (E+
m)pl iff there is a history {v−jl }j∈N such that v−jl ∈ Tpl
m−jC
n and
v−jl ≤ kµ0k−jkvlk.
If k0 is a subsequence of the natural numbers such that vk0(l)is a convergent sequence
in Cn, then vk0(l) is a bounded sequence. From the above inequality it follows that
vk0(l)
−1 also is a bounded sequence and has thus a convergent subsequence v k1(l)
−1 . If we
continue this way with induction, we can find a sequence of sequences: {{vkj(l)
−j }l∈N}j∈N
such that kj+1(l) is a subsequence of kj(l), and such that v kj(l) −j is a convergent se-quence. v1 v2 v3 v4 v5 v6 v7 ... v1 −1 v2−1 v−13 v−14 v−15 v−16 v7−1 ... v−21 v2−2 v−23 v−24 v−25 v−26 v7−2 ... v1 −3 v2−3 v−33 v−34 v−35 v−36 v7−3 ... vk0(1) vk0(2) vk0(3) vk0(4) vk0(5) vk0(6) vk0(7) ... → v vk1(1) −1 v k1(2) −1 v k1(3) −1 v k1(4) −1 ... → v−1 vk2(1) −2 v k2(2) −2 v k2(3) −2 ... → v−2 vk3(1) −3 ... → v−3
From continuity it follows that (dFm−j)(v−j) = v−j+1 and that v−j ≤ kµ
0 k−jkvk. So v ∈ (E+ m)p. Thus we have lim j→∞(E + m)xj ⊂ (E + m)x.
Since the real dimension of (E+
m)y = 2k for all points y ∈ Ck⊕ {0} we get:
lim j→∞(E + m)xj = (E + m)x.
Looking at the proof of the last claim, we see that if Em+(x) is a k-dimensional complex subspace of Cn then ϕ+ is complex differentiable in the point x and if
d(Fj)pj is a complex linear map for every j < m then we know from Claim 2.15 that
Em+(x) is a complex linear subspace. Looking at our original theorem, if pj ∈ U for
all j < m, ϕ+
m is complex differentiable in the point pm.
2.2.3 Extension to nearby points
Suppose al the hypotheses of the hadamard perron theorem are satisfied and that Fm = F for all m ∈ Z , for every point p ∈ Cn such that there is a history ˆp we can
define the sequence of maps: ˜
(Fm)p : Cn→ Cn: x → Fm(x + pm) − pm+1
This new sequence of maps satisfies again all the hypotheses of the Hadamard-perron theorem. So we apply all we know so far.
For this sequence of maps we have a sequence of functions ϕ+
m and ϕ−m. For a point
p with a history ˆp where p0 = p we define:
ϕ−p := ϕ−0 and ϕ+p := ϕ+0. Then ˜ (F0)p(graph(ϕ+p) = ϕ + F (p) and ˜ (F0)p(graph(ϕ−p) ⊂ ϕ − F (p).
We can also define the following maps
(ψp)+(x) := (ϕp)+(x − p) + F (p)
and
Then F (graph(ψp)+) = graphψF(p)+ and F (graph(ψp)−) ⊂ graphψF(p)−. For x ∈ graph(ψp)− d(F (x), F (p)) ≤ λ0d(x, p)
To construct the maps ϕ−p and to proof that they are differentiable,we only need the the maps ˜Fmp with m ≥ 0, to construct those maps we only need the pm with m ≥ 0.
To construct the sequence of maps ϕ+p and proof that they are differentiable we also need the negative history so we can do that only for the points with a history ˆp. Claim 2.19. The maps ϕ−p depends continuously on the point p. So the map
p → ϕ−p is continuous.
Proof. The fixed point of a contraction depends continuously on the contraction, and the sequence ˜Fm depends continuously on the point p
3 Skew products in C
2
3.1 Definitions
In this chapter we will introduce skew products. A skew product is a holomorphic function
F : U → C2 defined in neighborhood U of the origin such that:
F (t, x) = (p(t), f (t, x)).
Where p : P1(U ) → C and f : U → C are holomorphic functions. We are interested
in the case that
p(0) = 0 and |p0(0)| < 1. There are open neighborhoods A, B ⊂ C of 0 such that
A ⊕ B ⊂ U.
Since |p0(0)| < 1 there is a neighborhood V ⊂ A where all points converge to 0 under iteration from p. So for the function F the t-coordinate of all points in V ⊕ B ⊂ U converges to 0.
If the function F0 : C → C : x → F (0, x) is a polynomial the Fatou components of
the map F0 are very well understood, since polynomials are a special case of rational
functions and for rational functions the Fatou components are classified. Definition 3.1. If we have a skew product
F : C2 → C2 : (t, z) → (p(t), f (t, z))
with p(0) = 0. We say that a fatou component U ⊂ {0} ⊕ C of F0 bulges if there is
3.2 Bulging of Fatou components
First we will proof in theorem 3.3 that for specific functions the parabolic fatou component bulges, then we will proof it in theorem 3.4 using the same techniques for a more general class of functions.
Theorem 3.2. Suppose we have the following function: G : C2 → C2 : (t, z) →
(λt, z + z2+ ct) with |λ| < 1. Then
G0 : C → C : z → z + z2
has one attraction vector.
v0 = −1.
There is an open set BUv0 ⊂ C
2 such that:
Bv1 ⊂ BUv0
for z ∈ BUv0 lim
n→∞nzn= (0, −1)
And this convergence in locally uniformly. Proof. Lets consider the following sets:
C = {(t, z) ∈ C2; |z| < 1/4, |t| < 1 8|c||z|
2, z 6= 0}
∆0 = {(t, z) ∈ C2; z 6= 0}
Then on ∆0 we can define biholomorphic function
γ : B → B : (t, z) → (t, 1/z) So
γ−1 : γ(C) → C is holomorphic. For (t, z) ∈ C
z + z2+ ct 6= 0. So we can define the holomorphic function
γ ◦ G ◦ γ−1 : γ−1(C) → C2 : (t, u) → (λt, u
1 + (1/u + ctu)) (3.1) u
1 + (1/u + ctu) = u(1 − (1/u + ctu) + (1/u + ctu)
2− (1/u + ctu)3... (3.2)
And since |z| < 1/4 and |t| < 8c1|z|2
|u(ctu + (1/u + ctu)2....)| < |1/4| + (1 + 1/8)21
8 + (1 + 1/8)
3 1
82.... < |1/2|
Now we define the following set:
IB0 := {(t, u) ∈ γ−1(C); Re(u) < 8|c|} And clearly γ ◦ G ◦ γ−1(IB0) ⊂ IB0. Since if: (t1, u1) := γ ◦ G ◦ γ−1(t, u). Then for (t, u) ∈ H Re(u) − 11 2 < Re(u1) < Re(u) − 1/2. Since |t1| = |λ||t| < λ 8|c||u|2 ≤ 1 8|c|(|u| + 112)2. we can conclude: γ ◦ F ◦ γ−1(IB0) ⊂ IB0
So the orbit of every element in γ−1(IB0) converges to (0, 0) under G. This
conver-gence is uniformly. Just as the converconver-gence of lim n→∞nzn= (0, −1) Clearly Bv0 ⊂ BUv0 = ∞ [ n=1 G−n(IB0)
Corollary 3.3. If we consider the same function G : C2 → C2 All the Fatou
Now we will take a look at the general case, using the same techniques as with the previous proof.
Let U ⊂ C2 be an open neighborhood of 0. Let
F : U → C2 : (t, z) → (p(t), ψ(t, z)) be a skew product. With:
p(0) = 0, p0(0) = λ , |λ| < 1,
There is an open neighborhood V of 0, such that ψ can be written as a power series in V . So ψ(t, z) = α0(t) + α1(t)z1+ .... With: α0(0) = 0, α00(0) = c, α1(0) = 1, α2(0) = 0, ..., αk+1(0) = a
U0 = {(t, u) ∈ U |t = 0} Then ψ|U0 has k attraction vectors vj, and for each attraction
vector an open set Bvj ⊂ U0 .
Theorem 3.4. For every set Bvj there is a open set BUvj ⊂ U such that:
Bvj ⊂ BUvj.
Furthermore is for every element of z ∈ BUvj the orbit of z well defined and
lim
n→∞
k
√
nzn= (0, vj)
and this convergence is locally uniformly.
Proof. Let |p0(0)| < µ < 1 Then there is a δ1 > 0 such that |t| < δ1 ⇒ |p(t)| < |µt|.
And there is a β1 > 0 such that z < β1 ⇒ |ψ(t, z)| > 12|z|
α0(t) + α1(t)z + α2(t)z2... + αd(t)zd+ .. = (3.4)
α0(t) + α1(t)z + α2(t)z2... + αd(t)zd+ O(zk+1) (3.5)
there is a δ2 > 0 such that |t| < δ2 ⇒ |ψ(t, z) − α0(t)| > 14|z| this follows at once
from 3.3 and the fact that α1(0) = 1 and α0(0) = 0. And let δ3 > 0 such that
|t| < δ3 ⇒ |α0(t)| < 2(|c| + 1)|t| C := {(t, z) ∈ U ||z| < β1, |t| < M in{δ1, δ2, δ3}, |t| < |z| 4(|c| + 1)} (3.7) Claim 3.5. : If (t, z) ∈ C then ψ(t, z) 6= 0 Proof. : |ψ(t, z) − α0(t)| > |z| 4 And |α0| < |z| 4 . So: |ψ(t, z)| = |ψ(t, z) − α0(t) + α0(t)| > |(|ψ(t, z) − α0(t)| − |α0(t)|)| > 0
Let vj now be an attracting vector for ψ|U0 Let ∆j be the sector containing all
(t, z) ∈ C such that z = reiθv
j with r > 0 and |θ| < π/k let
γ : C ⊕ C∗ → C2 : (t, z) → (t, −1 kazk) γ|∆j : ∆j → γ(∆j) is biholomorphic. So is γ|∆j∩C : ∆j∩ C → γ(∆j∩ C). So let γj−1 : γ(∆j∩ C) → ∆j ∩ C. Now let Fj = γ ◦ F ◦ γj−1.
That this is a well defined function follows from the claim, since F ◦ γj−1(t, ω) ∈ C ⊕ C∗.
Now we will do some derivations: (t, ω) → (p(t), 1 γ(p(t), ψ(t, γ−1(t, ω)) = (p(t), c ψ(t,pc/ω)k k) c ψ(t,pc/ω)k k = c O(t) + α1(t)pc/ω + αk k+1pc/ωk k+1 + O(pc/ωk k+2 ) = ω (1 + O(t) + O(t)/pc/ω + αk k+1(t)c/ω + o(1/ω))k = ω
1 + kac/ω + o(1/ω) + O(tk)ω/c + O(t)
= ω
∞
X
n=0
(−1)n(kac/ω + o(1/ω) + O(tk)ω/c + O(t))
= ω − kac − O(t)ω − o(1) Now suppose that |t| < |1/ω2| then:
= ω − kac − O(t)ω − o(1) = ω + 1 − O(1/ω2)ω − o(1) = ω + 1 − o(1). (3.8) So there is a β2 > √2 3/21/λ−1 such that if |ω| > β2 and we write (t1, ω1) = G(t, ω) Then
Re(ω1) > Re(ω) + 1/2.
We now define the following set:
IBj := {(t, ω) ∈ γ−1(∆j ∩ C); |ω| >, Re(ω) > 2β2 − |Im(ω)|, |t| < |1/ω2|} (3.9) Claim 3.6. : Fj(IBj) ⊂ IBj Proof. : (t, ω) ∈ IBj ⇒ Re(ω1) > 2β2− |Im(ω1)| and |t1| ≤ λ|t| ≤ | 1 |ω| + 3/2| 2 If |ω| > 3/2 2 p1/λ − 1 and | 1 |ω| + 3/2| 2 ≤ 1 |ω1|2
So:
(t1, ω1) ∈ IBj
And thus for all (t, ω) ∈ IBj
lim n→∞F ◦n j (t, ω) = (0, ∞) So for (t, z) ∈ Hvj := γ −1 j (IBj) lim n→∞F ◦n (t, z) = (0, 0)
Further we have that
lim
k→∞
m
√
kµk|t| = 0
And by the same argument as in the Fatou flower theorem in chapter 2.2.3 we have that: lim k→∞ m √ kzk = vj So for (t, z) ∈ Hvj = γ −1(IB j) lim k→∞ m √ k(tk, zk) = (0, vj) And Bvj ⊂ BUvj := ∞ [ n=1 F−n(γ−1(IBj))
That the convergence on BUvj is locally uniformly follows from the fact that the
convergence on Hvj is uniformly.
Now we will give another proof of theorem 4.6, using the Hadamard-Perron theo-rem.
Proof. First we will calculate the Jacobian of F in the point zero.
J = δϕ(t) δt (0) 0 δψ(t,z) δt (0, 0) δψ(t,z) δz (0, 0) ! = λ 0 α00(0) α1(0) =µ 0 c 1
E^+
E^-Figure 3.1: The subspaces E+ and E−
This matrix admits a linear (|λ|, 1)-splitting with E+ = {0} ⊕ C
E− = the complex subspace space spanned by the vector (1 − µ, −c)
Now by the construction described in the stable manifolds section there is a δ > 0 and a function F0 : C2 → C2 such that
F0(x) = F (x) for kxk < δ F0 is C∞ and satisfies all the axioms of the Hadamard-Perron theorem. Let
A = C ⊕ {0} ∩ B(0, δ).
and let Bj ⊂ A be an attracting petal for the vector vj in which the orbits {zn}
of points z converge uniformly to 0 and for which the convergence of √knz
n to vj
is also uniformly. Then for every point x ∈ Bj there is a well defined stable graph
ϕ−x : E−→ E+. This way we can define the following set.
Sj =
[
x∈Bj
graph(ϕ−x) ∩ B(0, δ)
Then for every element (t, z) ∈ Sj, we have that there is an (0, z0) such that
(t, z) ∈ graphϕ−z0. And thus: d(F◦n(0, z0), F◦n(t, z)) < |λ 0 |nd((0, z 0), (t, z)) ≤ |λ 0 |n2δ.
E^-U
E^+
Figure 3.2: The set Sj
Now let > 0 then there is an N1 such that for n > N1 for every point (t, z)
d(F◦n(0, z0), F◦n(t, z)) <
2.
Furthermore since the convergence in the petal to 0 is uniformly, there is an N2 such
that for n > N2 d(Fn(0, z0), (0, 0)) < 2. ⇒ d(Fn(t, z), (0, 0)) < and d(√knFn(t, z), (0, v j)) < .
So all elements in Sj converge to (0, 0) from the direction (0, vj). If we show that Sj
contains an open subset V that contains Bj, then we can take:
BUvj :=
∞
[
n=1
F−n(V ),
and then we proved the theorem. We define this set in the following way: Vj = {z = (x, y + x −c 1 − µ) ∈ B(0, δ); d(y, B c j) > γkxk} Claim 3.7. Vj ⊂ Sj
E^-V
E^+
Figure 3.3: The set Vj
E^-E^+
gx
Figure 3.4: The function gx
Proof. From Claim 2.19 it follows that the function gx : Bj → C : y → ϕ−y(x). is
continuous.
Let z = (z1, z2) ∈ Vj so d(z2− z11−µ−c , Bjc) > βkz1k > γ
0
kz1k. Since the the
func-tions ϕ− are lipschitz continuous with constant γ0. We have for y ∈ Bj
gx(y) ∈ B(y + x −c 1 − µ, γkxk) (see figure 4.5) We have B(z1− z2 −c 1 − µ, βkxk) ⊂ Bj and C(z1− z2 −c 1 − µ, βkxk) ⊂ Bj. So gz1(C(z2− z1 −c
y+x (c/(1- )
y ||x||
Figure 3.5: The subspaces E+ and E−
Figure 3.6: gz1 sends C(z1− z2
−c
1−µ, βkxk) to a loop around z2
Figure 3.7: The homotopy between gz1(B(z1− z2
−c
1−µ, βkxk)) and S 1.
But if y /∈ Vγ this induces a non trivial homomorphism from π1(D) to π1(S1) but
that does not exist since π1(D) is trivial. So we have a contradiction.
As said before the theorem follows at once from this claim.
3.2.1 Attraction to (0,0) through parabolic fatou component
Looking at the proof at the last proof, we see that F admits a (λ, 1)-splitting in the point (0, 0). So for the function F0 there is by the Hadamard-Perron theorem a function:ϕ− : E− → E+
Such that if x ∈ graphϕ−then there is a λ < v < 1 such that d( F0◦n
(x), (0, 0)) < vnkxk.
So for the neighborhood U where F|U = F
0
|U we have that if
x ∈ WUs := U ∩ graphϕ− d(F◦n(x), (0, 0)) < vnkxk.
Theorem 3.8. : Let x ∈ U such that limn→∞F◦n(x) = (0, 0) and F◦m(x) 6∈ WUs.
Then there is a j ∈ {0, .., k − 1} such that the forward orbit of x ends up in U Bvj
eventually.
Proof. We will look at the Jacobian for F in the point (0, 0) again: and µ = ϕ0(0) 6= 0
J = δϕ(t) δt (0) 0 δψ(t,z) δt (0, 0) δψ(t,z) δz (0, 0) ! = µ 0 α00(0) α1(0) (3.10)
The determinant is unequal to 0 so from the inverse function theorem it follows that this function is invertible in a a neigborhood of 0, and that the inverse function is again holomorphic. Furthermore we know that the first component of the inverse function is only depended on the first variable and thus F−1 = (γ(t), χ(t, z)) and γ = ϕ−1. If we now look at the holomorphic function:
G = (ϕ(t), z)◦2◦ F−1 = (ϕ(t), χ(t, z)). Then:
Let vj be the attracting vectors for ψ0, and uj the attracting vectors for χ0. Then
by the proof of theorem 3.6. and the Fatou flower theorem there are sets Hvj and
Huj such that if we have the set
V = {(t, z) ∈ C2; z < , |t| < |z2|}
There is an > 0 such that:
F (Hvj) ⊂ Hvj
G(Huj) ⊂ Huj
V ⊂ Hv0 ∪ .. ∪ Hvk−1 ∪ Hu0 ∪ .. ∪ Huk−1
Claim 3.9. If limn→∞F◦n(x) = limn→∞(tn, zn) = (0, 0) and F◦n(x) /∈ WUs. Then
for every N ∈ R>0 there is an n > N such that: |tn| < |zn|2
Proof. Suppose towards contradiction that there is an N ∈ N such that n > N ⇒ |tn| > |zn|2. Since for n great enough |tn+1| < λ
0
|tn|, there is an C > 0 such that
|tn| < C|λ 0 |n|t|. Since |z n| <p|t2 n|, there is a λ 0 < v < 1 such that: |(tn, zn)| < Cvn|(t, z)|
But from (v) of the Hadamard-Perron theorem it then follows that there is an m such that F◦m(t, z) ∈ Ws
U, which is a contradiction.
Since (tn, zn) converges to (0, 0) there is an N > 0 such that for m > N
|(tm, zm)| < .
So from the claim it follows that for m > N (tm, zm) ∈ V.
Suppose towards a contradiction that (tn, zn) does not converge from any direction
vj, then (tn, zn) /∈ Hvj and thus for n > N there is an j ∈ {0, .., k − 1} such that:
(tn, zn) ∈ Huj.
But then there for every n > N there is a δn > 0 such that
|zn+1| > |zn| + δn.
4 Non wandering Fatou
components in C
2
Suppose we have a skew product
F : C2 → C2 : (t, z) → (p(t), ψ(t, z)),
where p is a polynomial with p(0) = 0 and |p0(0)| < 1 and ψ(t, z) = αd(t)zd+ ...α0(t),
with αi a holomorphic function, d > 1 and αd(0) 6= 0. If q is parabolic fixed point
of F0 : C → C : z → ψ(0, z) then we proved in the previous chapter that for every
attraction vector vj all the fatou components in the set Bvj bulge.
Since F0 is a polynomial with deg(f ) ≥ 2 we know from sullivans non wandering
theorem that for every Fatou component U of F0, there are distinct n, m such that
Fn
0(U ) = F0m(U ).
So if U is a Fatou component for F0, there are n, m such that Fn(U ) gets mapped
onto itself by Fm. Then by the classification theorem Fn(U ) is one of the following: (i) Fn(U ) is the immediate basin for an attracting fixed point.
(ii) Fn(U ) is the immediate basin for one attraction vector of a parabolic fixed
point.
(iii) Fn(U ) is a Siegel disc.
(iv) Fn(U ) is a Herman ring.
Now it is proven in [?] that in the cases (i), (iii) and (iv) the Fatou set for Fm
contains a component V such that Fn(U ) ⊂ V . It is also proven there that if
p0(0) = 0 and case (ii) holds , the Fatou set for Fm contains a component V such that Fn(U ) ⊂ V . From theorem 3.4. in chapter three we can conclude that in the
Since by theorem 1.13 .
F(F ) = F(Fm)
we have that V is a Fatou component for F and Fn(U ) thus bulges. The fatou
component U then also bulges since
U ⊂ F−n(V ) and
F−n(V ) ⊂ F(F ). So all the Fatou components for F0 bulge.
It is now an interesting question whether for all fatou component U of F for which P rojt(U ) ⊂ A
where A is the set such that for z ∈ U, limn→∞pn(x) = 0, U gets mapped into on of
those fattened Fatou components eventually. Since if that would be the case, there would be no wandering Fatou components in a neighborhood of the 0-fiber, because the fattened fatou components are pre periodic by Sullivan’s theorem.
Let U be a Fatou component with U ⊂ A {F|U◦n}
is a normal family. So there is a function G : U → C2 and a subsequence such that
lim
i→∞F ◦ni
|U = G.
Theorem 4.1. If U never ends up in one of the fattened fatou components, i.e. for every n ∈ N and every fattened Fatou component V.
Fn(U ) 6⊂ V Then there is a c ∈ J(F0) ⊂ C such that:
Proof. Clearly G(U ) ⊂ C0. Suppose towards a contradiciton that G is non constant,
then G(U ) is an open subset of C0. Since we know that the Julia set has an empty
interior. G(U ) has a non empty intersection with the Fatou set of F0. There is thus
a Fatou component V of F0 such that G(U ) ∩ V 6= ∅. Let b ∈ G(U ) ∩ V , then there
is an > 0 such that B(b, ) is in the fattened Fatou component of V . There is an z ∈ U such that G(z) = b and so there has to be an n > 0 such that Fn(z) ∈ B(b, ),
which implies that Fn(U ) is a subset of the fattened Fatou component of V. So we
have our contradiction, and thus is G a constant function. By the same argument the element c cannot be in the Fatou component, so c ∈ J(F0).
We will now make a distinction between p0(0) = 0 and 0 < |p0(0)| < 1. In the case that p0(0) = 0 there is the following theorem due to Lilov [3].
Theorem 4.2. (Lilov,2003). Let t1 ∈ A, and let D be an open one-dimensional disk
lying in the t1-fiber. Then the forward orbit of D must intersect one of the fattened
Fatou components of F0.
Since every Fatou component is open the following is an immediate consequence. Corollary 4.3. If U is a Fatou component for F such that P rojt(U ) ⊂ A then there
is an n and a fattened Fatou component V such that Fn(U ) ⊂ V
In the case that 0 < p0(0) < 1 theorem 4.2 does not hold, a counterexample is given in [4]. The counterexample given in [4] is not a counterexample against corollary 4.3, and t is still an open question whether in this case all Fatou components U with P rojt(U ) ⊂ A end up eventually in one of the fattened Fatou components.
