Size of the Julia set of a structurally finite transcendental entire function

Masahiko Taniguchi

Graduate School of Science, Kyoto University 606 Kyoto, Japan

Abstract In this note, we will show that, for every structurally finite transcendental entire function, the Hausdorff dimension of its Julia set is two.

1 Introduction and main results

In [8] we defined structurally finite entire functions and proved that every such f(z) of type (p, q) can be written as an indefinite integral

f(z) = ∫ ^{z} P (t)e^{Q(t)}dt

with polynomials Q(z) of degree q and P (z) of degree p. For the definitions and details, see [8] and [9].

Since we are interested in dynamical properties of such an f(z), we may assume, taking conjugation by a similarlity if necessary, that

f(z) = a ∫_{0}^{z} P (t)e^{Q(t)}dt + b

with monic polynomials P and Q of degrees p and q. Here we exclude the case of linear polynomials (the case that p = 0, q = 0). In the sequel, we denote by f^{k}(z) the k-th iteration of f(z) and by J_{f} the Julia set of f(z). See for instance [3] for the basic facts on dynamical properties of entire functions.

Now in this note, we give a proof of the following

Theorem 1 For every transcendental structurally finite entire function f(z), the Hausdorff dimension of J_{f} is two.

Remark Compare with a theorem of Stallard ([5] II): For every transcendental entire function f(z) with a bounded set of singular values, the Hausdorff dimension of J_{f} is greater than 1. (Here we say that α is a singlar value of f(z) if every neighborbood of α is not evenly covered under f : ℂ → ℂ.)

Also, we note the following

Proposition 2 Let f(z) be a (not necessarily transcendental) structurally finite entire function. If f(z) is hyperbolic, namely if the union S^{+}(f) of the orbits of all singular values is relatively compact in ℂ and the distance between S^{+}(f) and J_{f} is positive, then the Julia set J_{f} has vanishing area.

Devaney and Keen proved in [1] that, if the Schwarzian derivative of a meromorophic f(z) is a polynomial (in particular, if f(z) is a structurally finite entire function) and if f(z) is hyperbolic, then the Julia set J_{f} has vanishing area. The proof of Proposition 2 is essentially the same as those in [1] and [2]. Hence we omit the proof. Also see [6].

Acknowledgment: The author would like to express profound thanks to Professor Yasuhiro Gotoh for valuable comments.

2 Proof of the main theorem

Now to prove theorem 1, it suffices to show that the Hausdorff dimension of the set of escaping points of f(z) in

Π = \{z = re^{iθ} | r > 0, |θ| ≤ \frac{π}{4q} \}

is two, because f(z) is in the Speiser class (cf. for instance, [3]).

Fix a positive constant ϵ (< 1/4). Next set

Q_{θ}(t) = Re Q(e^{iθ}t) = cos(qθ) t^{q} + ⋯

and

R_{θ}(t) = |(\frac{P′Q′ − P Q′′}{(Q′)^{3}} )′ (e^{iθ}t)|^{2} |e^{Q(e^{iθ}t)}|^{2}

for every θ with |θ| ≤ π/4q. Then, unless R_{θ}(t) equals identically to 0, we can write R_{θ}(t) as

\frac{A_{θ}(t)}{B_{θ}(t)} e^{2Q_{θ}(t)}

with real polynomials A_{θ}(t) and B_{θ}(t) of degree at most 2(p + 4q − 6) (≥ 0 unless R_{θ} equals identically to 0) and of degree 2(6q − 6) ≥ 0, respectively. Since the term of the highest degree of A_{θ}(t) and that of B_{θ}(t) have positive coefficients independent of θ, we can find a finite t_{0} such that

R′_{θ}(t) > 0

unless R_{θ}(t) equals identically to 0, and 2.

R_{θ}(t) ≤ 2c_{∞} t^{2(p−2q)}e^{2Q_{θ}(t)}

for every θ with |θ| ≤ π/4q and every t not less than t_{0}, where we set

c_{∞} = max \{1, lim_{t→+∞} t^{−2(p−2q)} \frac{A_{θ}(t)}{B_{θ}(t)}\} ,

which is independent of θ as noted above. (Here the first condition is satisfied for every sufficiently large t, since

R′_{θ}(t) = \frac{(A′_{θ}B_{θ} − A_{θ}B′_{θ}) + 2A_{θ}B_{θ}Q′_{θ}}{(B_{θ})^{2}} e^{2Q_{θ}(t)},

and the term of highest degree of

(A′_{θ}B_{θ} − A_{θ}B′_{θ}) + 2A_{θ}B_{θ}Q′_{θ}

is that of 2A_{θ}B_{θ}Q′_{θ} which has a positive coefficient depending only on f(z) and cos qθ.)

Also fix an R so large that R satisfies the following inequalities:

1. R > \frac{5}{2} q (32q/π) (> 8πq > 16q), R > \frac{1}{2} e^{6^{q}/2}

2. R > 4q/ϵ,

3. R > t_{0}, and

4. for every z ∈ Π with |z| ≥ R, (a)

max \{\frac{q}{2}(|z| + 1)^{q−1}, 1\} < |Q′(z)| < 2q(|z|−1)^{q−1}, |\frac{Q′′(z)}{Q′(z)}| < \frac{2q}{|z| + 1}

(b)

|\frac{P′(z)}{P (z)} + Q′(z)| < 2q(|z| − 1)^{q−1},

(c) |z|^{p−q}|e^{Q(z)}| > 1,

\frac{|a|}{2q} |z|^{p−q+1}|e^{Q(z)}| > max \{e^{|z|^{q}/2}, \frac{4|aP (z)e^{Q(z)}|}{|z|^{q}}\} ,

\frac{|aP (z)e^{Q(z)}|}{(|z| + 1)^{q}} > 4π (> 1).

Now, set

S = Π ∩ {|z| ≥ R + 1},

and

T = S ∩ f^{−1}(Π).

Definition We say that the image f(B) of a disk B is nearly disk-shaped with respect to ϵ if f′(z_{B}) ≠ 0 and the distortion function

V (f, B) = sup_{z∈B,z≠z_{B}} |\frac{f(z) − f(z_{B})}{z − z_{B}} − f′(z_{B})|

of f(z) on B is bounded by ϵ|f′(z_{B})|. Here and in the sequel, ”disks” always mean closed disks and z_{B} is the center of a disk B.

Lemma 3 For every disk B with center z_{B} ∈ S and radius |z_{B}|^{−q}, the image f(B) is nearly disk-shaped (with respect to ϵ).

In particular, f(B) contains a disk with center f(z_{B}) and radius

\frac{3}{4} |f′(z_{B})||z_{B}|^{−q},

and is contained in a disk with the same center and radius

\frac{5}{4} |f′(z_{B})||z_{B}|^{−q}.

Proof. First we note that, if z ∈ B, then |z − z_{B}| ≤ |z_{B}|^{−q} (< 1). Hence by the condition 4-(b) for R, we have

|N(f)(z)| (= |\frac{f′′(z)}{f′(z)}|) = |\frac{P′(z)}{P (z)} + Q′(z)| < 2q|z_{B}|^{q−1}

on B, which implies by the condition 1 for R that

|log \frac{f′(z)}{f′(z_{B})}| = |∫_{z_{B}}^{z} \frac{f′′(t)}{f′(t)} dt| < 2q|z_{B}|^{q−1} · |z_{B}|^{−q} = 2q|z_{B}|^{−1}(< 1/8),

and hence that

|\frac{f′(z)}{f′(z_{B})} − 1| < 2 |log \frac{f′(z)}{f′(z_{B})}| < 4q|z_{B}|^{−1} < \frac{1}{4}.

Thus we conclude that

f(z) − f(z_{B}) = f′(z_{B}) ((z − z_{B}) + ∫_{z_{B}}^{z} ϵ(t)dt)

with an error function ϵ(t) such that (by the condition 2 for R)

|ϵ(t)| < 4q|z_{B}|^{−1} < ϵ

on B, which implies that the image f(B) is nearly disk-shaped with respect to ϵ. In particular, since |ϵ(t)| < 1/4, we conclude the second assertion.

Lemma 4 For every z ∈ S, we can represent f(z) as

\frac{a}{q}z^{p−q+1}(1 + e_{f}(z))e^{Q(z)}

with an error function e_{f}(z) satisfying

|e_{f}(z)| ≤ \frac{C_{1}}{|z|}

for every z with |z| ≥ R, where (and in the proof below) C_{k} are constants depending only on f(z) and R.

Proof. First, fix a z ∈ S. By the condition 4-(a) for R, |Q′(z)| > 1 on the ray

ℓ_{z} = {te^{iθ} | R ≤ t ≤ |z|, θ = arg z}.

Integrating on this ray, we have

f(z) = a ∫_{ℓ_{z}} P (t)e^{Q(t)}dt + C(θ) + b
= a [\frac{P (t)}{Q′(t)} e^{Q(t)}]_{R}^{z} − a ∫_{ℓ_{z}} (\frac{P}{Q′} )′ (t)e^{Q(t)}dt + C(θ) + b
= a\frac{P (z)}{Q′(z)} e^{Q(z)} − a (\frac{P′Q′ − P Q′′}{(Q′)^{3}} ) (z)e^{Q(z)} + I(z)
+C(θ) + C̃(θ) + b,

where

I(z) = a ∫_{ℓ_{z}} (\frac{P′Q′ − P Q′′}{(Q′)^{3}} )′ (t)e^{Q(t)}dt,

C(θ) = a ∫_{0}^{R} P (re^{iθ})e^{Q(re^{iθ})}e^{iθ}dr,

and

C̃(θ) = −a\frac{P (Re^{iθ})}{Q′(Re^{iθ})} e^{Q(Re^{iθ})} + a (\frac{P′Q′ − P Q′′}{(Q′)^{3}} ) (Re^{iθ})e^{Q(Re^{iθ})}.

Now for the first term in the right hand side, we have an estimate

|a\frac{P (z)}{Q′(z)} − \frac{a}{q}z^{p−q+1}| ≤ C_{2}|z|^{p−q}

and similarly the second term is bounded by C_{3}|z^{p−2q+1}e^{Q(z)}|, on S with suitable C_{2} and C_{3}. (Here we assume that C_{3} = 0 if p = 0 and q = 1.)

Next, by the definition of t_{0} and the condition 3 for R, R_{θ}(t) is strictly increasing on [t_{0}, +∞) for every θ with |θ| ≤ π/4q. Hence, we conclude that

|I(z)| ≤ |a| ∫_{R}^{|z|} \sqrt{R_{θ}(t)}dt ≤ C_{4}|z^{p−2q+1}e^{Q(z)}|

with a suitable C_{4}. (We assume that C_{4} = 0 if p ≤ 1 and q = 1.) Finally, |C(θ)| and |C̃(θ)| are continuous on [−π/(4q), π/(4q)], and hence are bounded by a constant C_{5} for every such θ. Thus, by noting the condition 4-(c) for R, we have a desired estimate by setting C_{1} = C_{2} + C_{3} + C_{4} + 2C_{5} + |b|.

Now we take another R_{1} > R + 2π such that

1.

\frac{C_{1}}{R_{1}} < sin \frac{π}{8q} (< \frac{1}{2})

and, if p − q + 1 ≠ 0, then

2.

R_{1} sin \frac{π}{8|p − q + 1|} > 4π.

In particular, Lemma 4 and the condition 1 for R_{1} give that

\frac{|a|}{2q} |z|^{p−q+1}|e^{Q(z)}| < |f(z)| < \frac{2|a|}{q} |z|^{p−q+1}|e^{Q(z)}|

for every z with |z| ≥ R_{1}. Furthermore, fixing a value A of arg a, we have by Lemma 4 a continuous branch of the arg f(z) on S such that

| arg f(z) − [(p − q + 1)θ + Im Q(z) + A]| ≤ sin^{−1} (\frac{C_{1}}{|z|}) .

Hence we have the following

Lemma 5 T contains

T′ = \{z ∈ Π | |z| ≥ R_{1}, |(p − q + 1)θ + Im Q(z) + A − 2mπ| ≤ \frac{π}{8q} , m ∈ ℤ\}.

Proof. If z ∈ T′, the condition 1 for R_{1} gives that

| arg f(z) − 2mπ| < sin^{−1} (\frac{C_{1}}{|z|}) + \frac{π}{8q} < \frac{π}{4q}

with some m ∈ ℤ, which implies that f(z) ∈ Π. Thus T contains T′.

If p − q + 1 = 0, set

T′′ = \{z ∈ Π | |z| ≥ R_{1}, |Im Q(z) + A − 2mπ| ≤ \frac{π}{16q} , m ∈ ℤ\},

which is clearly contained in T′.

If p − q + 1 ≠ 0, for every j = −2|p − q + 1|, ⋯, 2|p − q + 1| − 1, set

B_{j} = \frac{(2j + 1)π}{16q|p − q + 1|},

and

S_{j} = \{re^{iθ} ∈ Π | r ≥ R_{1}, B_{j} − \frac{π}{16q|p − q + 1|} ≤ θ ≤ B_{j} + \frac{π}{16q|p − q + 1|}\} .

Then we have the following

Lemma 6 If p − q + 1 ≠ 0, then

⋃_{j=−2|p−q+1|}^{2|p−q+1|−1} S_{j}

is coincident with S′ = S ∩ {|z| ≥ R_{1}}, and T′ ∩ S_{j} contains

T′′_{j} = \{z ∈ S_{j} | |(p − q + 1)B_{j} + Im Q(z) + A − 2mπ| ≤ \frac{π}{16q} , m ∈ ℤ\}.

Proof. The first assertion is clear. Next if z ∈ T′′_{j}, then

|(p − q + 1)θ + Im Q(z) + A − 2mπ|
< \frac{π}{16q} + |(p − q + 1)B_{j} + Im Q(z) + A − 2mπ| ≤ \frac{π}{8q} ,

which implies that z ∈ T′.

Thus set

T′′ = ⋃_{j=−2|p−q+1|}^{2|p−q+1|−1} T′′_{j},

if p − q + 1 ≠ 0, and we conclude that T′′ ⊂ T′.

Next set

η_{1} = \frac{π}{32q}

and T_{∗} be the subset of the image Q(T′′) of T′′ under Q(z) consisiting of all such z that the disk with center z and radius η_{1} is contained in Q(T′′). Further set

T^{∘} = Q^{−1}(T_{∗}) ∩ S′.

Then we have the following

Lemma 7 Every disk B with center z_{B} ∈ T^{∘} and radius |z_{B}|^{−q} is contained in T′′.

Proof. By the condition 4-(a) for R, we have (as in the proof of Lemma 3) that

|log \frac{Q′(z)}{Q′(z_{B})}| ≤ |z_{B}|^{−q} max_{z∈B} |N(Q)(z)| < max_{z∈B} |N(Q)(z)| < 2q|z_{B}|^{−1}

for every z ∈ B, which implies that the distortion of Q on B satisfies

V (Q, B) < 4q|z_{B}|^{−1} < ϵ.

Thus Q(B) is nearly disk-shaped (with respect to ϵ), and in particular, Q(B) is contained in a disk with center Q(z_{B}) ∈ T_{∗} and radius

\frac{5}{4} |Q′(z_{B})||z_{B}|^{−q} < \frac{5}{2} q|z_{B}|^{−1} < η_{1}

by the conditions 4-(a) and 1 for R. Hence we conclude that Q(B) ⊂ Q(T′′), which means that B ⊂ T′′.

Here, we recall the following well-known covering lemma. See for instance,

[7] Lemma in 1.6 of Chapter I.

Lemma 8 Let X be a measurable set in ℂ, and B = {B_{j}} be a countable covering of X by disks with bounded radii. Then we can find an absolute constant C (depending neither on X nor on B) and a subset {B_{j_{k}}} of B such that B_{j_{k}} are mutually disjoint and

∑_{k} m(B_{j_{k}}) ≥ Cm(X),

where m(E) is the the area ∫_{E} dxdy for every measurable set E in ℂ.

Lemma 9 There is a δ > 0 such that, for every disk D with center z_{D} ∈ S′ and radius r(D) greater than 2π, we have

ρ(D ∩ T^{∘}, D) ≥ δ.

Here and in the sequel, the density of a measurable subset X of a mesurable set Y is defined by

ρ(X, Y) = \frac{m(X)}{m(Y)}.

Proof. First note that T_{∗} contains the intersections

E = ⋃_{j} S_{j}^{∗} ∩ T_{j}^{strips},

of

S_{j}^{∗} = {z ∈ Q(S_{j}) | d(z, ℂ − Q(S_{j})) ≥ η_{1}}

(where d(z, E) is the distance from z to the set E) and periodic parallel strips with period 2π;

T_{j}^{strips} = \{|(p − q + 1)B_{j} + Im z + A − 2mπ| ≤ \frac{π}{32q} (= \frac{π}{16q} − η_{1}) , m ∈ ℤ\}

for every j. Hence we can find a δ′ > 0 such that

ρ(D^{∗} ∩ E, D^{∗}) ≥ δ′

for every disk D^{∗} with center in Q(S′) and radius greater than 2π. Indeed, for every z ∈ Q(S′) and r ≥ 2π, set

F (z, r) = ρ(D(z, r) ∩ E, D(z, r)),

where D(z, r) is the disk with center z and radius r. Then F (z, r) is always positive by the condition 2 for R_{1}, and continuous on Q(S′)×[2π, +∞). Also it is easy to see that F (z, r) is bounded away from 0 when |z| or r tends to +∞.

In particular, if q = 1 (and hence Q(z) = z), then the assertion of Lemma 9 is clear from the definition of T^{∘} with δ = δ′. So we consider the case that q > 1. Let D′ be the disk with center z_{D} and radius r(D) − 1/2. Take a covering of the compact set D′ ∩ S′ by a finite number of open disks with center in D′ ∩ S′ and radius 1/2. Then by the covering lemma (Lemma 8), we can find a finite set {B_{m}}_{m=1}^{M} consisting of disjoint disks with center z_{m} in D′ ∩S′ and radius 1/2 satisfying that D̃ = ∪_{m=1}^{M} B_{m} is contained in D and

m(D̃) ≥ C m(D′ ∩ S′).

On the other hand, it is clear from the shape of S′ that there is a positive constant δ_{0} depending only on S′ such that

ρ(D′′ ∩ S′, D′′) ≥ δ_{0}

for every D′′ with center in S′ and radius greater than 2π − 1/2. Hence we conclude that

ρ(D̃, D) ≥ Cρ(D′ ∩ S′, D) ≥ δ_{1} = (1 − (\frac{1}{4π} ))^{2} Cδ_{0}.

Now for every B_{m}, we can show as in the proof of Lemma 7 that Q(B_{m}) is nearly disk-shaped (with respect to ϵ), and in particular, Q(B_{m}) contains a disk with center Q(z_{m}) and radius

\frac{3}{4} |Q′(z_{m})| · \frac{1}{2} > \frac{3}{16}q|z_{m}|^{q−1} > \frac{3q}{16}R (> \frac{3q}{16}8πq > 3qπ) > 2π

by the conditions 1 and 4-(a) for R, and is contained in a disk with the same center and radius (5/8)|Q′(z_{B})|. (Here recall that q ≥ 2.) Hence, we conclude that

ρ(Q(B_{m}) ∩ E, Q(B_{m})) ≥ (\frac{3}{5})^{2} δ′.

Since

|\frac{Q′(z)}{Q′(z_{m})} − 1| < \frac{1}{4}

on D_{m}, we conclude that

ρ(B_{m} ∩ T^{∘}, B_{m}) ≥ (\frac{3}{5})^{4} δ′.

Thus in the case that q > 1, we have the assertion with

δ = (\frac{3}{5})^{4} δ′δ_{1}.

Now we give a proof of the main theorem.

For this purpose, take a disk B with center z_{B} ∈ T′ and radius |z_{B}|^{−q} arbitrarily. Then we have shown in Lemma 3 that f(B) is nearly disk-shaped (with respect to ϵ), and contains a disk B′ with center f(z_{B}) ∈ S′, whose absolute value can be written as

\frac{|a|}{q}|z_{B}|^{p−q+1}|(1 + e_{f}(z_{B}))e^{Q(z_{B})}|

with |e_{f}(z_{B})| < 1/2 by Lemma 4 and the condition 1 for R_{1}, and radius

r(B′) = \frac{3}{4} |f′(z_{B})||z_{B}|^{−q} − \frac{1}{8} > \frac{5}{8} |f′(z_{B})||z_{B}|^{−q}

by the condition 4-(d) for R, and is contained in a disk B′′ with the same center and radius

\frac{5}{4} |f′(z_{B})||z_{B}|^{−q}.

Note that r(B′) is greater than 2π by the condition 4-(d) for R and that the distance between B′ and ℂ − f(B) is greater than 1/8 by the definition.

In particular, Lemma 9 gives that

ρ(B′ ∩ T^{∘}, B′) ≥ δ,

which implies that

ρ(B′ ∩ T^{∘}, B′′) ≥ \frac{δ}{4}.

Thus we conclude the following

Lemma 10 There is a positive constant δ′′ depending only on f(z) and R_{1} such that, for every disk B with center z_{B} ∈ T′ and radius |z_{B}|^{−q}, there exists a packing {B_{m}}_{m=1}^{M} of T′ ∩ f(B) with a finite number of disjoint disks B_{m} in T′′(⊂ T′) with center z_{m} in T^{∘} and radius |z_{m}|^{−q} such that

ρ ( ⋃_{m=1}^{M} B_{m}, f(B)) ≥ δ′′.

Moreover, for every m,

|f(z_{m})| ≥ \frac{|f(z_{B})|}{2}.

Proof. First, we cover T^{∘} by open disks with center z ∈ T^{∘} and radius |z|^{−q}. Then since the above B′ is compact, we can select a finite cover of B′ ∩ T^{∘}. Then by Lemmas 7 and 8, we can find a packing {B_{m}}_{m=1}^{M} of T′′ by a finite number of disks B_{m} with center z_{m} ∈ T^{∘} and radius |z_{m}|^{−q} such that

ρ ( ⋃_{m=1}^{M} B_{m}, f(B)) ≥ Cρ(B′ ∩ T^{∘}, B′′) ≥ \frac{1}{4}Cδ.

Here note that, since |z_{m}|^{−q} < 1/8 by the condition 1 for R, every such disk B_{m} is contained in f(B), and hence {B_{m}}_{m=1}^{M} is actually a packing of T′′ ∩ f(B). Thus we have the first assertion with δ′′ = Cδ/4.

Next, every point z in f(B), Lemmas 3, 4 and the conditions 4-(d) for R, 1 for R_{1} imply that

|f(z)| > |f(z_{B})| − \frac{5}{4} |f′(z_{B})||z_{B}|^{−q} > \frac{|f(z_{B})|}{2},

which shows the second assertion.

Now we define a family E_{k} consisting of a finite number of disjoint compact sets inductively. For every B as in Lemma 10, we denote by P(f(B)) the packing of f(B) obtained in Lemma 10. Fix such a disk B_{∗} and set E_{0} = {B_{∗}}. Define

E_{k} = {G | G ⊂ F ∈ E_{k−1}, f^{k}(G) ∈ P(f^{k}(F))}.

Then

E_{∞} = ⋂_{k=1}^{∞} ( ⋃_{G∈E_{k}} G)

consists of escaping points whose orbit is contained in S, and Lemma 10 implies that

ρ (( ⋃_{G∈E_{k},G⊂F} f^{k}(G)) , f^{k}(F)) ≥ δ′′

for every k and F ∈ E_{k−1}.

Here we note the following distortion lemma.

Lemma 11 (Distortion Lemma) Let {D_{j}}_{j=0}^{n} be a sequence of disks with radii ℓ_{j} and {f_{j}} be a sequence of univalent functions on neighborhoods of D_{j}. Assume that ℓ_{j} ≤ ℓ_{0} < 1/5,

f_{j}(D_{j}) ⊃ D_{j+1}, ℓ_{j} · |f′_{j}(z)| > 1

on D_{j}, and

ℓ_{j} · max_{z∈D_{j}} |N(f_{j})(z)| < ϵ

for every j. Then letting

ϕ = ϕ_{n,0} = (f_{n} ∘ ⋯ ∘ f_{0})^{−1} : D = D_{n+1} → D′_{0} = ϕ(D_{n+1}),

we have

|\frac{ϕ′(z)}{ϕ′(z_{D})} − 1| < ϵ

on D, and hence the distortion V (ϕ, D) of ϕ(z) is bounded by ϵ. Proof. Let w_{0} be the center of D, and on D we set

ϕ_{n,j} = (f_{n} ∘ ⋯ ∘ f_{j})^{−1} (j = 0, ⋯, n)

and D′_{j} = ϕ_{n,j}(D). Then since

|ϕ′_{n,j}(z)| < ℓ_{j} ⋯ ℓ_{n}

on D, the diameter d(D′_{j}) of D′_{j} is not greater than

ℓ_{j} ⋯ ℓ_{n}(2ℓ_{n+1}) ≤ ℓ_{0}^{n−j+1}(2ℓ_{j}).

On the other hand,

\frac{ϕ′(w)}{ϕ′(w_{0})} = ∏_{j=0}^{n} \frac{(f_{j})′(ϕ_{n,j}(w_{0}))}{(f_{j})′(ϕ_{n,j}(w))},

and each factor on D has the following estimate:

|log \frac{(f_{j})′(ϕ_{n,j}(w_{0}))}{(f_{j})′(ϕ_{n,j}(w))}| = |∫_{ϕ_{n,j}(w)}^{ϕ_{n,j}(w_{0})} \frac{f′′_{j}(t)}{f′_{j}(t)}dt|

≤ d(D′_{j}) max_{z∈D_{j}} |N(f_{j})(z)| < 2ℓ_{0}^{n−j+1}ϵ.

Thus we have

|log \frac{ϕ′(w)}{ϕ′(w_{0})}| < ∑_{j=0}^{n} 2ℓ_{0}^{n−j+1}ϵ < \frac{2ℓ_{0}ϵ}{1 − ℓ_{0}} ,

which implies the assertion as in the proof of Lemma 3, since 2ℓ_{0}/(1 − ℓ_{0}) < 1/2 if ℓ_{0} < 1/5.

Lemma 12 We can find a δ_{∗} > 0 independent of k such that

ρ (( ⋃_{G∈E_{k}} G) ∩ F, F ) ≥ δ_{∗},

for every F ∈ E_{k−1}.

Proof. Fix k and F ∈ E_{k−1} arbitrarity. Then for every G_{0} ∈ E_{k} such that G_{0} ⊂ F, we can find a sequence of elements G_{k−j} of E_{j} (j = 0, ⋯, k) such that D_{k} = B_{∗}, and for every other j, D_{j} = f^{j}(G_{k−j}) is a disk in P(f^{j}(G_{k+j−1})). Let z_{j} be the center of D_{j}, (and hence |z_{j}|^{−q} is the radius of D_{j}). Then by the condition 4-(d) for R gives that

|z_{j}|^{−q}|f′(z)| (> |f′(z)|(|z| + 1)^{−q}) > 1

and the conditions 2 and 4-(b) for R give that

|z_{j}|^{−q}|N(f(z))| < 2q|z_{j}|^{−1} < ϵ

for every z ∈ D_{j}.

Thus by the above distortion lemma, we conclude that

\frac{3}{4}|(f^{−k})′(z_{k})| < |(f^{−k})′(z)| < \frac{5}{4}|(f^{−k})′(z_{k})|

on D_{k} = f^{k}(G_{0}) for every G_{0} ∈ E_{k} contained in F ∈ E_{k−1}. Hence we have the assertion with δ_{∗} = (3/5)^{2}δ′′.

Finally, set

d_{k} = \frac{2}{ψ^{k+1}(6)}

with ψ(x) = \frac{1}{2} e^{x^{q}/2}. Note that ψ(x) > x and ψ′(x) > 1 on [6, +∞). And for the diameter d(G) of every G ∈ E_{k}, we have the following

Lemma 13 For every G ∈ E_{k},

d(G) < d_{k}.

Proof. First, |z_{B_{∗}}| > R > ψ(6) by the condition 1 for R. Next since, by Lemma 10, every point in f(B) has an absolute value greater than |f(z_{B})|/2 for every disk B with center z_{B} in S′ and radius |z_{B}|^{−q}, we can see inductively that the inequality

|z_{f^{k−1}(F)}| > ψ^{k}(6)

for the center z_{f^{k−1}(F)} of every disk f^{k−1}(F) with F ∈ E_{k−1} implies

|z_{f^{k}(G)}| > \frac{|f(z_{f^{k−1}(F)})|}{2} > ψ(|z_{f^{k−1}(F)}|) > ψ^{k+1}(6)

by the condition 4-(d) for R, and hence

r(f^{k}(G)) = |f(z_{f^{k}(G)})|^{−q} < 1/ψ^{k+1}(6),

for every f^{k}(G) ∈ P(f^{k}(F)). Since clearly |(f^{−k})′(z)| < 1 on f^{k}(G), we conclude the assertion.

Summing up, we have obtained a sequence {E_{k}} such that every G ∈ E_{k} is contained in a F ∈ E_{k−1} and every G ∈ E_{k} contains at least one element of E_{k+1}. We have also found constants d_{k} and δ_{∗} such that

ρ (( ⋃_{G∈E_{k}} G) ∩ F, F ) ≥ δ_{∗} > 0, d(G) ≤ d_{k},

d_{k} → 0, \frac{k}{log(1/d_{k})} =→ 0,

for every k and G in E_{k}. Hence the Fundamental lemma in [2] (also cf. [3] Lemma 3.2.7) shows that the Hausdorff dimension of E_{∞} is two.

References

[1] R.L. Devaney and L. Keen, Dynamics of meromorphic maps with polynomial Schwarzian derivative, Ann. Sci. École Norm Sup 22, (1989), 55–81.

[2] C.T. McMullen, Area and Hausdorff dimension of Julia sets of entire functions, Trans. A.M.S., 300, (1987), 329-342.

[3] S. Morosawa, Y. Nishimura, M. Taniguchi, and T. Ueda, Holomorphic Dynamics, Cambridge Univ. Press, 1999.

[4] G.M. Stallard Entire functions with Julia sets of zero measure, Math Proc. Camb. Phil. Soc. 108 (1990) 551-557.

[5] G.M. Stallard The Hausdorff dimension of Julia sets of entire functions I,II,III, I: Ergod. Th. and Dynam. Sys. 11 (1991), 769–777; II: Math. Proc. Camb. Phil. Soc. 119 (1996), 513–536; III: ibid. 122 (1997), 223– 244.

[6] G.M. Stallard The Hausdorff dimension of Julia sets of hyperbolic meromorphic functions, Math. Proc. Camb. Phil. Soc. 127 (1999), 271–288.

[7] E.M. Stein Singular integrals and differentiability properties of functions, Princeton University Press, 1970

[8] M Taniguchi, Explicit representations of structurally finite entire functions, Proc. Japan Acad. 77 (2001), 69-71.

[9] M. Taniguchi, Synthetic deformation spaces of an entire function, to appear.

14, July, 2001

Dear the Editor of Mathematical Proceedings of the Cambridge Philosophical Society

I enclose my paper tilted ”Size of the Julia set of a structurally finite transcendental entire function”.

I wish this paper to be accepted to Mathematical Proceedings of the Cambridge Philosophical Society.

Sincerely yours,

Masahiko Taniguchi

Graduate School of Science,

Kyoto University, Kyoto 606, Japan

e-mail address: tanig@kusm.kyoto-u.ac.jp