Complex Dynamics II Walter Bergweiler July 19, 2016 Lecture at CAU Kiel Summer term 2016 Version of July 19, 2016 This lecture is a continuation of the lecture “Complex dynamics I” from the winter term 2015/16. While that lecture covered the basic results of the theory, which can be found in standard textbooks on the subject such as [7, 14, 26, 36], and which are largely due to Fatou and Julia already, the present lecture is devoted to more specialised (and also more recent) aspects of the theory. Contents 1 Iteration of entire functions 3 1.1 The escaping set ................................ 3 1.2 Curves in the escaping set ........................... 11 1.3 The Eremenko-Lyubich class .......................... 18 1.4 Multiply connected Fatou components .................... 24 2 Hausdorff dimension of Julia sets and escaping sets 41 2.1 Hausdorff measure and Hausdorff dimension ................. 41 2.2 Julia sets of rational functions ......................... 49 2.3 Julia sets and escaping sets of entire functions ................ 51 1
Transcript
Complex Dynamics II
Walter Bergweiler
July 19, 2016
Lecture at CAU KielSummer term 2016
Version of July 19, 2016
This lecture is a continuation of the lecture “Complex dynamics I” from the winterterm 2015/16. While that lecture covered the basic results of the theory, which can befound in standard textbooks on the subject such as [7, 14, 26, 36], and which are largelydue to Fatou and Julia already, the present lecture is devoted to more specialised (andalso more recent) aspects of the theory.
2 Hausdorff dimension of Julia sets and escaping sets 412.1 Hausdorff measure and Hausdorff dimension . . . . . . . . . . . . . . . . . 412.2 Julia sets of rational functions . . . . . . . . . . . . . . . . . . . . . . . . . 492.3 Julia sets and escaping sets of entire functions . . . . . . . . . . . . . . . . 51
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1 Iteration of entire functions
General references for the iteration theory of entire functions are the articles [8, 33] andthe lecture notes [12].
1.1 The escaping set
We recall that the Fatou set F (f) of a rational or entire function f is the set where theiterates fn form a normal family (in the sense of Montel). The Julia set J(f) is the
complement of the Fatou set with respect to the Riemann sphere C (if f is rational) orthe complex plane C (if f is entire). Thus for entire f we have J(f) = C\F (f). Asapparent from part 1 of this lecture in the previous semester, these sets are the mainobjects studied in complex dynamics.
For the iteration theory of entire functions there is a further set that will turn out tobe important.
Definition 1.1. Let f be an entire functions. Then
I(f) = z ∈ C : fn(z)→∞ as n→∞
is called the escaping set of f .
For polynomials the escaping set is nothing else than the basin of attraction of thesuperattracting fixed point ∞.
Theorem 1.1. Let f be a transcendental entire function. Then I(f) 6= ∅.
The proof requires a number of lemmas. Here and in the following we use the notationD(a, r) = z ∈ C : |z − a| < r and D = D(0, 1).
Lemma 1.1. (Bohr’s theorem) For all ρ ∈ (0, 1) there exists a positive constant c suchthat if h : D→ C is holomorphic with h(0) = 0 and
max|z|=ρ|h(z)| ≥ 1,
then ∂D(0, R) ⊂ h(D) for some R ≥ c.
Proof. Suppose that such a constant c does not exist for some ρ ∈ (0, 1). Then thereexists a sequence (hn) of holomorphic functions hn : D→ C with hn(0) = 0 and
max|z|=ρ|hn(z)| ≥ 1
such that
cn = supR > 0: ∂D(0, R) ⊂ hn(D)
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satisfies cn → 0 as n → ∞. Without loss of generality we may thus assume that cn < 1for all n. Thus there exists an, bn ∈ C with |an| = 1 and |bn| = 2 such that hn(z) 6= anand hn(z) 6= bn for all z ∈ D. The functions gn : D→ C defined by
gn(z) =h(z)− anbn − an
thus satisfy gn(z) 6= 0 and gn(z) 6= 1 for all z ∈ D. By Montel’s theorem, the gnform a normal family. Passing to a subsequence if necessary, we may thus assume that(gn) converges. Passing to further subsequences we may also assume that (an) and (bn)converge. Thus hn → h for some holomorphic function h : D→ C with h(0) = 0 and
max|z|=ρ|h(z)| ≥ 1.
In particular, h is non-constant. This implies that h(D) is open and thus there existsR > 0 such that D(0, R) ⊂ h(D). Hurwitz’s theorem yields that D(0, R/2) ⊂ hn(D) forall large n, contradicting cn → 0 as n→∞.
Remark 1.1. Let cρ be the supremum of the set of all c that satisfy the conclusion ofLemma 1.1. One can show that
cρ =(1− ρ)2
4ρ.
For
h(z) = 4cρz
(1− z)2
we have
max|z|=ρ|h(z)| = 1
and ∂D(0, R) ⊂ h(D) for 0 < R < cρ, but ∂D(0, R) 6⊂ h(D) for R ≥ cρ.The function k given by
(1.1) k(z) =z
(1− z)2
is called the Koebe function. It maps D bijectively onto C\(−∞,−14]. It is extremal for
various problems in function theory.
For a bounded domain G in C we denote by U(G) the unbounded component of C\G.We call T (G) = C\U(G) the topological hull of G. Thus T (G) is the union of G andthe bounded components of its complement. Informally, T (G) is obtained from G by“filling the holes”; cf. Figure 1. A (bounded) domain G is simply connected if and only if
T (G) = G. In the literature, the notation G is often used instead of T (G).
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G T (G)
Figure 1: A bounded domain G and its topological hull T (G).
For each bounded component K of C\G there exists a (piecewise smooth) simpleclosed curve γ in G such that K is contained in the interior int(γ) of γ. Here the interioris defined by
int(γ) = z ∈ C : n(γ, z) 6= 0,
with the winding number
n(γ, z) =1
2πi
∫γ
dζ
ζ − z.
Thus
T (G) =⋃γ
int(γ),
where the union is taken over all (piecewise smooth) closed curves γ in G.Finally, we put
µ(G) = min|z| : z ∈ U(G).
Then, for µ(G) > 0,
µ(G) = supR > 0: D(0, R) ∈ T (G).
Lemma 1.2. Let f be entire, r > 0 and 0 < ρ < 1. Let c > 0 be as in Lemma 1.1. Then
(1.2) µ(f(D(0, r)) ≥ c max|z|=ρr
|f(z)| − (1 + c)|f(0)|.
Proof. We may assume that the right hand side of (1.2) is positive since otherwise theconclusion is trivial. The function h given by
h(z) =f(rz)− f(0)
max|z|=ρr |f(z)− f(0)|
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satisfies the hypotheses – and thus the conclusion – of Lemma 1.1. Hence there existsR ≥ c such that ∂D(0, R) ⊂ h(D). This implies that
∂D
(f(0), R max
|z|=ρr|f(z)− f(0)|
)⊂ f(D(0, r))
and hence
T (f(D(0, r)) ⊃ D
(f(0), R max
|z|=ρr|f(z)− f(0)|
)⊃ D
(0, c max
|z|=ρr|f(z)− f(0)| − |f(0)|
)⊃ D
(0, c max
|z|=ρr|f(z)| − (1 + c)|f(0)|
)from which the conclusion follows.
Lemma 1.3. Let f be entire (and non-constant) and let G be a bounded domain in C.Then
(1.3) f(T (G)) ⊂ T (f(G))
and
(1.4) ∂T (f(G)) ⊂ f(∂T (G)).
Proof. For a (piecewise smooth) closed curve γ in G and a ∈ C with f(a) ∈ C\ tr(f γ),where tr(·) denotes the trace of a curve, we have
n(f γ, f(a)) =1
2πi
∫fγ
dw
w − f(a)=
1
2πi
∫γ
f ′(ζ)dζ
f(ζ)− f(a)=
∑z∈int(γ)∩f−1(f(a))
n(γ, z)mz,
where mz denotes the multiplicity of an f(a)-point z. For a simple closed curve γ anda ∈ int(γ) this yields f(a) ∈ int(f γ) and hence
f(int(γ)) ⊂ int f(γ),
form which the first claim (1.3) follows.Next we show that
(1.5) ∂f(G) ⊂ f(∂G).
In order to do so, let w0 ∈ ∂f(G). Then there exists a sequence (wk) in f(G) such thatwk → w0. We have wk = f(zk) for some zk ∈ G and without loss of generality we mayassume that zk → z0 for some z0 ∈ G. Thus w0 = f(z0). If z0 ∈ G, then w0 = f(z0) ∈ Gand thus w0 /∈ ∂f(G) since f(G) is open. Hence z0 ∈ ∂G and thus w0 ∈ f(∂G). Thisproves (1.5).
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It follows from (1.5) that
(1.6) ∂T (f(G)) ⊂ f(∂G).
For a bounded component A of C\G we have A ⊂ T (G) and hence
f(∂A) ⊂ f(A) ⊂ f(T (G)) ⊂ T (f(G))
by the inclusion (1.3) already proved. Thus
f(∂G\∂T (G)) ⊂ T (f(G)).
and hence
(1.7) f(∂G\∂T (G)) ∩ ∂T (f(G)) = ∅.
Now (1.4) follows from (1.6) and (1.7).
Proof of Theorem 1.1. Since f is transcendental, we have
M(r, f) := max|z|=r|f(z)| ≥ r2
for large r. This follows from the Cauchy inequalities saying that if f(z) =∑∞
k=1 akzk,
then |ak|rk ≤M(r, f) for all k ≥ 0 and all r > 0. It follows that
for large r, say for r ≥ r0. For a bounded domain G satisfying µ(G) ≥ r0 we thus find,since T (f(G)) ⊂ f(T (G)) ⊃ D(0, µ(G)) by Lemma 1.3 and the definition of µ(G), that
For z ∈ A and n ≥ 0 we have fn(z) ∈ ∂T (Gn) and thus |fn(z)| ≥ 2nr0. Hence fn(z)→∞,so A ⊂ I(f).
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Remark 1.2. Theorem 1.1 is due to Eremenko [17], the above proof is due to Domın-guez [16].
Let Gn be as in the above proof and z ∈ A ⊂ I(f) as there. Then
fn(z) ∈ ∂T (Gn) ⊂ U(Gn).
Let Bn be the component of f−n(U(Gn)) containing z and put Cn = Bn ∪∞. Then Cnis a compact, connected subset of C. Moreover, Cn ⊂ Cn−1 for n ≥ 1. Thus
C :=∞⋂n=0
Cn 6= ∅
is a compact, connected subset of C. Moreover, z ∈ C, ∞ ∈ C and C\∞ ⊂ I(f).It seems intuitively clear that the component of C\∞ which contains z is unbounded.
The following lemma, which is left as an exercise in [29, p. 84], implies that this is indeedthe case.
Lemma 1.4. Let X be compact, connected metric space, let Y be a compact subset of Xand let A be a component of X\Y . Then the closure of A intersects Y .
Applying this with X = C and Y = ∞ then yields the following result.
Theorem 1.2. Let f be a transcendental entire function. Then I(f) has an unboundedconnected component.
Remark 1.3. Theorem 1.2 is due to Rippon and Stallard [31]. A question of Ere-menko [17], whether all components of I(f) are unbounded, is still open.
One consequence of Theorem 1.1 is a comparatively easy proof that the Julia set J(f)of a transcendental entire function f is not empty.
Theorem 1.3. Let f be a transcendental entire function. Then J(f) 6= ∅.
Proof. An application of Picard’s theorem to the function
h(z) =f(f(z))− zf(z)− z
shows that f 2 has a fixed point z1. (This was an exercise in part 1 of this course.) Onthe other hand, there exists z2 ∈ I(f) by Theorem 1.1. Since z1 and z2 cannot be in thesame component of F (f), the conclusion follows.
More generally, we have the following result.
Theorem 1.4. Let f be a transcendental entire function. Then card J(f) =∞.
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Proof. Clearly I(f) 6= ∅ implies that card I(f) = ∞. The conclusion follows as in theprevious proof if we show that card(C\I(f)) =∞. This is clear if f 2 has infinitely manyfixed points. Thus suppose that f 2 has only finitely many fixed points. On the otherhand, as noted in the previous proof, f 2 has at least one fixed point. Without loss ofgenerality we may assume that 0 is a fixed point of f 2 since this can be achieved byconjugation; that is, f 2(0) = 0. We consider the function
g(z) =f 2(z)
z.
Since f 2 has only finitely many fixed points, g takes the value 1 only finitely often. ByPicard’s theorem, g has infinitely many zeros. These zeros are also zeros of f 2 and thus,since f 2(0) = 0, they are not contained in I(f).
Theorem 1.5. Let f be a transcendental entire function. Then J(f) = ∂I(f).
Proof. It is easy to see that ∂I(f) ⊂ J(f). Moreover, ∂I(f) is backward invariant; thatis, f−1(∂I(f)) ⊂ ∂I(f). The conclusion now follows from the result proved in part 1 ofthis course that if A is a closed, backward invariant subset of the Julia set which containsat least three points, then J(f) = A. (This result follows from an application of Montel’stheorem to the complement B of A.)
Before discussing two examples, we state one lemma which will be used in the discus-sion of the first example and which will also be used frequently in the sequel.
Lemma 1.5. (Koebe’s one-quarter theorem) Let f : D → C be holomorphic andinjective with f(0) = 0 and f ′(0) = 1. Then f(D) ⊃ D
(0, 1
4
).
Remark 1.4. We will not prove Koebe’s theorem here. We note, however, that the weakerresult that f(D) ⊃ D(0, K) for some absolute constant K was an exercise in part 1 ofthis course. This weaker result would suffice for all applications of Koebe’s theorem thatwe will be discussing.
The Koebe function defined by (1.1) shows that the choice K = 14
as given above isoptimal.
If f : D(a, r)→ C is holomorphic and injective, then the function g defined by
(1.8) g(z) =f(a+ rz)− f(a)
f ′(a)r
satisfies the hypothesis of Koebe’s theorem and the conclusion g(D) ⊃ D(0, 14) translates
into
f(D(a, r)) ⊃ D(f(a), 14|f ′(a)|r).
Example 1.1. Let f(z) = λez where λ ∈ C\0.We show that I(f) ⊂ J(f). In order to do so, suppose that z0 ∈ I(f) ∩ F (f). Then
there exists r0 > 0 such that fn(z) → ∞ as n → ∞ uniformly for z ∈ D(z0, r0). Hencethere exists n0 such that |fn(z)| ≥ max8, |λ| for n ≥ n0 and z ∈ D(z0, r0).
(1.9) fn(D(z0, r0)) ∩ x+ i((2k + 1)π − arg λ) : x ∈ R, k ∈ Z = ∅
for n ≥ n0.Let now zn = fn(z0) and rn > 0 maximal such that D(zn, rn) ⊂ fn(D(z0, r0)). It
follows from (1.9) that rn ≤ π for all n ≥ n0. This implies that f is injective in D(zn, rn).Koebe’s theorem now yields that
f(D(zn, rn)) ⊃ D(zn+1,
14|f ′(zn)|rn
).
Since |f ′(zn)| = |f(zn)| ≥ 8 for n ≥ n0 this implies that
f(D(zn, rn)) ⊃ D(zn+1, 2rn)
and hence rn+1 ≥ 2rn for n ≥ n0. This is a contradiction to rn ≤ π for all n. ThusI(f) ⊂ J(f).
We consider the case λ = 1 in more detail; that is, the case f(z) = ez. For x ∈ R wehave f(x) ≥ x + 1 and thus fn(x) ≥ x + n for n ∈ N. Hence R ⊂ I(f). By completeinvariance, we thus have O−(R) ⊂ I(f).
It is known that J(exp) = C. This was conjectured by Fatou [20] and proved by Mis-iurewicz [27] in 1981. We also note that Rempe [30] has shown that I(exp) is connected.
Example 1.2. Let f(z) = z + 1 + e−z. For Re z > δ > 0 we have
Re f(z) = Re z + 1 + Re(e−z) ≥ Re z + 1− |e−z| = Re z + 1− e−Re z > Re z + 1− e−δ
and hence
Re fn(z) > Re z + n(1− e−δ
).
It follows that
z ∈ C : Re z > 0 ⊂ I(f) ∩ F (f).
In other words, the right half-plane is contained in a Baker domain of f . This exampledue to Fatou [20] was already discussed in part 1 of this course.
4 sin z and f3(z) = e−z + z + 1. Therange shown is | Im z| ≤ 5π/2 in all pictures, those for the real part are different.
1.2 Curves in the escaping set
We have seen in Theorem 1.2 that the escaping set of a transcendental entire functionalways has at least one unbounded component, and noted that it is conjectured that allcomponents are unbounded. This is made plausible by examples for which the escapingset actually consists of curves tending to ∞.
We discuss only one example, but note that the techniques extend to more generalfunctions; see [15, 34]. We consider the functions E(z) = Eλ(z) = λez where 0 < λ < 1/e.Then E(0) = λ > 0, E(1) = λe < 1 and E(x)/x → ∞ as x → ∞. The intermediatevalue theorem implies that there exists fixed points α and β of E satisfying 0 < α < 1and β > 1. It is easy to show that α and β are unique; for example, this follows fromthe convexity of E|R. Since E ′(α) = E(α) = α and E ′(β) = E(β) = β we see that α isattracting while β is repelling; see Figure 3
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α β x
y
y = x
y = 14ex
Figure 3: The graph of f(x) = 14ex.
For Re z < β we have
|E(z)| = λeRe z < λeβ = E(β) = β.
Thus H := z ∈ C : Re z < β satisfies E(H) ⊂ H and hence H ⊂ F (E). Since Hcontains the attracting fixed point α we actually have H ⊂ A(α), with the attractingbasin A(α) = z ∈ C : En(z)→ α of α.
For x ∈ R and n ∈ Z we have E(x+ (2n+ 1)πi) = −λex ∈ H and thus also
(1.10) H ∪ x+ (2n+ 1)πi : x ∈ R, n ∈ Z ⊂ A(α).
Theorem 1.6. J(E) = C\A(α).
Proof. Clearly J(E) ⊂ C\A(α). Suppose now that z ∈ C\A(α) but z /∈ J(E). Thenthere exists a Fatou component U with U ∩ A(α) = ∅ such that z ∈ U . Choose r > 0such that D(z, r) ⊂ U . Using Koebe’s theorem we see as in Example 1.1 that
Ek(D(z, r)) ⊃ D(Ek(z), 1
4|(Ek)′(z)|r
).
For Re ζ ≥ β we have
|E ′(ζ)| = |E(ζ)| = λeRe z ≥ λeβ = β.
It follows that
Ek(D(z, r)) ⊃ D(Ek(z), 1
4βkr).
On the other hand, it follows from (1.10) that Ek(D(z, r)) does not contain a disk ofradius greater than π. Thus 1
4βkr ≤ π for all k, a contradiction.
Theorem 1.6 says that with
Xn = z ∈ C : ReEk(z) ≥ β for 0 ≤ k ≤ n
we have J(E) =⋂∞n=1Xn. Figure 4 shows the sets X1, . . . , X5 for E(z) = 1
4ez. Actually,
while we said that Figure 2 shows J(E), it does in fact only show one of the sets Xn.
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Figure 4: The sets X1, . . . , X5 for E(z) = 14ez.
For n ∈ Z we put
Tn = x+ iy : x ≥ β, (2n− 1)π < y < (2n+ 1)π,
cf. Figure 5.
T−2
T−1
T0
T1
T2
Figure 5: The sets Tn.
It follows from Theorem 1.6 and (1.10) that
J(E) ⊂⋃n∈Z
Tn.
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Thus for z ∈ J(E) and k ∈ N0 there exists a unique sk ∈ Z such that Ek(z) ∈ Tsk . Thesequence s = s(z) = (sk)k≥0 is called the itinerary (or external address) of z. For a givensequence s = (sk)k≥0 in Z we denote by H(s) set of all z with itinerary s. We also denoteby Σ the set of all integer-valued sequences; that is, the set of all sequences (mk)k≥0 suchthat mk ∈ Z for all k ≥ 0. With this Theorem 1.6 takes the following form.
Theorem 1.7. J(E) =⋃s∈Σ H(s).
The following two theorems give a description of the sets H(s).
Theorem 1.8. Let s = (sk)k≥0 ∈ Σ. If H(s) 6= ∅, then there exists t ∈ R such that
(1.11) lim supk→∞
|sk|Ek(t)
<∞.
Proof. Let z = x + iy ∈ H(s). Then (2|sk| − 1)π ≤ | ImEk(z)| ≤ |Ek(z)| ≤ Ek(x), fromwhich the conclusion follows.
We say that s = (sk)k≥0 ∈ Σ is admissible (or exponentially bounded) if there existst > 0 such that (1.11) holds. We denote by ts the infimum of all t > 0 for which (1.11)holds. Noting that if β < t′ < t′′ <∞, then
(1.12) limk→∞
Ek(t′′)
Ek(t′)=∞,
we see that
lim supk→∞
|sk|Ek(t)
=∞ for β < t < ts
and
limk→∞
|sk|Ek(t)
= 0 for t > ts.
Theorem 1.9. Suppose that s = (sk)k≥0 ∈ Σ is admissible. Then there exists an injectivecurve gs : [ts,∞)→ Ts0 satisfying gs(t)→∞ as t→∞ such that tr(gs) = H(s).
The curves gs are called hairs (or external rays). The point gs(ts) is called the endpointof the hair.
We will not give a complete proof of Theorem 1.9, but only show how gs can be definedon the open interval (ts,∞). We will not show that it extends to an injective, continuousfunction on the closed interval [ts,∞) and that tr(gs) = H(s). We refer to the papersmentioned as well as [9].
We denote by log the principal branch of the logarithm, mapping C\(−∞, 0] to z ∈C : | Im z| < π and, for m ∈ Z, put Λm(z) = log z − log λ+ 2πim. Then Λm is a branchof E−1. If Re z ≥ β, then
Theorem 1.9 will then be a consequence of the following three lemmas whose proofswe omit.
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Lemma 1.7. The sequence (gs,k) has a subsequence which converges uniformly on [ts,∞)and thus gs extends to a continuous map gs : [ts,∞)→ H.
Lemma 1.8. gs is injective on [ts,∞).
Lemma 1.9. Let z ∈ H(s). Then there exist t ≥ ts with z = gs(t).
The map
σ : Σ→ Σ, (sk)k≥0 7→ (sk+1)k≥0,
that is, σ(s0s1s2 . . .) = s1s2s3 . . ., is called the shift map. It follows from (1.15) that
E(gs,k(t)) = gσ(s),k−1(E(t))
Thus E(gs(t)) = gσ(s)(E(t)) and
(1.17) Ek(gs(t)) = gσk(s)(Ek(t))
for t ≥ ts and k ∈ N.It follows from (1.16) that
(1.18)∣∣gσk(s)(E
k(t))− Ek(t)− 2πisk∣∣ ≤ M
β − 1.
We deduce that if t > ts, then gs(t) ∈ I(f). In other words, except possibly for theendpoint, the hairs are contained in the escaping set. The endpoints may or may not bein I(f).
In some sense, Theorems 1.7, 1.8 and 1.9, together with (1.17), give a complete de-scription of the dynamics of E on the Julia set. The Julia set consists of curves tendingto ∞. These curves are permuted by E according to the shift map, and except possiblyfor the endpoints, the iterates tend to ∞ on these curves.
It is apparent that similar methods work for other functions, for example one maytreat functions of the form λ sin z with 0 < λ < 1 in almost the same way; see Figure 2for pictures of the corresponding Julia sets.
We sketch an alternative method to prove that H(s) is the trace of a curve; see [32].Here one does not construct an explicit parametrization of the curve as in the proof ofLemma 1.6, but instead uses a general topological result. The advantage of this methodis that it works in much more general contexts.
We introduce some terminology. The image of a compact interval under a continuous,injective map is called an arc. A compact, connected metric space is called a continuum.Since continuous images of connected sets are connected, an arc is a continuum.
Let X be a continuum and let x ∈ X. Then x is called a cut point if X\x isdisconnected and non-cut point otherwise. For the interval [a, b], the points a and b arenon-cut points while all other points are cut-points.
The topological result needed is the following [28, Theorem 6.17].
Lemma 1.10. A continuum is an arc if and only if it has exactly two non-cut points.
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The idea is now to apply this lemma to H(s)∪∞. In order to do so, one introducesan order ≺ on H(s) by saying that u ≺ v for u, v ∈ H(s) if v = ∞ and u 6= ∞, or ifu, v 6=∞ and there exists N ∈ N such that |En(v)| ≥ |En(u)|+ 10 for all n ≥ N .
It is clear that ≺ is transitive and thus to show that ≺ is indeed a strict order we haveto show that if u 6= v, then u ≺ v or v ≺ u. So let u, v ∈ H(s)\∞ with u 6= v. It followsfrom (1.14) that
|En(u)− En(v)| = |Lsn(En+1(u))− Lsn(En+1(v))| ≤ 1
β|En+1(u)− En+1(v)|
and thus
|u− v| ≤ 1
βn|En(u)− En(v)|.
It follows that there exists N ∈ N such that |EN(u) − EN(v)| > 10. Since EN(u) andEN(v) are both contained in TsN we have | ImEN(u)− ImEN(v)| < 2π. Hence
Induction now shows that |En(v)| ≥ |En(u)|+ 10 for all n ≥ N . Thus ≺ is indeed a strictorder.
To apply Lemma 1.10 we have to show that H(s) ∪ ∞ has exactly two non-cutpoints. In order to do so, one shows that the minimum of H(s) ∪ ∞ with respect to≺ is a non-cut point, and so is ∞, which is the maximum with respect to ≺. All otherpoints are cut points. Indeed, if u ∈ H(s) is not the minimum, then v ∈ H(s) : v ≺ uand v ∈ H(s) : u ≺ v ∪ ∞ is a partition of (H(s) ∪ ∞)\u into two disjoint,non-empty, open sets. We omit the details.
Remark 1.5. Let X be the set of endpoints of the hairs that form J(E). In the termi-nology of Theorem 1.9, X is thus the set of all gs(ts) with admissible s. Mayer [24] has
shown that X ∪ ∞ is a connected subset of C, while X is totally disconnected; that is,every connected component of X consists of only one point.
17
1.3 The Eremenko-Lyubich class
Example 1.1 shows that both the Fatou and Julia set may intersect the escaping set. Wewill consider a class of functions for which the escaping set is contained in the Julia set.
Recall (from part 1 of this course) that sing(f−1) denotes the set of singularities ofthe inverse of an entire function f . It consists of the critical and asymptotic values of f .
Definition 1.2. The set of all transcendental entire functions f for which sing(f−1) isbounded is called the Eremenko-Lyubich class and denoted by B.
Examples of functions in B are f(z) = λez with sing(f−1) = 0 or f(z) = sin z withsing(f−1) = −1, 1.
An example for which sing(f−1) is infinite but bounded is given by f(z) = (sin z)/z.One can show that all critical points of f are real and it is apparent from the graph (cf.Figure 6) that the set of the corresponding critical values is bounded. Moreover, the onlyasymptotic value of this function is 0. An example of an entire function f /∈ B is givenby f(z) = z sin z; see again Figure 6.
Figure 6: The graphs of f(x) = 10sinx
xand g(x) = x sinx.
As indicated by the name, the class B was introduced to complex dynamics by Ere-menko and Lyubich [18], who also proved the results below about this class.
Theorem 1.10. Let f ∈ B. Then I(f) ⊂ J(f).
We immediately obtain the result proved in Example 1.1. The proof of the abovetheorem will use similar ideas based on Koebe’s theorem, as well as the following result.
Theorem 1.11. Let f ∈ B and let R > 0 be such that sing(f−1) ⊂ D(0, R). Let A be acomponent of f−1(z ∈ C : |z| > R). Then A is simply connected.
Let H = z ∈ C : Re z > logR. Then every branch of log f defined in some subdo-main of A can be extended holomorphically to a biholomorphic map from A to H.
Proof. Let a ∈ A and let U ⊂ A be a neighborhood of a in which some branch φ oflog f is defined. It is clear that for every a ∈ A there exists such a neighborhood, and
18
we have φ(a) = b for some b ∈ H with eb = f(a). Moreover, by restricting to a smallerneighborhood U if necessary we can achieve that there exists a neighborhood V ⊂ H ofb such that φ : U → V is biholomorphic.
Let ψ : V → U be the inverse function of φ : U → V ; cf. Figure 7. Assuming V is
f
exp
logR
R
H
∆A
φ
ψ
Figure 7: Illustration of the proof of Theorem 1.11.
sufficiently small, it is of the form ψ = τ exp with some branch τ of f−1. Since thereare no singularities of f−1 in expH, ψ can be continued along any curve in H. By theMonodromy theorem, ψ has a holomorphic continuation to H. Clearly this continuation(which we denote again by ψ) maps H to A.
We will show that ψ is injective. Suppose that this is not the case, say ψ(c1) = ψ(c2)where c1, c2 ∈ H, c1 6= c2. Then exp c1 = f(ψ(c1)) = f(ψ(c2)) = exp c2 and thusc1 = c2 + 2πik for some k ∈ Z. In a neighborhood of d := ψ(c1) = ψ(c2) there existbranches φj of log f satisfying φj(d) = cj. We conclude that φ1 = φ2 + 2πik. For theinverse functions we obtain ψ1(z) = ψ2(z − 2πik), first only in a neighborhood of c1, butby the identity theorem then also in H.
19
For R′ > R and H ′ = z ∈ C : Re z > logR′ we thus find that
ψ(∂H ′) = ψ(logR′ + iy : 0 ≤ y ≤ 2πk
is bounded. As in (1.5) we have ∂ψ(H ′) ⊂ ψ(∂H ′). Thus ∂ψ(H ′) is bounded. On theother hand, since A is unbounded, we see that ψ(H ′) is unbounded. Choosing S > 0with ∂ψ(H ′) ⊂ D(0, S) we find that A ⊃ ψ(H ′) ⊃ z ∈ C : |z| > S. This yields that|f(z)| > R′ for |z| > S, which leads to a contradiction since f is transcendental.
We conclude that ψ is injective. Noting that |f(ψ(z)| = |ez| → R as Re z → logRwe see that ψ(z) → ∂A as z → ∂H, and vice versa. From this we can conclude thatψ : H → A is indeed biholomorphic. In particular, A is simply connected.
Suppose now that, in addition to the hypotheses of Theorem 1.11, we have R > |f(0)|.Then 0 /∈ A and, since A is simply connected, we can define branches of the logarithmin A. The function log ψ, with ψ as in the above proof, is then holomorphic and injective,for any branch of the logarithm.
We obtain the following result.
Theorem 1.12. Let f ∈ B and let R > |f(0)| be such that sing(f−1) ⊂ D(0, R). Let
W = exp−1(f−1(z ∈ C : |z| > R)
)and
H = z ∈ C : Re z < logR
Then every component of W is simply connected and there exists a holomorphic functionF : W → H such that
expF (z) = f(exp z) for z ∈ W.
Moreover, for every component U of W the restriction F |U is a biholomorphic mapfrom U to H.
We say that the function F is obtained from f by a logarithmic change of variable; cf.Figure 8.
Theorem 1.13. Let f ∈ B and let R,W,H and F be as in Theorem 1.12. Let U be acomponent of W and let G : H → U be the inverse function of F |U : U → H. Then
|G′(w)| ≤ 4π
Rew − logRfor w ∈ H
and
|F ′(z)| ≥ ReF (z)− logR
4πfor z ∈ U.
20
F
f
exp exp
logR
R
H
∆
U
A
W
Figure 8: The logarithmic change of variable.
Proof. Obviously U does not contain any disk of radius greater than π. On the otherhand, Koebe’s theorem, applied to the restriction of G to the disk D(w,Rew − logR),yields that U = G(H) contains a disk of radius 1
4|G′(w)|(Rew − logR). Thus we obtain
14|G′(w)|(Rew − logR) ≤ π, from which the first conclusion follows.
With w = F (z) and thus z = G(w) we have F ′(z) = 1/G′(F (z)) = 1/G′(w). Thisyields the second conclusion.
Remark 1.6. One may express the estimate for F ′ also in terms of f ′. With F (ζ) =log f(eζ) we have
F ′(ζ) =eζf ′(eζ)
f(eζ)
and thus ∣∣∣∣zf ′(z)
f(z)
∣∣∣∣ ≥ log |f(z)| − logR
4πfor |f(z)| > R.
21
Proof of Theorem 1.10. Let f ∈ B and let R,W,H and F be as in Theorem 1.12. Asmentioned, we proceed similarly as in Example 1.1. Suppose that z0 ∈ I(f)∩F (f). Thenthere exists r0 > 0 such that fn(z) → ∞ as n → ∞ uniformly for z ∈ D(z0, r0). Hencethere exists n0 ≥ 0 such that |fn(z)| > R for n ≥ n0 and z ∈ D(z0, r0). Without loss ofgenerality we assume that n0 = 0 since otherwise we may replace z0 by fn0(z0).
Let w0 ∈ W with expw0 = z0. Then expD(w0, δ) ⊂ D(z0, r0) for some δ > 0. Since
expF (w) = f(expw)
for w ∈ W and hence
expF n(w) = fn(expw)
for w ∈ D(w0, δ) and n ∈ N we see that
ReF n(w) = log |fn(ew)| → ∞
as n→∞, uniformly for w ∈ D(w0, δ).By Theorem 1.13 we have
|(F n)′(w0)| =n−1∏k=0
|F ′(F k(w0))| ≥n−1∏k=0
ReF k(w0)− logR
4π.
Combining the last two equations we see that
(1.19) |(F n)′(w0)| → ∞
as n → ∞. Noting that F is injective on any component of W and thus F n is injectiveon D(w0, δ) we deduce from Koebe’s theorem that F n(D(w0, δ)) contains a disk of radius14|(F n)′(w0)|δ. On the other hand, F n(D(w0, δ)) ⊂ W , and W does not contain any disk
of radius greater than π. We conclude that
1
4|(F n)′(w0)|δ ≤ π,
contradicting (1.19).
Remark 1.7. We saw in section 1.2 that for f(z) = ez the escaping set consists of curvestending to ∞. This result can be extended to many – but not all – functions in theEremenko-Lyubich class. For example, it holds for f ∈ B satisfying M(r, f) ≤ exp(rµ)for some µ > 0 and all large r; see [6, 32].
We recall the classification of invariant Fatou components. Let f be entire and let Ube a component of F (f) which is invariant; that is, f(U) ⊂ U . Then U is of one of thefollowing types:
• U contains an attracting fixed point z0. Then fn|U → z0 as n→∞, and U is calledthe immediate attractive basin of z0.
22
• ∂U contains a fixed point z0 of multiplier 1, and fn|U → z0 as n→∞. In this case,U is called a Leau domain.
• There exists a biholomorphic map φ : U → D such that φ(f(φ−1(z))) = e2πiαz forsome α ∈ R\Q. In this case, U is called a Siegel disk.
• fn|U → z0 as n→∞. In this case, U is called a Baker domain.
The immediate attracting basing is also called Bottcher or Schroder domain dependingon whether z0 is superattracting or not.
For rational functions there is the additional possibility of a Herman ring. However,we had seen in part 1 of this course that these do not exist for entire functions; seeTheorem 1.14 below.
Of course, the above classification can be extended to periodic domains in an obviousway.
The following is an obvious consequence of Theorem 1.10.
Corollary 1.1. Let f ∈ B. Then f has no Baker domains.
Remark 1.8. The subset of B consisting of all entire functions f for which sing(f−1)is finite is called the Speiser class. It was shown by Eremenko and Lyubich [18] andGoldberg and Keen [22] that functions in S do not have wandering domains. Thus – insome sense – such functions are similar to rational functions.
It was shown only recently by Bishop [11] that functions in B may have wanderingdomains.
The result quoted above that entire functions do not have Herman rings follows fromthe following result (which was an exercise in part 1 of the course, but whose proof weinclude for completeness).
Theorem 1.14. Let f be a transcendental entire functions and let U be a multiply con-nected component of F (f). Then U ⊂ I(f).
Proof. Suppose that U 6⊂ I(f). Then there exists a sequence (nk) tending to∞ such thatfnk |U → φ for some holomorphic function φ : U → C. Given a closed curve γ in U wehave
n(γ, z)fnk(z) =1
2πi
∫γ
fnk(ζ)
ζ − zdζ
for z ∈ C\ tr(γ). Since (fnk) converges on tr(γ), we conclude that (fnk) converges on theinterior
int(γ) = z ∈ C : n(γ, z) 6= 0of γ. So the sequence (fn) has a subsequence which converges in int(γ), and this impliesthat int(γ) ⊂ F (f). (The last argument was an exercise in part 1. One uses that otherwiseint(γ) would contain a repelling periodic point.) Thus for every curve γ in U we haveint(γ) ⊂ U . Hence U is simply connected, contradicting the hypothesis.
Together with Theorem 1.10, Theorem 1.14 yields the following result.
Corollary 1.2. Let f ∈ B. Then all components of F (f) are simply connected.
23
1.4 Multiply connected Fatou components
Theorem 1.14 says that multiply connected Fatou components are contained in the es-caping set. We want to show that such components actually exist. The examples showingthis will be defined using infinite products.
Therefore we first collect the basic definitions and results about those. The obvi-ous definition of the convergence of an infinite product
∏∞k=1 pk would be that the limit
limn→∞∏n
k=1 pk exists. However, in this case we would have convergence as soon as onefactor is 0. Also, we have
n∏k=1
1
k=
1
n!→ 0,
as n → ∞, meaning that the infinite product could be zero even though all factors aredifferent from 0.
In order to obtain the result that a product is equal to 0 if and only if a factor is equalto 0 one defines convergence as follows.
Definition 1.3. Let (pk)k∈N be a sequence in C. Then∏∞
k=1 pk is called convergent ifthere exists N ∈ N such that pk 6= 0 for k ≥ N and if
limn→∞
n∏k=N
pk
exists and is different from 0. We then put
∞∏k=1
pk =N∏k=1
pk · limn→∞
n∏k=N
pk.
It is then easy to see that∏∞
k=1 pk = 0 if and only if there exists k ∈ N such thatpk = 0.
If a series∑∞
k=1 ak converges, then limk→∞ ak = 0. Analogously we find here that ifan infinite product
∏∞k=1 pk converges, then limk→∞ pk = 1. Therefore we often write the
factors pk in the form pk = 1 + ck.
Definition 1.4. An infinite product∏∞
k=1(1 + ck) is called absolutely convergent if theinfinite product
∏∞k=1(1 + |ck|) converges.
Theorem 1.15. An absolutely convergent infinite product converges.
Sketch of proof. Let n > m ≥ N . For a suitable set X of multiindices we have∣∣∣∣∣n∏
k=N
(1 + ck)−m∏
k=N
(1 + ck)
∣∣∣∣∣ =
∣∣∣∣∣∣∑
(k1,...,kl)∈X
ck1 · · · · · ckl
∣∣∣∣∣∣≤
∑(k1,...,kl)∈X
|ck1 · · · · · ckl |
=n∏
k=N
(1 + |ck|)−m∏
k=N
(1 + |ck|),
24
and the conclusion follows from Cauchy’s criterion.
Theorem 1.16. The infinite product∏∞
k=1(1 + ck) converges absolutely if and only if theseries
∑∞k=1 ck converges absolutely.
Proof. Without loss of generality we may assume that ck ≥ 0 for all k. If either productor series converges, we have limk→∞ ck = 0. Thus ck ≤ 1 for large k, say for k ≥ N .
It is easy to see that if 0 ≤ x ≤ 1, then x log 2 ≤ log(1 + x) ≤ x; see Figure 9. It
1x
y y = x
y = x log 2
y = log(1 + x)
Figure 9: Illustration of the inequality x log 2 ≤ log(1 + x) ≤ x.
follows that
log 2n∑
k=N
ck ≤n∑
k=N
log(1 + ck) = log
(n∏
k=N
(1 + ck)
)≤
n∑k=N
ck,
from which the conclusion follows.
Various concepts like uniform convergence or locally uniform convergence carry overfrom infinite series to infinite products almost literally. The argument in the above proofyields the following result.
Theorem 1.17. Let G be a domain and let (gk) be a sequence of functions from G toC. Suppose that
∑∞k=1 |gk| converges (locally) uniformly. Then
∏∞k=1(1 + gk) converges
(locally) uniformly.
The following version of Weierstraß’s theorem for products follows directly from theversion for sequences.
Theorem 1.18. Let G be a domain and let (gk) be a sequence of holomorphic functionsfrom G to C. If
∏∞k=1(1 + gk) converges locally uniformly, then the function f : G→ C,
f(z) =∞∏k=1
(1 + gk(z)),
is holomorphic in G. Moreover, f(z) = 0 if and only if there exists k ∈ N such thatgk(z) = −1.
25
With these preparations we can now give an example of a function with a multiplyconnected Fatou component.
Example 1.3. Let (ak) be the sequence defined recursively by a1 = 1, a2 = 2 and
(1.20) an+1 =ann∏n−1k=1 ak
for n ≥ 2. Then
f(z) = z∞∏k=1
(1− z
ak
)defines an entire function f with a multiply connected Fatou component.
Verification of the properties of the example. Writing (1.20) as
(1.21) an+1 = an
n−1∏k=1
anak
we easily see by induction that an ≥ 2n−1 for all n ∈ N. This implies that the infiniteproduct defining f converges locally uniformly. Thus f is an entire function and the zerosof f are precisely the an.
It also follows from (1.21) that an/an−1 →∞ as n→∞. Thus an > 9an−1 for large n.In fact, it is easy to check that this holds for n ≥ 5. For such n we put
An =z ∈ C : 3an−1 < |z| < 1
3an.
We will show that
(1.22) f(An) ⊂ An+1
for large n. This implies that An ⊂ F (f) for large n. In fact, the proof will show thatthis holds for n ≥ 5. Moreover, the component Un of F (f) containing An is contained inI(f). Since f(0) = 0 we have 0 /∈ Un. Thus Un is multiply connected.
It remains to prove (1.22). Since f has no zeros in An this follows if we can prove thefollowing four inequalities (for n ≥ 5):
(a) |f(z)| ≤ 13an+1 for |z| = 1
3an,
(b) |f(z)| ≤ 13an+1 for |z| = 3an−1,
(c) |f(z)| ≥ 3an for |z| = 13an,
(d) |f(z)| ≥ 3an for |z| = 3an−1.
26
To prove (a) we note that if |z| = 13an, then
|f(z)| = 1
3an
∞∏k=1
∣∣∣∣1− z
ak
∣∣∣∣≤ 1
3an
∞∏k=1
(1 +
an3ak
)
=1
3an
n−1∏k=1
(1 +
an3ak
)· 4
3·∞∏
k=n+1
(1 +
an3ak
).
Since an > 9an−1 we have 1 ≤ an/(9ak) for k ≤ n− 1 and thus
1 +an3ak≤ 4an
9akfor 1 ≤ k ≤ n− 1.
Since ak > 9ak−1 for k > n we also have
1 +an3ak≤ 1 +
1
3 · 9k−nfor k ≥ n− 1.
Thus
|f(z)| ≤ 1
3an
n−1∏k=1
(4an9ak
)· 4
3·∞∏
k=n+1
(1 +
1
3 · 9k−n
)=
(4
9
)nan+1
∞∏j=1
(1 +
1
3 · 9j
).
The product on the right converges. In fact, we have
log
(∞∏j=1
(1 +
1
3 · 9j
))=∞∑j=1
log
(1 +
1
3 · 9j
)≤ 1
3
∞∑j=1
1
9j=
1
24
and thus
∞∏j=1
(1 +
1
3 · 9j
)≤ e1/24 ≤ 2.
Altogether thus
|f(z)| ≤ 2
(4
9
)nan+1 ≤
1
3an+1
for |z| = 13an, which is (a). Part (b) follows from (a) by the maximum principle.
27
To prove (c) we note that if |z| = 13an, then
|f(z)| = 1
3an
∞∏k=1
∣∣∣∣1− z
ak
∣∣∣∣≥ 1
3an
n−1∏k=1
(an3ak− 1
)· 2
3·∞∏
k=n+1
(1− an
3ak
).
≥ 1
3an · 2n−1 · 2
3
∞∏j=1
(1− 1
3 · 9j
).
Again the product on the right side converges. In fact, noting that log(1− x) ≥ −2x for0 ≤ x ≤ 1
2, we have
∞∏j=1
(1− 1
3 · 9j
)= exp
(∞∑j=1
log
(1− 1
3 · 9j
))≥ exp
(−2
3
∞∑j=1
1
9j
)= e−1/12 ≥ 1
2
and thus
|f(z)| ≥ 1
92n−1an ≥
1
3an
for |z| = 13an.
The proof of (d) is similar. In fact, if |z| = 3an−1, then
|f(z)| = 3an−1
∞∏k=1
∣∣∣∣1− z
ak
∣∣∣∣≥ 3an−1
n−2∏k=1
(3an−1
ak− 1
)· 2 ·
∞∏k=n
(1− 3an−1
ak
)
≥ 3an−1
n−2∏k=1
(2an−1
ak
)· 2 ·
∞∏j=1
(1− 1
3 · 9j
)
= 6 · 2n−1an
∞∏j=1
(1− 1
3 · 9j
)and thus
|f(z)| ≥ 3 · 2n−1an ≥ 3an
This is (d), which completes the proof.
Figure 10 shows points which are mapped to the annulus A5 under iteration (and fromthere to A6, A7, etc.) in light gray. Points which are mapped to An for n = 6, 7, 8 withoutbeing mapped to An−1 first are shown in darker levels of gray. All other points are inblack, including those mapped to the Leau domain at 0.
28
Figure 10: The function from Example 1.3.
The computations are actually done with
f(z) = z8∏
k=1
(1− z
ak
),
but it should be noted that a8 = 22957 and a9 = 223117. The ranges shown are |Re z| ≤ 100and | Im z| ≤ 100 in the left, −36 ≤ Re z ≤ 52 and | Im z| ≤ 44 in the middle and−4 ≤ Re z ≤ 8 and | Im z| ≤ 6 in the right picture.
Since f(32) = 0, one would expect to see a preimage of the Leau domain at 0 in thefirst two pictures. However, this preimage is so small that it cannot be seen there. A close-up of the range |Re z − 32.006| ≤ 0.0018 and | Im z| ≤ 0.0018 is almost indistinguishablefrom the right picture in Figure 10.
In the above proof, we have shown that Un is multiply connected by noting that 0 /∈ Unbut 0 ∈ T (Un); that is, we showed that 0 is in a bounded component of the complementof Un. In particular, Un contains a curve surrounding 0.
We will show that this is always the case for multiply connected components.
Theorem 1.19. Let f be a transcendental entire function and let U be a multiply con-nected component of F (f). For k ∈ N denote by Uk the Fatou component containingfk(U). Then Uk is also multiply connected.
Let further γ be a simple closed curve in U which is not null-homologous in U ; thatis, n(γ, a) 6= 0 for some a ∈ C\U . Let b ∈ C. Then n(fk γ, b) 6= 0 for all large k.
Proof. Without loss of generality we may assume that the orientation of γ is positive. Forc ∈ int(γ) we then have n(γ, c) > 0 and thus, as in the proof of Lemma 1.3,
n(fk γ, fk(c)) =1
2πi
∫fkγ
dw
w − fk(c)
=1
2πi
∫γ
(fk)′(ζ)dζ
fk(ζ)− fk(c)
=∑
z∈int(γ)∩f−k(fk(c))
n(γ, z)mz,
≥ n(γ, c) > 0
29
which means that fk(c) ∈ int(fk γ).Since int(γ) contains a point a ∈ C\U , it also contains a point of ∂U . Hence we may
choose c ∈ ∂U ⊂ J(f). Since J(f) is completely invariant, we then have fk(c) ∈ J(f).Thus Un is multiply connected.
In fact, since J(f) is the closure of the set of repelling periodic points of f we mayalso choose c as a repelling periodic point of f . This implies that there exists R > 0 suchthat |fk(c)| < R for all k. On the other hand, since fk|U →∞ by Theorem 1.14, we have
tr(fk γ) ⊂ z ∈ C : |z| > maxR, |b|
for large k. It follows that
n(fk γ, b) = n(fk γ, fk(c)) 6= 0
as claimed.
Our next aim is to show the Fatou components Un in the above example are wanderingdomains. In fact, we shall show that multiply connected Fatou components of transcen-dental entire functions are always wandering. These results of Baker [1, 2, 3] where thefirst examples of wandering domains.
In order to do so, we need some results about harmonic functions. We recall that areal- or complex-valued, twice continuously differentiable function h defined on a domainin C is called harmonic if
∆h :=∂2h
∂x2+∂2h
∂y2= 0.
The differential operator ∆ is called the Laplace operator. It follows easily from theCauchy-Riemann equations that if h is holomorphic, then Reh, Imh and h are harmonic.
Conversely, if u : D→ R is harmonic, then there exists a harmonic function v : D→ Rsuch that f = u+iv is holomorphic. (This can be seen by defining v =
∫uxdy−uydx.) The
function v is called the harmonic conjugate of u. It is unique up to an additive constant.Instead of the unit disk one can take any simply connected domain here, but not anarbitrary domain. The function u : C\0 → R, u(z) = log |z|, is harmonic, and in simplyconnected subdomains of C\0 the harmonic conjugate v is given by v(z) = arg z, butthis cannot be defined harmonically (or even continuously) in the whole domain C\0.
As a further example of a harmonic function we note that if f is a non-vanishingholomorphic function, then log |f | is harmonic. This can be seen using the definition, ornoting that locally log |f | = Re(log f) for some branch of the logarithm.
Lemma 1.11. (Poisson integral formula) Let f be harmonic in a domain containingthe disk D(0, R). Then
h(z) =1
2πi
∫|ζ|=R
Re
(ζ + z
ζ − z
)h(ζ)
dζ
ζ=
1
2π
∫ 2π
0
Re
(Reit + z
Reit − z
)h(Reit)dt
for z ∈ D(0, R).
30
Proof. There exists S > R such that h is harmonic in D(0, S). Hence h is of the formh = Re f for some function f holomorphic in D(0, S). Cauchy’s integral formula yields
(1.23) f(z) =1
2πi
∫|ζ|=R
f(ζ)
ζ − zdζ =
1
2πi
∫|ζ|=R
ζ
ζ − zf(ζ)
dζ
ζ
for z ∈ D(0, R). In particular,
(1.24) f(0) =1
2πi
∫|ζ|=R
f(ζ)dζ
ζ.
Applying the last formula to the function given by
ζ → R2
R2 − zζf(ζ)
yields, noting that R2 = ζζ for |ζ| = R,
(1.25) f(0) =1
2πi
∫|ζ|=R
R2
R2 − zζf(ζ)
dζ
ζ=
1
2πi
∫|ζ|=R
ζ
ζ − zf(ζ)
dζ
ζ.
Combining (1.23), (1.24) and (1.25) we obtain
f(z) = f(z) + f(0)− f(0) =1
2πi
∫|ζ|=R
K(z, ζ)f(ζ)dζ
ζ
with
K(z, ζ) =ζ
ζ − z+
ζ
ζ − z− 1 = 2 Re
(ζ
ζ − z
)− 1 = Re
(2ζ
ζ − z− 1
)= Re
(ζ + z
ζ − z
).
Combining the last two equations the conclusion follows by taking real parts.
The term
K(z, ζ) = Re
(ζ + z
ζ − z
)occurring in the above formulas is called the Poisson kernel. We note that
K(z, ζ) =1
2
(ζ + z
ζ − z+ζ + z
ζ − z
)=
1
2
((ζ + z)(ζ − z) + (ζ + z)(ζ + z)
(ζ − z)(ζ − z)
)=|ζ|2 − |z|2
|ζ − z|2
Lemma 1.12. (Harnack inequality) Let h be harmonic and positive in D(a,R) andlet 0 < ρ < 1. Then
1− ρ1 + ρ
h(a) ≤ h(z) ≤ 1 + ρ
1− ρh(a) for z ∈ D(a, ρR).
31
Proof. Without loss of generality we may assume that a = 0. We may also assume thath is actually harmonic in a domain containing D(0, R), since otherwise we replace R byr ∈ (0, R) and consider the limit as r → R. For z ∈ D(0, ρR) we then have
h(z) =1
2π
∫ 2π
0
K(z,Reit)h(Reit)dt =1
2π
∫ 2π
0
Re
(Reit + z
Reit − z
)h(Reit)dt
where
K(z,Reit) =R2 − |z|2
|Reit − z|2≤ R2 − |z|2
(R− |z|)2=R + |z|R− |z|
≤ R + ρR
R− ρR=
1 + ρ
1− ρ
and analogously
K(z,Reit) =R2 − |z|2
|Reit − z|2≥ R2 − |z|2
(R + |z|)2=R− |z|R + |z|
≥ 1− ρ1 + ρ
.
This yields
h(z) ≤ 1 + ρ
1− ρ
∫ 2π
0
h(Reit)dt =1 + ρ
1− ρh(0),
that is, the right inequality claimed. The left inequality follows by the same argument.
Corollary 1.3. Under the hypothesis of Lemma 1.12, we have
max|z−a|≤ρR
h(z) ≤(
1 + ρ
1− ρ
)2
min|z−a|≤ρR
h(z).
Corollary 1.4. Let G ⊂ C be a domain and let K ⊂ G be compact. Then there existsC > 1 such that for any positive harmonic function h : G→ R we have
maxz∈K
h(z) ≤ C minz∈K
h(z).
Sketch of proof. We may assume that K is connected. Since K is compact, we can coverK by finitely many disks of the form D(ak, Rk/2) such D(ak, Rk) ⊂ G for all k. ApplyingCorollary 1.3 with ρ = 1
2to each of these disks the conclusion follows with C = 9m, where
m is the number of disks.
Theorem 1.20. A multiply connected Fatou component of a transcendental entire func-tion is a wandering domain.
Proof. Suppose there exists a transcendental entire function f with a multiply connectedpreperiodic Fatou component U . Then U is mapped by some iterate of f to a periodicFatou component V . By Theorem 1.19, V is also multiply connected. Without loss ofgenerality we may assume that V is invariant; that is, f(V ) ⊂ V .
Let now γ be a simple closed curve in V which is not null-homologous in V . Thenf γ is also a closed curve in V , since f(V ) ⊂ V . We put K = tr(γ) ∪ tr(f γ). Then
32
K is compact and there exists a neighborhood W of K such that fn(z)→∞ as n→∞uniformly for z ∈ W . Without loss of generality we may assume that |fn(z)| > 1 for alln ∈ N and all z ∈ W . Thus, by Corollary 1.4, there exists C > 1 such that
(1.26) maxz∈K
log |fk(z)| ≤ C minz∈K
log |fk(z)|
for all k ∈ N. With γk = fk γ we have fk(K) = tr(γk) ∪ tr(γk+1). By Theorem 1.19we have n(0, γk) 6= 0 for large k. For such k let Rk be maximal with D(0, Rk) ⊂ int(γk).Then Rk →∞ and
minz∈K
log |fk(z)| ≤ minz∈tr(γ)
log |fk(z)| = logRk.
It follows that
maxz∈K
log |fk(z)| ≤ C logRk.
In particular,
maxz∈tr(γk)
log |f(z)| = maxz∈tr(γ)
log |fk+1(z)| = maxz∈tr(fγ)
log |fk(z)| ≤ C logRk.
Since D(0, Rk) ⊂ int(γk) this implies that
logM(Rk, f) = max|z|≤Rk
log |f(z)| ≤ C logRk = logRCk
and hence
M(Rk, f) ≤ RCk .
The Cauchy inequalities now yield that f is a polynomial of degree at most C, a contra-diction.
Corollary 1.5. A multiply connected Fatou component of a transcendental entire functionis bounded.
Proof. Let U be a multiply connected Fatou component of the transcendental entire func-tion f and let Uk be the component containing fk(U). With γk and Rk as in the previousproof we have D(0, Rk) ⊂ int(γk) for large k, meaning that D(0, Rk) lies in a boundedcomponent of C\Uk. Choosing k so large that U intersects D(0, Rk) we see that U alsolies in a bounded component of C\Uk. Thus U is bounded.
Corollary 1.6. Let f be a transcendental entire function. Then J(f) contains a non-degenerate continuum.
An alternative proof of Theorem 1.20 can be given with the hyperbolic metric, whichwe briefly introduce. Define
ρD : D→ (0,∞), ρD(z) =2
1− |z|2.
33
For a (rectifiable) curve γ in D we call
`D(γ) =
∫γ
ρD(z)|dz|.
the hyperbolic length of γ in D. Finally, for a, b ∈ D the hyperbolic distance of a and b isdefined by
λD(a, b) = infγ`D(γ),
where the infinimum is taken over all curves connecting a with b.It is not difficult to see that λD is indeed a metric on D. It is called the hyperbolic
metric or Poincare metric on D.For a ∈ D and θ ∈ R the map T = Ta,θ given by
(1.27) T (z) = Ta,θ(z) = eiθz − a1− az
is a biholomorphic map from D onto D. In turn, every biholomorphic map T : D→ D hasthis form. A computation shows that if T has this form, then
(1.28)|T ′(z)|
1− |T (z)|2=
1
1− |z|2
which we may also write as
ρD(T (z))|T ′(z)| = ρD(z).
This implies that
`D(T γ) = `D(γ)
for every rectifiable curve γ in D and thus
λD(a, b) = λD(T (a), T (b))
for a, b ∈ D.It is not difficult to see that, for a ∈ D, among the curves connecting 0 with a the one
of minimal hyperbolic length is given by the straight line segment connecting 0 with a.This implies that
λD(0, a) =
∫ |a|0
2
1− t2dt = log
1 + |a|1− |a|
= 2 artanh |a|.
For a, b ∈ D and T as above we thus have
λD(a, b) = λD(T (a), T (b)) = 2 artanh |T (b)| = 2 artanh
∣∣∣∣ b− a1− ab
∣∣∣∣ .So we have an explicit expression for the hyperbolic metric (but this will not be usedin the sequel). We also note that the shortest curve with respect to the hyperbolic
34
metric that connects a and b is the image of a straight line segment under the Mobiustransformation T . It follows from this that this curve is an arc of a circle perpendicularto the unit circle.
Lemma 1.13. (Schwarz’s lemma) Let f : D → D be holomorphic satisfying f(0) = 0.Then |f(z)| ≤ z for all z ∈ D and |f ′(0)| ≤ 1.
If |f(z)| = z for some z 6= 0 or if |f ′(0)| = 1, then f has the form f(z) = eiθz forsome θ ∈ R.
The proof is left as an exercise.Given a holomorphic function f : D → D and z ∈ D, put w = f(z). We can apply
Schwarz’s lemma to T0,w f T−10,z , with Tθ,a as in (1.27). Then |(Tw,0 f T−1
z,0 )′(0)| ≤ 1,with equality only if f is biholomorphic. This is equivalent to
|Tw,0(w)f ′(z)| = |(Tw,0 f)′(z)| ≤ |T ′z,0(0)|.
A computation using (1.28) now yields the following result.
Lemma 1.14. (Schwarz-Pick lemma) Let f : D→ D be holomorphic. Then
|f ′(z)|1− |f(z)|2
≤ 1
1− |z|2.
If equality holds for some z ∈ D, then f is biholomorphic.
Let U, V be domains and f : U → V be holomorphic. Then f is called a covering mapif for every v ∈ V there exists a neighborhood N of v such that for every component Dof f−1(N) the map f |D : D → N is bijective. (More generally, this definition is made forsurjective continuous maps between topological spaces.) It can be shown that f : U → Vis a covering if every branch of f−1 defined in some subdomain of V can be continuedanalytically along any path in V .
Let U be a domain in C such that C\U contains at least two points. A domain withthis property is called hyperbolic.
Since biholomorphic maps are in particular covering maps, the following result is asubstantial generalization of the Riemann mapping theorem.
Theorem 1.21. (Uniformization theorem) Let U ⊂ C be a hyperbolic domain. Thenthere exists a holomorphic covering map h : D→ U .
Essentially the same arguments that we used to show that a function f ∈ B can be“lifted” to a function F by a logarithmic change of variable yield the following result; seeFigure 11.
Theorem 1.22. Let U, V be hyperbolic domains with covering maps φ : D → U andψ : D → V . Let f : U → V be holomorphic. Then there exists a holomorphic functionF : D→ D such that ψ F = f φ.
Moreover, given z1, z2 ∈ D with f(φ(z1)) = ψ(z2) the map F can be chosen such thatF (z1) = z2.
35
F
f
φ ψ
U V
DD
Figure 11: The lift F : D→ D of a map f : U → V
We call F the lift of f . We will show that for a covering map h as in Theorem 1.21the map ρU : U → (0,∞) defined by
ρU(h(z))|h′(z)| = ρD(z)
is well-defined; that is, if w ∈ U , if h1, h2 : D → U are covering maps and if z1, z2 ∈ Dsatisfy h1(z1) = h2(z2) = w, then ρD(z1)/|h′1(z1)| = ρD(z2)/|h′2(z2)|, and this commonvalue is defined to be ρU(w).
So let h1, h2 : D→ U be covering maps. Applying Theorem 1.22 with U = V , h1 = φ,h2 = ψ and f(z) = z we see that there exists a holomorphic map T : D → D such thath2 T = h1 and T (z1) = z2. It follows that T is also a covering and thus biholomorphic.Hence T is of the form (1.27) so that (1.28) holds. This yields
ρD(z1)
|h′1(z1)|=
ρD(z1)
|(h2 T )′(z1)|=
ρD(z1)
|h′2(z2)| · |T ′(z1)|=
ρD(z2)
|h′2(z2)|.
So ρU is indeed well-defined. The map ρU is called the density of the hyperbolic metricAs in the special case U = D discussed before, we call, for a (rectifiable) curve γ in U ,
`U(γ) =
∫γ
ρU(z)|dz|
36
the hyperbolic length of a curve γ in U . For a, b ∈ U the hyperbolic distance of a and b isthen defined by
λU(a, b) = infγ`U(γ),
where the infinimum is taken over all curves connecting a with b.As before one easily checks that λU is indeed a metric on U , called the hyperbolic
metric or Poincare metric.Let now f : U → V be holomorphic. Let φ : D → U and ψ : D → V be coverings and
let F : D→ D be the lift of f as in Theorem 1.22. Thus ψ F = f φ. Given z ∈ U wechoose ζ ∈ D with φ(ζ) = z so that
ρU(z) =ρD(ζ)
|φ′(ζ)|
and
ρV (f(z)) =ρD(F (ζ))
|ψ′(F (ζ))|=ρD(F (ζ))|F ′(ζ)||(ψ F )′(ζ)|
=ρD(F (ζ))|F ′(ζ)||f ′(z)| · |φ′(ζ)|
.
The Schwarz-Pick Lemma yields that ρD(F (ζ))|F ′(ζ)| ≤ ρD(ζ), with equality only if F isbiholomorphic. This yields the following result.
Lemma 1.15. If U and V are hyperbolic domains and f : U → V is holomorphic, then
(i) ρV (f(z))|f ′(z)| ≤ ρU(z) for all z ∈ U ,
(ii) `V (f γ) ≤ `U(γ) for every (rectifiable) curve γ in U ,
(iii) λV (f(a), f(b)) ≤ λU(a, b) for all a, b ∈ U .
Equality occurs in (i) or (ii) if and only if f is a covering, assuming that γ is non-degenerate in (ii). If we have equality in (iii) for a 6= b, then f is a covering.
We may have strict inequality in (iii) also if f is not a covering. In fact, if f is anon-injective covering, there exists a 6= b with f(a) = f(b), so that the left side is equalto 0. A specific example is exp: zRe z > 0 → z : |z| > 1, a = 1, b = 1 + 2πi.
The choice f(z) = z in Lemma 1.15 leads to the following result.
Lemma 1.16. If U and V are hyperbolic domains, U ⊂ V , then
(i) ρV (z) ≤ ρU(z) for all z ∈ U ,
(ii) `V (γ) ≤ `U(γ) for every (rectifiable) curve γ in U ,
(iii) λV (a, b) ≤ λU(a, b) for all a, b ∈ U .
Equality can occur, assuming a 6= b in (iii), only if f is a covering.
Lemma 1.17. (i) Let U be simply connected hyperbolic domain. Then
1
2 dist(z, ∂U)≤ ρU(z) ≤ 2
dist(z, ∂U)
for z ∈ U .
37
(ii) There exist a, b, c > 0 such that if |z| > c, then
a
|z| log |z|≤ ρC\0,1(z) ≤ b
|z| log |z|
for z ∈ C\0, 1.
Sketch of proof. Let r = dist(z, ∂U). Then D(z, r) ⊂ U and thus ρU(z) ≤ ρD(z,r)(z) byLemma 1.16. Since ζ → z + rζ is a covering (in fact a biholomorphic map) from D toD(z, r) one easily finds that ρD(z,r)(z) = 2/r. This proves the right inequality of (i).
The left inequality of (i) uses the Koebe one quarter theorem. Note that a coveringmap from D to U is in fact biholomorphic. Taking such a biholomorphic map h : D→ Uwith h(0) = z. Then h(D) ⊃ D(h(0), |h′(0)|/4) = D(z, |h′(0)|/4) and thus r ≥ |h′(0)|/4.We obtain
from which the left inequality of (i) follows.The upper bound in (ii) can be obtained by using Lemma 1.16 with U = C\D, noting
that z 7→ exp((z − 1)/(z + 1)) is a covering from D to C\D. We refer to the problemsessions for the computation of ρC\D.
We omit the proof of the lower bound in (ii).
To give an alternative proof of Theorem 1.20 we proceed as there and assume that thetranscendental entire function f has a multiply connected invariant Fatou component V .As there we choose a simple closed curve γ in V and put K = tr(γ) ∪ tr(f γ).
In the proof of Theorem 1.20, Harnack’s inequality was used to obtain (1.26), whichis equivalent to
(1.29) log maxz∈K|fk(z)| ≤ C log min
z∈K|fk(z)|.
To give a proof of (1.29) using the hyperbolic metric, we choose uk, vk ∈ K such thatak = fk(uk) and bk = fk(vk) satisfy |ak| = minz∈K |fk(z)| and |bk| = maxz∈K |fk(z)|.Lemma 1.16 yields that
λV (ak, bk) ≤ λV (uk, vk) ≤M := maxu,v∈K
λV (u, v).
On the other hand, assuming without loss of generality that 0, 1 ∈ J(f), Lemmas 1.16and 1.17 yield that
λV (ak, bk) ≥ λC\0,1(ak, bk) ≥ a
∫ |bk||ak|
dt
t log t= a log
(log |bk|log |ak|
).
With C = exp(M/a) we obtain
log |bk|log |ak|
≤ C
from which (1.29) follows directly.As another example of a result which is used using the hyperbolic metric we mention
the following result.
38
Theorem 1.23. Let f be entire and let U be a Baker domain of f . Then there exists acurve γ tending to ∞ in U and C > 0 such that |f(z)| ≤ C|z| for z ∈ tr(γ).
Proof. Let z0 ∈ U and let γ0 : [0, 1] → U be a curve connecting z0 and f(z0); that is,γ0(0) = z0 and γ0(1) = f(z0). For k ∈ N let γk = fk γ and let γ be the concatenationof the curves γk. One may parametrize γ as γ : [0,∞)→ U , γ(t) = f [t](γ0(t− [t])), where[t] denotes the integer part of t. Then
by Lemma 1.15, (ii).By Theorem 1.20, U is simply connected. Without loss of generality we may assume
that 0 ∈ J(f). We may also assume that |f(z)| > |z|, since otherwise the conclusion istrivial. Thus, by Lemma 1.17, (i),
λU(z, f(z)) ≥ 1
2
∫ |f(z)|
|z|
dt
t=
1
2log|f(z)||z|
.
The conclusion follows from the last two inequalities with C = e2K .
Next we use the hyperbolic metric to prove a result of J.-H. Zheng [37].
Theorem 1.24. Let f be a transcendental entire function and let U be a multiply con-nected component of F (f).
Then there exists sequences (rk) and (Rk) tending to ∞ satisfying Rk/rk → ∞ suchthat
z : rk < |z| < Rk ⊂ fk(U)
for large k.
Previously we always denoted by Uk the component of F (f) containing fk(U). Wenote that if U is multiply-connected, then U is wandering by Theorem 1.20 and thusbounded by Corollary 1.5. This easily implies that Uk = fk(U). In fact, fk : U → Uk is aproper map.
Before starting with the proof we note that z 7→ 1/z is a biholomorphic self-map ofC\0, 1. Lemma 1.17 now yields that
(1.30)a
|z| log1
|z|
≤ ρC\0,1(z) ≤ b
|z| log1
|z|
for |z| < 1/c. It follows that if 0 < r < 1/|c| and if γr is (a parametrization of) thecircle of radius r around zero, then `C\0,1(γr) ≤ 2π/ log(1/r) and thus `C\0,1(γr) → 0as r → 0. The next lemma says that this is different for curves γ which besides 0 alsosurround some other point of positive distance from 0.
39
Lemma 1.18. For each ε > 0 there exists δ > 0 such that if γ is a closed curve in C\0, 1whose interior contains 0 and a further point z with |z| ≥ δ, but does not contain 1, then`C\0,1(γ) ≥ ε.
The proof is left as an exercise. We note that the proof does not require an explicitlower bound for ρC\0,1(z) such as the one given in (1.30).
Proof of Theorem 1.24. Let γ be a positively oriented simple closed curve in U which isnot null-homologous in U ; that is, n(γ, z) = 1 for z ∈ int(γ) and int(γ)∩ (C\U) 6= ∅. Letγk = fk γ.
By Problem 1 from Problem Sheet 10 we have nk := n(γk, 0) → ∞ as k → ∞. Wecan split the curve γk into nk curves σ1, . . . , σnk
such that n(σj, 0) = 1 for 1 ≤ j ≤ nk.(We omit the details of this argument.) We have
`Uk(γk) =
nk∑j=1
`Uk(σj)
and thus there exists jk with
`Uk(σjk) ≤ 1
nk`Uk
(γk).
Let now ak, bk ∈ J(f), with ak in the interior and bk in the exterior of σjk . Supposethat bk/ak 6→ ∞, so that – on some subsequence – the point ck = −ak/(bk − ak) satisfies|ck| ≥ δ for some δ > 0. Let T : C → C, T (z) = (z − ak)/(bk − ak). Then T (ak) = 0,T (bk) = 1 and T (0) = ck. Now
`Uk(σjk) ≥ `C\ak,bk(σjk) ≥ `C\0,1(T σjk) ≥ ε
by Lemma 1.16 and Lemma 1.18 while
`Uk(γk) ≤ `u(γ)
by Lemma 1.15. The last three inequalities yield that
`u(γ) ≤ ε
nk,
which is a contradiction since nk →∞. Thus bk/ak →∞. Choosing ak, bk as above with|ak| maximal and |bk| minimal the conclusion follows.
40
2 Hausdorff dimension of Julia sets and escaping sets
2.1 Hausdorff measure and Hausdorff dimension
A general reference for the material of this subsection is [19].We denote by diamA = supx,y∈A ‖x−y‖2 the (Euclidean) diameter of a subset A of Rd
and by B(a, r) = z ∈ Rd : ‖x − a‖ < r the ball of radius r around a point a ∈ Rd. Sofor a ∈ R2 = C we have B(a, r) = D(a, r).
Definition 2.1. Let A ⊂ Rd and δ > 0. A sequence (Aj) of subsets of Rd is called aδ-cover of A if diamAj < δ for all j ∈ N and
A ⊂∞⋃j=1
Aj.
For s > 0 let
Hδs (A) = inf
∞∑j=1
(diamAj)s : (Aj) is δ-cover of A
.
Then
Hs(A) = limδ→0
Hδs (A)
is called the s-dimensional Hausdorff measure of A.
Note that Hδs (A) is a non-increasing function of δ, so the limit defining Hs(A) exists.
We include the possibility that Hδs (A) = ∞, meaning that the series
∑∞j=1(diamAj)
s
diverges for all δ-covers (Aj) of A.If 0 < s < t, then Hs(A) ≤ δt−sHt(A) for all δ > 0. Hence Hs(A) = 0 if Ht(A) <∞.
Definition 2.2. Let A ⊂ Rd. Then
dimA = infs > 0: Hs(A) <∞
is called the Hausdorff dimension of A.
If follows that Hs(A) =∞ for 0 < s < dimA and Hs(A) = 0 for s > dimA.
Example 2.1. (Cantor middle third set) Let C0 = [0, 1] and for n ∈ N let Cn consistsof the intervals obtained by removing the middle third of each interval from Cn−1. Thus
41
C1 = [0, 13] ∪ [2
3, 1] and C2 = [0, 1
9] ∪ [2
9, 1
3] ∪ [2
3, 7
9] ∪ [8
9, 1]. It follows that Cn consists of 2n
intervals of length 1/3n. The set
C =∞⋂n=0
Cn
is called the Cantor middle third set. Since, by construction, the Cn are compact setssatisfying Cn ⊂ Cn−1, the set C is also compact. Given δ > 0, the set Cn is a δ-cover ofC if 1/3n < δ. Since Cn consists of 2n intervals of length 1/3n this yields
Hδs (C) ≤ 2n
(1
3n
)s= exp(n(log 2− s log 3)).
For s = (log 2)/(log 3) the right hand side is equal to 1 and thus Hs(C) ≤ 1. In particular,
dimC ≤ log 2
log 3.
The main tool to give lower bounds for the Hausdorff dimension is the followingtheorem.
Theorem 2.1. (Mass distribution principle) Let A ⊂ Rd and let µ be a Borel prob-ability measure on A such that there exist c, t, r0 > 0 with
µ(B(a, r)) ≤ crt
for all a ∈ A and all r ∈ (0, r0]. Then dim A ≥ t.
Proof. Let δ > 0 and let (Aj) be a δ-cover of A. We may assume that the Aj are closed(and thus in particular measurable) and that Aj ∩A 6= ∅ for all j, say aj ∈ A∩Aj. Then
1 = µ(A) ≤∑j
µ(Aj) ≤∑j
µ(B(aj, 2 diamAj)) ≤ c∑j
(2 diamAj)t
This implies that
Ht(A) ≥ 1
2tc
and hence dimA ≥ t.
Remark 2.1. In the above proved we used that Aj ⊂ B(aj, 2 diamAj) for a ∈ Aj andthus diamAj ≤ 2 diamB(aj, 2 diamAj). This shows that in the definition of the Hausdorffdimension it suffices to consider coverings by balls Aj. Similarly, it suffices to considercoverings by cubes.
We will apply the mass distribution principle (that is, Theorem 2.1) to measures µobtained as a weak limit of a sequence of measures. Here we say that a sequence (µn) of
42
probability measures on compact metric space X converges weakly to the measure µ onX if ∫
X
g dµn →∫X
g dµ
for all continuous functions g : X → R.We will use the following lemma.
Lemma 2.1. Let (µn) be a sequence of probability measures on compact metric space X.Then (µn) has a weakly convergent subsequence.
The proof uses the Riesz representation theorem for linear functional. Let C(X,R)be the space of continuous functions from X to R, endowed with the norm ‖ · ‖∞. Acontinuous linear functional L : C(X,R)→ R is called positive, if L(g) ≥ 0 for all g withg(x) ≥ 0 for all x ∈ X.
Lemma 2.2. Let L : C(X,R) → R be a positive linear functional. Then there exists ameasure µ such that
L(g) =
∫X
g dµ
for g ∈ C(X,R). Conversely, every measure µ on X defines a positive linear functionalL this way.
We mention that a corresponding result holds for linear functionals which are notpositive, using signed measures.
Sketch of proof of Lemma 2.1. Let (gk) be a dense sequence in C(X,R). Let (µ1k) be a
subsequence of (µn) such that (∫X
g1 dµ1k
)converges. Let (µ2
k) be a subsequence of (µ1k) such that(∫
X
g2 dµ2k
)converges. Inductively we choose a subsequence (µmk ) of (µm−1
k ) such that(∫X
gm dµmk
)converges. Choosing (µnk
) as the “diagonal sequence”, that is, µnk= µkk, we find that(∫
X
gm dµnk
)
43
converges for all m ∈ N. Since (gm) is dense in C(X,R) we conclude that
L(g) := limk→∞
∫X
g dµnk
converges for all g ∈ C(X,R). It is not difficult to see that L : C(X,R)→ R is a positive,continuous, linear functional. The conclusion now follows from Lemma 2.2.
Example 1.1 (continued). We will show that for the Cantor middle third set C weactually have
dimC =log 2
log 3.
Let µ0 be the Lebesgue measure restricted to C0 = [0, 1] and let µn the the Lebesguemeasure on Cn, scaled by a factor (3/2)n so that µn is a probability measure; that is,
µn(A) =
(3
2
)nmeas(A ∩ Cn)
with the Lebesgue measure meas(·). Note that if I is one of the 2n intervals that formCn, then
µn(I) =1
2n.
In fact, we have
µk(I) =1
2n
for all k ≥ n. Let now µ a the limit measure obtained from Lemma 2.1. (One can showthat (µn) actually converges and not only has a convergent subsequence, but we do notneed this result.) It follows that
µ(I) =1
2n
Let now x ∈ C and 0 < r ≤ 1. Choose n ∈ N such that 3−n < r ≤ 31−n. The length ofthe interval B(x, r) = (x− r, x+ r) is 2r and thus less or equal to 2 · 31−n = 6/3n. Sincethe intervals that form Cn have length 1/3n, we find that B(x, r) intersects at most 7 ofthese intervals. It follows with s = (log 2)/(log 3) that
µ(D(x, r)) ≤ 7
2n=
7
3sn≤ 7rs
and hence dimC ≥ s by the mass distribution principle (Theorem 2.1) as claimed.
Remark 2.2. The Cantor set C is related to iteration as follows: Let f : R → R be amap satisfying f(x) = 3x for 0 ≤ x ≤ 1/3 and f(x) = 3(1−x) for 2/3 ≤ x ≤ 1. Moreover,suppose that f(x) /∈ [0, 1] for x ∈ R\ ([0, 1/3] ∪ [2/3, 1]). For example, we may take
f(x) =3
2− 3
∣∣∣∣x− 1
2
∣∣∣∣ .44
Then Cn = x ∈ R : fk(x) ∈ [0, 1] for 0 ≤ k ≤ n and
C = x ∈ R : fk(x) ∈ [0, 1] for all k ∈ N.
The graph of the polynomial p(x) = λx(1 − x) with λ > 4 looks similar to that of f . Itcan be shown (and this is essentially Problem 1 of Problem set 7 of part 1 of this course)that
J(p) = x ∈ R : fp(x) ∈ [0, 1] for all k ∈ N.
One can estimate dim J(p) by modifying the methods of the above example.
The following result [25] is based on the same ideas that were used to compute theHausdorff dimension of the Cantor middle third set, but applied in a more general setting.In order to state it, consider for l ∈ N0 a collection El of finitely many pairwise disjointcompact subsets of Rn such that the following two conditions are satisfied:
(a) every element of El+1 is contained in a unique element of El;
(b) every element of El contains at least one element of El+1.
We say that the El satisfy the nesting conditions ; see Figure 12.
E0
E1
E2
Figure 12: Nested sets E0, E1, . . . .
For measurable sets X, Y ⊂ Rn with measY > 0 the density of X in Y is defined by
dens(X, Y ) =meas(X ∩ Y )
meas(Y ).
Theorem 2.2. For l ∈ N0 let El be a collection of finitely many pairwise disjoint compactsets satisfying the nesting conditions. Denote by Cl the union of all elements of El andput C =
⋂∞l=0Cl. Suppose that (∆l) and (dl) are sequences of positive real numbers such
that if F ∈ El, thendens(Cl+1, F ) ≥ ∆l
45
anddiamF ≤ dl.
Then
lim supl→∞
∑lj=0 | log ∆j|| log dl|
≥ n− dimC.
In the example of the Cantor middle third set we have ∆l = 23, dl = 3−l and n = 1.
Proof of Theorem 2.2. The idea is to apply Lemma 2.1 to a recursively defined sequence(µk) of measures. Here µk is supported on Ck and has the property that if F ∈ Ek−1,then µk|F∩Ck
is the suitably scaled Lebesgue measure on F ∩ Ck, scaled in such a waythat µk(F ) = µk−1(F ). In other words, assuming that µk−1 has been defined, µk−1(F ) is“distributed” onto the components of F ∩ Ck according to their size.
For the technical definition, we assume without loss of generality that measC0 = 1and and define µ0 as the Lebesgue measure on C0; that is, µ0(A) = meas(A∩C0). Assumethat µk−1 has been defined, so in particular µk−1 is a scaled Lebesgue measure on everyelement of Ek−1, say µk−1(A) = αF meas(A) if A ⊂ F ∈ Ek−1. We then put
µk(A) =µk−1(A ∩ Ck)dens(Ck, F )
.
Then
µk(A) =measF
meas(Ck ∩ F )µk−1(A ∩ Ck) =
µk−1(F )
µk−1(Ck ∩ F )µk−1(A ∩ Ck)
and hence µk(F ) = µk−1(F ) as claimed.Let now µ be the (weak) limit of convergent subsequence of (µk). We note here that
the sequence (µk) actually converges, but we do not need this result. Since µk is supportedon Ck, we find that the support of µ is contained in C.
Let now B ∈ Ek. Then µl(B) = µk(B) for l ≥ k and thus
(2.1) µ(B) = µk(B) for B ∈ Ek.
Denoting by Bk−1 the element of Ek−1 containing B ∈ Ek we find for A ⊂ B that
µk(A) =µk−1(A ∩ Ck)
dens(Ck, Bk−1)=
µk−1(A)
dens(Ck, Bk−1)≤ µk−1(A)
∆k−1
.
Induction yields that
µk(A) ≤ µ0(A)∏k−1j=0 ∆j
=measA∏k−1j=0 ∆j
for A ⊂ B ∈ Ek.
Together with (2.1) this yields that
µ(B) ≤ measB∏k−1j=0 ∆j
for B ∈ Ek.
46
We may assume that the sequence (dk) is decreasing. One may also assume thatdk → 0 and d0 < 1. For A ⊂ Rn with diamA < d0 we choose k such that
dk+1 ≤ diamA < dk
and put
A′ =⋃
B∈Ek+1
B∩A 6=∅
B.
Then A ∩ C ⊂ A ∩ Ck+1 ⊂ A′ and
diamA′ ≤ diamA+ 2dk+1 ≤ 3 diamA.
It follows that
µ(A) = µ(A ∩ C) ≤ µ(A′) =∑
B∈Ek+1
B∩A 6=∅
µ(B) =1∏k
j=0 ∆j
∑B∈Ek+1
B∩A 6=∅
measB =1∏k
j=0 ∆j
measA′.
Since A′ is contained in a cube of sidelength diamA′, we have measA′ ≤ (diamA′)n. Itfollows that
µ(A) ≤ 1∏kj=0 ∆j
(diamA′)n ≤ 3n(diamA)n∏k
j=0 ∆j
.
It follows that
(2.2) µ(A) ≤ 3n(diamA)n−s
if
k∏j=0
∆j ≥ (diamA)s
and thus in particular if
k∏j=0
∆j ≥ dsk.
The last inequality is equivalent to
k∑j=0
| log ∆j| = −k∑j=0
log ∆j = − log
(k∏j=0
∆j
)≤ − log dsk = | log dsk| = s| log dk|
and hence satisfied for large k if
(2.3) s > lim supk→∞
∑kj=0 | log ∆j|| log dk|
.
47
We deduce that (2.2) holds if s satisfies (2.3) and diamA is sufficiently small. Theconclusion now follows from the mass distribution principle.
Remark 2.3. In the nesting conditions the elements of El are assumed to be pairwisedisjoint. The proof of Theorem 2.2 shows that the conclusion also holds if we only requirethat if A,B ∈ El, then meas(A ∩B) = 0.
Theorem 2.3. Let X ⊂ Rn be compact and let f : X → X be Lipschitz continuous withLipschitz constant L > 1. Let ε > 0 and m ∈ N. Suppose that each y ∈ X has mpreimages x1, . . . , xm satisfying ‖xj − xk‖ ≥ ε for j 6= k. Then
dimX ≥ logm
logL.
Proof. For each finite subset E0 of X we can choose a finite subset E1 of f−1(E0) suchthat each point in E0 has m preimages in E1, with ‖x − x′‖ ≥ ε if x, x′ ∈ E1 withf(x) = f(x′). Clearly, cardE1 = m cardE0. Beginning with E0 = y for some fixedy ∈ X and performing this process repeatedly we obtain a sequence (Ek) of sets withcardEk = mk such that each point in Ek−1 has m preimages in Ek, with ‖x− x′‖ ≥ ε ifx, x′ ∈ Ek with f(x) = f(x′). We denote by δx the Dirac measure at a point x which isdefined by
δx(A) =
1 if x ∈ A,0 if x /∈ A.
For k ≥ 0 we define the measure µk by
µk :=1
mk
∑x∈Ek
δx.
By Lemma 2.1, the sequence (µk) has a (weakly) convergent subsequence, say µkj → µ.For x ∈ X we have f(B(x, r)) ⊂ B(f(x), Lr). For 0 < r < ε/2 we thus have
µk(B(x, r)) ≤ 1
mµk−1(f(B(x, r))) ≤ 1
mµk−1(B(f(x), Lr)).
Iterating this we obtain
µk(B(x, r)) ≤ 1
mlµk−l
(B(f l(x), Llr
))≤ 1
ml
for l ≤ k as long as Ll−1r < ε/2. Given r < ε/2 we choose j such that Lj−1r < ε/2 ≤ Ljr.Then
µk(B(x, r)) ≤ 1
mj
for k ≥ j and hence, with s = (logm)/(logL),
µ(B(x, r)) ≤ 1
mj= e−j logm = e−js logL ≤ exp
(s(
log r − logε
2
))=
(2
ε
)srs.
from which the conclusion follows by the mass distribution principle (Theorem 2.1).
48
2.2 Julia sets of rational functions
Identifying C with the unit sphere S2 in R3 via the stereographic projection ϕ : S2 → Cwe can consider the Julia set of a rational function as a compact subset of R3. However,we can also consider J(f)\∞ as a subset of C = R2. It is not difficult to see that thedimension is the same in both cases.
We will prove the following result [21].
Theorem 2.4. Let f be a rational function (of degree at least 2). Then dim J(f) > 0.
In order to apply Theorem 2.3 we show that rational functions are Lipschitz continu-ous.
Lemma 2.3. Let f : C → C be rational. Then f is Lipschitz continuous with respect tothe chordal metric.
Proof. Recall from part 1 of this lecture that for a function f meromorphic in a domaincontaining the trace of a curve γ : [a, b] → C the spherical length (f γ) of the curvef γ satisfies
(f γ) =
∫γ
f ](ζ)|dz| ≤ supζ∈tr(γ)
f ](ζ)
∫γ
|dz| = `(γ) supζ∈tr(γ)
f ](ζ)
with the Euclidean length `(γ) of γ. Considering
f ∗(z) = limζ→z
χ(f(ζ), f(z))
χ(ζ, z)= f ](z)
1 + |z|2
2= |f ′(z)| 1 + |z|2
1 + |f(z)|2
instead of the spherical derivative
f ](z) = limζ→z
χ(f(ζ), f(z))
|ζ − z|=
2|f ′(z)|1 + |f(z)|2
we find that
(f γ) =
∫γ
2f ∗(ζ)
1 + |z|2|dz| ≤ sup
ζ∈tr(γ)
f ∗(ζ)
∫γ
2
1 + |z|2|dz| = (γ) sup
ζ∈tr(γ)
f ∗(ζ).
We conclude that for a rational function f the spherical metric σ(·, ·), defined by
σ(a, b) = infγ
(γ),
with the infimum taken over all curves connecting a and b, satisfies
σ(f(a), f(b)) ≤ Lσ(a, b)
with
L = supζ∈C
f ∗(ζ).
Thus f is continuous with respect to the spherical metric and hence, since the sphericalmetric and the chordal metric are equivalent, also with respect to the chordal metric.
49
Remark 2.4. If ∞ ∈ F (f), which we can assume without loss of generality in the proofof Theorem 2.4, then f |J(f) is also Lipschitz continuous with respect to the Euclideanmetric. Of course, this follows from Lemma 2.3. However, it can be also proved directly:choose R > 0 with J(f) ⊂ D(0, R), let p1, . . . , pm be the poles of f and choose ε > 0such the disks D(pj, ε) are pairwise disjoint and do not intersect the Julia set. Then twopoints a, b ∈ J(f) can be connected by a curve γ in G := D(0, R+ ε)\
⋃mj=1D(pj, ε) with
`(γ) ≤ π|a−b|. With M = supz∈G |f ′(z)| we then have |f(a)−f(b)| ≤M`(γ) ≤Mπ|a−b|.
Proof of Theorem 2.4. We apply the previous lemma to X = J(f) ⊂ S2 ⊂ R3. As shown
in part 1 of this lecture, the exceptional set E(f) consisting of all z ∈ C for which thebackward orbit is finite is contained in F (f); that is, E(f) ∩ J(f) 6= ∅. The argumentsused there actually show that card f−3(z) ≥ 2 for all z ∈ J(f). Since J(f 3) = J(f) wemay thus assume that card f−1(z) ≥ 2 for all z ∈ J(f).
We claim that there exists ε > 0 such that for every z ∈ J(f) there exist a, b ∈ f−1(z)
such χ(a, b) ≥ ε. (Note that if a′, b′ ∈ S2 are the points corresponding to a, b ∈ C viastereographic projection, then χ(a, b) = ‖a− b‖ by the definition of the chordal metric.)
Assume that such an ε > 0 does not exist. Then there exist a sequence (zk) in J(f)and a sequence (εk) of positive real numbers tending to 0 such that f−1(zk) is containedin a ball of radius εk, say f−1(zk) ⊂ B(ak, εk). Passing to subsequences if necessary wemay assume that the sequences (zk) and (ak) converge, say zk → z and ak → a. Thenf−1(z) = a, contradicting card f−1(z) ≥ 2 for all z ∈ J(f). Thus ε > 0 as above existsand the conclusion follows from Theorem 2.3.
There are examples of rational functions f for which dim J(f) = 2. This is obviously
the case if J(f) = C, but there are also other examples.On the other hand, as shown by the following example, nothing more than dim J(f) >
0 can be said in general.
Example 2.2. For λ > 0 let fλ(z) = λ(z2 − 1). Then dim J(fλ)→ 0 as λ→∞.
Proof. Put ε = 4/λ. We may assume that λ > 8 so that ε < 12. For z ∈ C\(D(1, ε) ∪
D(−1, ε)) both |z+ 1| and |z−1| are greater than or equal to ε while at least one of theseterms is greater than or equal to 1, so that
|fλ(z)| = λ|(z + 1)(z − 1)| ≥ λε = 4
for z ∈ C\(D(1, ε) ∪D(−1, ε)). And if |z| ≥ 4, then
It follows that fnλ (z)→∞ as n→∞ if z ∈ C\(D(1, ε) ∪D(−1, ε)).This implies that J(fλ) ⊂ D(1, ε)∪D(−1, ε). By the complete invariance of J(fλ) we
thus have J(fλ) ⊂ f−nλ (D(1, ε) ∪D(−1, ε)) for all n ∈ N.There are two branches ϕ± of f−1
λ defined in D(1, ε), one mapping to D(1, ε) and onemapping to D(−1, ε). For both branches we have
|ϕ′±(z)| = 1
|f ′λ(ϕ(z))|=
1
2λ|ϕ(z)|)≤ 1
2λ(1− ε)=
1
2λ− 8<
1
λ.
50
For the branches ψ± of f−1λ defined in D(−1, ε) we get analogously |ψ′±(z)| ≤ 1/λ. Every
branch σn of f−nλ defined on one of the disks D(1, ε) and D(−1, ε) is the composition ofn such branches ϕ± and ψ±. Thus |σ′n(z)| ≤ 1/λn and hence diamσn(D(±1, ε)) ≤ 2ε/λn.We conclude that f−nλ (D(1, ε) ∪D(−1, ε)) and hence J(fλ) can be covered by 2 · 2n setsof diameter at most 2ε/λn.
For s > 0 and δ > 0 we thus have Hδs (J(fλ)) ≤ 2 · 2n · (2ε/λn)s = 21+sεs(2/λs)n if n is
large. For s = (log 2)/(log λ) we obtain Hs(J(fλ)) <∞. Thus
dim J(fλ) ≤log 2
log λ,
from which the conclusion follows.
2.3 Julia sets and escaping sets of entire functions
The analogue of Theorem 2.4 for transcendental entire function is the following resultwhich is an immediate consequence of Corollary 1.6.
Theorem 2.5. Let f be a transcendental entire function. Then dim J(f) ≥ 1.
Bishop [13] has recently shown that this result is sharp; that is, there exists a transcen-dental entire function f such that dim J(f) = 1. This example has multiply connectedwandering domains as discussed in §1.4, and the boundaries of these domains consist ofrectifiable curves.
McMullen used Theorem 2.2 to prove the following result.
Theorem 2.6. Let λ ∈ C\0. Then dim I(λez) = 2.
It follows from Theorem 1.10 that also dim J(λez) = 2.Besides Theorem 2.2, we shall also need the following theorem of Koebe, closely related
to the one quarter theorem.
Lemma 2.4. (Koebe distortion theorem). Let f : D→ C be holomorphic and injec-tive with f(0) = 0 and f ′(0) = 1. Then
|z|(1 + |z|)2
≤ |f(z)| ≤ |z|(1− |z|)2
and1− |z|
(1 + |z|)3≤ |f ′(z)| ≤ 1 + |z|
(1− |z|)3
for z ∈ D.
Again we have equality for the Koebe function. Note that |z|/(1+|z|2)→ 14
as |z| → 1.This can be used to deduce the Koebe one quarter theorem (Lemma 1.5) from the Koebedistortion theorem.
Using the function defined by (1.8) we can again transfer the statements to injectiveholomorphic functions defined on an arbitrary disk.
51
Lemma 2.5. Let a ∈ C, r > 0 and f : D(a, r) → C holomorphic and injective. Let0 < % < 1 and z ∈ D(a, %r). Then
1
(1 + %)2≤ |f(z)− f(a)||f ′(a)| · |z − a|
≤ 1
(1− %)2
and1− %
(1 + %)3≤ |f
′(z)||f ′(a)|
≤ 1 + %
(1− %)3.
After the statement of Koebe’s one quarter theorem we remarked that a normal familyargument shows that the conclusion holds with 1
4replaced by an unspecified constant K.
Similarly, a normal family argument shows that the quantities appearing Lemmas 2.4and 2.5 have lower and upper bounds depending only on |z| and ρ, respectively. Forexample, there exists constants C1 and C2 depending only on ρ such that if f is as inLemma 2.5, then C1 ≤ |f ′(z)/f ′(a)| ≤ C2 for z ∈ D(a, %r). These weaker estimates wouldsuffice for our applications.
Applying to Lemma 2.5, or to the weak version of it just mentioned, the same argumentthat we used to deduce Corollary 1.4 from the Harnack inequality (Lemma 1.12) yieldsthe following result.
Lemma 2.6. Let G ⊂ C be a domain and let K ⊂ G be compact. Then there existsC > 1 such that for any injective holomorphic function f : G→ R we have
maxz∈K|f ′(z)| ≤ C min
z∈K|f ′(z)|.
The constant C in Lemma 2.6 depends on both G and K. However, if G′ is a domainwith a compact subset K ′ such that there exists a biholomorphic map g : G → G′ withK ′ = g(K), then we can take the same constant C for G′ and K ′. This will be used inthe following proof in the special case that g is linear.
Proof of Theorem 2.6. Let f(z) = λez. For m,n ∈ Z we consider the square
P (T ) = R ∈ Q : R ⊂ T, dist(R, ∂T ) ≥ 10 and P (T ) =⋃
R∈P (T )
R,
and passing to a smaller value of η and a larger value of M if necessary, we may achievethat
dens(P (f(Qm,n)), f(Qm,n)
)≥ η for m ≥M ;
see Figure 13 for a sketch of P (f(R)) for a square R ∈ Q.
f
R
f(R)
Figure 13: A square R ∈ Q, its image f(R) and the squares in P (f(R)).
For such M we put R = QM,0,
E0 = R
and
E1 =f−1(S) : S ∈ P (f(R))
.
For T ∈ E1 we then have f(T ) ∈ P (f(R)). We now put
E2 =f−2(S) ∩ T : S ∈ P (f 2(T )), T ∈ E1
.
53
and, inductively,
Ek =f−k(S) ∩ T : S ∈ P (fk(T )), T ∈ Ek−1
.
By definition, the Ek satisfy the nesting conditions.For S ∈ Q the function f is actually injective in S ′, the open square which has the
same center but twice the sidelength as S, and
f(S ′) =reit : e2m−π/2 ≤ r ≤ e2m−π/2, |t| ≤ π
2
Moreover, if T ∈ P (f(S)), then 2T ∈ f(S), since dist(T, ∂f(S)) ≥ 10.
This shows that if T ∈ Ek so that S = fk(T ) ∈ Q, then fk+1 : T → f(S) is injective,and the inverse function g : f(S)→ T extends injectively to f(S ′). Applying Lemma 2.6with K = f(S) and G = f(S ′), we conclude that
dens(Ek+1, T
)= dens(g(P (f(S)), g(f(S))) ≥ 1
C2dens(P (f(S)), f(S)).
On the other hand
dens(P (f(S)), f(S)) ≥ η.
Thus
dens(Ek+1, T
)≥ η
C2dens(P (f(S)), f(S)),
so that we can apply Theorem 2.2 with ∆l = ∆ = η/C2.To estimate dl, we note that if z ∈ Q, then | arg z| ≤ π/4 and hence
Re f(z) ≥ 1
2
√2|f(z)| = |λ|
2
√2eRe z.
With c = |λ|√
2/2 and h(x) = cex we thus have Re f(z) ≥ h(Re z) and hence
Re f l(z) ≥ hl(Re z)
for l ∈ N. It follows that if T ∈ El and z ∈ T , then
|(f l)′(z)| =l∏
k=1
|fk(z)| ≥ hl(Re z) ≥ hl(M),
with hl(M)→∞ as l→∞, provided M is chosen large enough, so that the initial squareR is sufficiently far to the right. This implies that
diamT ≤ diam f l(T )
hl(z)=
π√
2
2hl(M)≤ 3
hl(M).
So we can take dl = 3/hl(M) in Theorem 2.2. It is easy to see that (log hl(M))/l → ∞as l→∞. The conclusion now follows from Theorem 2.2.
54
An entire function is said to be of finite order if there exists constants K and R suchthat |f(z)| ≤ exp(|z|K) for |z| ≥ R. The infimum of the set of all K such this holdsfor some R is called the order of f and denote by ρ(f), with inf ∅ = ∞, meaning thatρ(f) =∞ if there is no such K. It follows that
ρ(f) = lim supr→∞
log logM(r, f)
log r.
For example, we have ρ(exp) = 1. More generally, ρ(ep) = deg p if p is polynomial. Heredeg p denote the degree of a polynomial.
The following result proved in [6] and [35] is a substantial generalization of McMullen’sresult (Theorem 2.6) for the exponential function.
Theorem 2.7. Let f ∈ B be of finite order. Then dim I(f) = 2.
The idea of the proof of Theorem 2.6 was to consider a (small) square and its image,which is a (big) quarter annulus, and then to “pack” this quarter annulus with squareswhich are again mapped to quarter annuli. This procedure is then iterated. The choice tostart with a square is arbitrary and motivated only by the fact that the image of a squareis easy to write down explicitly. Instead, we could also have taken disks or other domains,and in turn we could also have fixed the image domains, instead of quarter annuli takingfor example (large) squares, disks or other domains. In the proof of Theorem 2.7 belowit will be convenient to consider domains which are mapped onto large squares. (Thearguments are from [10], simplifying those of [6, 35].)
The main tool used when dealing with the Eremenko-Lyubich class will again be thelogarithmic change of variable.
Lemma 2.7. Let f ∈ B and R > 0 with sing(f−1) ⊂ D(0, R) and R > |f(0)|. LetF : W → H = z : Re z > logR be the functions obtained from f by logarithmic changeof variable as in Theorem 1.10.
Let x0 = infRe z : z ∈ W and define h : [x0,∞)→ [logR,∞),
h(x) = maxy∈R
ReF (x+ iy).
Then h is increasing and convex. Moreover, if h is differentiable at x and zx ∈ W withh(x) = ReF (zx), then
h′(x) = F ′(zx).
Proof. The maximum principle implies that h is increasing. To prove that h is convex,let c ∈ R and put G : W → C, G(z) = F (z)− cz. Then
k(x) := maxy∈R
ReG(x+ iy) = h(x)− cx.
Let x0 < x1 < x < x2. The maximum principle yields that k(x) = maxk(x1), k(x2).We now choose c ∈ R such that k(x1) = k(x2); that is,
Let zx = x+ yx. Then y 7→ ReF (x+ iy) has a local maximum at yx and thus
∂ ReF
∂y(zx) = 0.
The Cauchy-Riemann equations yield that
F ′(zx) =∂ ReF
∂x(zx).
For the right derivative
h′+(x) = limt→0+
h(x+ t)− h(x)
t
we have
h′+(x) ≥ limt→0+
ReF (zx + t)− ReF (zx)
t=∂ ReF
∂x(zx) = F ′(zx).
The same argument yields that the left derivative satisfies
h′−(x) ≤ limt→0−
ReF (zx + t)− ReF (zx)
t= F ′(zx),
from which the conclusion follows.
Remark 2.5. For a convex function h defined on an interval I, the left and right deriva-tives h′+(x) and h′−(x) exist for all x ∈ I. This follows since the difference quotient(h(x+ t)−h(x))/t is a monotone function of t for t > 0 and t < 0. Moreover, the convex-ity implies that the left and right derivatives are non-decreasing and that h′−(x) ≤ h′+(x)for all x ∈ I. It follows that h is differentiable except possibly at a countable subset of I.
Remark 2.6. In terms of f , the convexity of h means that
(2.4) logM(r, f) ≤ log r2 − log r
log r2 − log r1
logM(r1, f) +log r − log r1
log r2 − log r1
logM(r2, f).
for r1 < r < r2. This actually holds for every entire function f , not only for f ∈ B,and can be deduced essentially by the same argument as above, applying the maximumprinciple to z 7→ log |f(z)|− c log |z|. This function is subharmonic, and to such functionsthe maximum principle applies. The inequality (2.4) is known has the Hadamard threecircles theorem. It says that logM(r, f) is a convex function of log r.
56
The hypothesis in Theorem 2.7 that f has finite order says that there exists K > 0such that
log |f(z)| ≤ |z|K
if |z| is large. In terms of the function F obtained from the logarithmic change of variableand the function h introduced in Lemma 2.7 this means that
h(x) = maxRe ζ=x
ReF (ζ) = maxRe ζ=x
log |f(eζ)| ≤= maxRe ζ=x
|eζ |K = eKx.
Lemma 2.8. Let h : [x0,∞) → R be a continuously differentiable, increasing functionsatisfying h(x) ≤ eKx for some K > 0. Let ε > 0. Then there exists E ⊂ [x0,∞)satisfying
(2.5) lim supx→∞
meas(E ∩ [x0, x])
x≤ ε
such that
h′(x) ≤ K
εh(x) for x /∈ E.
Proof. Let
E =
x ≥ x0 : h′(x) >
K
εh(x)
.
Then
meas(E ∩ [x0, x]) =
∫E∩[x0,x]
dt
≤∫E∩[x0,x]
εh′(t)
Kh(t)dt
≤ ε
K
∫ k
x0
h′(t)
h(t)dt
=ε
K(log h(x)− log h(x0))
≤ ε
K
(log eKx − log h(x0)
)= εx− ε
Klog h(x0),
form which the conclusion follows.
Remark 2.7. We want to apply Lemma 2.8 to the function h defined in Lemma 2.7. Butthis function is not continuously differentiable. However, the continuity was only usedto deduce for g = log h that g(x) − g(x0) =
∫ xx0g′(t)dt. In the standard version of the
fundamental theorem of calculus one assumes here that g′ is Riemann integrable. In our
57
application the function g is not differentiable, but the left and right derivatives exist andare Riemann integrable (since the light and left derivatives of h exist and are monotone).Moreover, they coincide except possibly for a countable set, so their integrals are equal.From this the desired conclusion follows.
Sketch of the proof of Theorem 2.7. Let F be the function obtained from f by the loga-rithmic change of variable and let R, H = z : Re z > logR, h and zx be as in Lemma 2.7.For w0 ∈ H we consider the square
(2.6) S(w0) =
w ∈ C : |Rew − Rew0| ≤
1
4Rew0, | Imw − Imw0| ≤
1
4Rew0
.
For x > x0 so large that S(F (zx)) ⊂ H we denote by G the branch of F−1 satisfyingG(F (zx)) = zx and we put
Q(zx) = G(S(F (zx))).
For large x the half-plane H actually contains the square around F (zx) which has twicethe sidelength of S(F (zx)), and G is univalent in that larger square.
F logR
H
U
W
S(F (zx))
F (zx)
zx
Q(zx)
Q(zx)−2πi
Figure 14: The points zx and F (zx), the square S(F (zx)) and its preimage Q(zx).
We can proceed as in the proof of McMullen’s Theorem 2.6 if, for any w0 ∈ H withsufficiently large real part, there exist x1, . . . , xn such that the density of the union ofthe domains Q(zx1), . . . , Q(zxn), together with their translates by multiples of 2πi, isbounded below. More precisely, denoting by P (S(w0)) the union of all domains of the formT = Q(zxj)+2πik with 1 ≤ j ≤ n and k ∈ Z for which T ⊂ S(w0) and dist(T, S(w0)) ≥ 10,we want dens(P (S(w0)), S(w0)) ≥ ∆ for some ∆ > 0.
Such an estimate would follow if Q(zx) ⊃ D(zx, δ) for some δ > 0 and all large x. Bythe Koebe one quarter theorem, we have
Q(zx) ⊃ D
(zx,
1
4|G′(F (zx))| ·
1
4ReF (zx)
)= D
(zx,
ReF (zx)
16|F ′(zx)|
)= D
(zx,
h(x)
16h′(x)
).
58
Lemma 2.8 now shows that given ε > 0 there exists E ⊂ [x0,∞) satisfying (2.5) such that16h′(x)/h(x) ≤ 1/δ and thus Q(zx) ⊃ D(zx, δ) for x /∈ E.
As in section 1.2, we now consider E(z) = λez where 0 < λ < 1/e. By Theorem 1.7and the results following it we have J(E) =
⋃s∈Σ H(s), where H(s) is either empty or of
the trace of some injective curve gs : [ts,∞)→ C satisfying gs(t)→∞ as t→∞. Such acurve gs is called a hair and gs(ts) is called its endpoint.
Theorem 2.6 yields that dim J(E) = 2. Karpinska [23] proved the surprising resultthat this changes if one removes the endpoints of the hairs from J(E).
Theorem 2.8. Let X be set of endpoints of the hairs of J(E). Then dim J(E)\X = 1.
For the proof we will use the following lemma, which will be treated in Problem Set 9.
Lemma 2.9. Let K ⊂ Rn be bounded, R > 0 and r : K → (0, R]. Then there exists an atmost countable subset L of K such that B(x, r(x)) ∩ B(y, r(y)) = ∅ for x, y ∈ L, x 6= y,and ⋃
x∈K
B(x, r(x)) ⊂⋃x∈L
B(x, 4r(x))
We shall deduce the following result from Lemma 2.9.
Lemma 2.10. Let K ⊂ Rn be bounded and let ρ > 1. Suppose that for all x ∈ K andδ > 0 there exist r(x) ∈ (0, 1), d(x) ∈ (0, δ) and N(x) ∈ N satisfying N(x)d(x)ρ ≤ r(x)n
such that B(x, r(x)) ∩ K can be covered by N(x) sets of diameter at most d(x). ThendimK ≤ ρ.
Proof. Choose R > 0 such that K ⊂ B(0, R) and let δ > 0. Since
K ⊂⋃x∈K
B
(x,
1
4r(x)
),
Lemma 2.9 yields the existence of an at most countable subset L of K such that
K ⊂⋃x∈L
B(x, r(x))
while
B
(x,
1
4r(x)
)∩B
(y,
1
4r(y)
)= ∅ for x, y ∈ L, x 6= y.
For each x ∈ L, let A1(x), A2(x), . . . , AN(x)(x) be the sets of diameter at most d(x) whichcover B(x, r(x)) ∩K so that N(x)d(x)ρ ≤ r(x)n. Then
K ⊂⋃x∈L
N(x)⋃j=1
Aj(x).
Now ∑x∈L
N(x)∑j=1
(diamAj(x))ρ ≤∑x∈L
N(x)d(x)ρ ≤∑x∈L
r(x)n.
59
Since r(x) ≤ δ we have B(x, 1
4r(x)
)≤ B
(0, R + 1
4δ)
for all x ∈ L. Since the ballsB(x, 1
4r(x)
), x ∈ L, are pairwise disjoint, this yields
∑x∈L
(1
4r(x)
)n≤(R +
1
4δ
)n.
We obtain ∑x∈L
N(x)∑j=1
(diamAj(x))ρ ≤ (4R + δ)n.
Thus the ρ-dimensional Hausdorff measure of K is finite. In particular, dimK ≤ ρ.
The key lemma to prove Theorem 2.8 is the following.
Lemma 2.11. Let δ > 0 and z ∈ J(E)\X. Then z ∈ I(E) and | ImEk(z)| ≤(ReEk(z)
)δfor all large k.
Proof. We have z = gs(t) for some admissible sequence s ∈ Σ and some t > ts. Chooset′ ∈ (ts, t). By (1.12) we have
limk→∞
Ek(t)
Ek(t′)=∞.
Choose c1 > max1/δ, 2. For large k we have Ek−1(t) ≥ c1Ek−1(t′) and hence
(2.7) Ek(t) = λ expEk−1(t) ≥ λ(expEk−1(t′)
)c1= λ1−1/c1Ek(t′)c1 ≥ λ1/2Ek(t′)c1 .
By the definition of ts and (1.12) we have
limk→∞
skEk(t′)
= 0.
The last two inequalities yield that
(2.8) |sk| ≤ Ek(t′)
for large k.It follows from the discussion after Lemma 1.9, in particular (1.17) and (1.18), that
with c2 = M/(β − 1) we have∣∣Ek(gs(t))− Ek(t)− 2πisk∣∣ ≤ c2.
This implies that z = gs(t) ∈ I(E). Moreover,∣∣ImEk(z)∣∣ ≤ 2π |sk|+ c2
and
Ek(t) ≥ ReEk(z) + c2.
60
Combining the last two inequalities with (2.7) and (2.8) yields∣∣ImEk(z)∣∣ ≤ 2πEk(t′) + c2 ≤ λ−1/2c1Ek(t)1/c1 + c2 ≤ λ−1/2c1
(ReEk(z) + c2
)1/c1+ c2,
from which the conclusion follows since c1 > 1/δ.
Sketch of proof of Theorem 2.2. Let M > 0 be large. For 0 < δ < 1 we consider Ωδ =x + iy : x > M, |y| < |x|δ. Note that if z ∈ J(E)\X, then Ek(z) ∈ Ωδ for large k byLemma 2.11.
Let Y = z ∈ Ωδ : Ek(z) ∈ Ωδ for all k ∈ N. For z0 ∈ Ωδ and k ∈ N we putzk = Ek(z0) and xk = Re zk. As in (2.6) we consider the squares
S(zk) =
z ∈ C : |Re z − Re zk| ≤
1
4xk, | Im z − Im zk| ≤
1
4xk
.
We can cover S(zk) ∩ Ωδ by Nk = Nk(z0) squares of sidelength(
52xk)δ
, where
Nk ≤12xk(
52xk)δ + 1 ≤ x1−δ
k .
Let g be the branch of (Ek)−1 which maps zk to z0. This extends to z ∈ C : Re z > β.Put tk(z0) = |g′(zk)| = 1/|(Ek)′(z0)|. Noting that (Ek)′(z0) =
∏kj=1E
j(z0) and zk =
Ek(z0) we find that 1/x1+δk ≤ tk(z0) ≤ 1/xk for large k.
Similarly as before we deduce from Koebe’s theorems that
g(Sk) ⊃ D
(z0,
1
16tk(z0)xk
)⊃ D
(z0,
1
16xδk
)⊃ D
(z0,
1
xδ/2k
)
and that the images of the squares of sidelength(
52xk)δ
under g have diameter at mostdk = dk(z0) where
dk = 12tk(z0)√
2
(5
2xk
)δ≤ 60xδ−1
k ≤ x2δ−1k .
Withrk = x
δ/2k ,
and Nk and dk as before, we thus see that D(z0, rk) can be covered by Nk sets of diameterat most dk. For ρ = 1 + 3δ and 0 < δ < 1
6we have
Nkdρk ≤ x1−δ
k x(2δ−1)ρk = x−2δ+6δ2
k ≤ xδk = r2k.
Lemma 2.10 now implies that every bounded subset of Y has Hausdorff dimension atmost ρ. Thus dimY ≤ ρ = 1 + 3δ. Since δ can be chosen arbitrarily small, this yieldsthat dimY ≥ 1. Since Y contains curves to ∞, we actually have dimY = 1.
This implies that dimE−1(Y ) = 1 and in fact that dimE−k(Y ) = 1 for all k ∈ N.Hence dim
(⋃∞k=0E
−k(Y ))
= 1. Since J(E)\X ⊂⋃∞k=0E
−k(Y ) by Lemma 2.11, theconclusion follows.
61
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