Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 26, 2001, 391–399

ON LIMITING DIRECTIONS OF JULIA SETS

Jianyong Qiao

China University of Mining & Technology, Center of Mathematics, Beijing 100083 Chinese Academy of Sciences, Academy of Mathematics & System Sciences, Beijing 100080

P. R. China; qjy@mail.cumtb.edu.cn

Abstract. We deal with the iteration of transcendental entire functions, and prove some properties on the Julia sets.

1. Introduction and main results

Let f: C → C be a transcendental entire function; we define the iterated sequence of f by f^{0}(z) = z, f^{n+1}(z) = f ∘ f^{n}(z), n = 1, 2, …. The Fatou set and the Julia set are defined by N(f) = {z ∈ C | {f^{n}} is normal at z} and J(f) = C \ N(f) respectively. Qiao ([7]) proved that the Julia set of a transcendental entire function of finite order has infinitely many limiting directions; here a limiting direction of J(f) means a limit of the set {arg z_{n} | z_{n} ∈ J(f) is an unbounded sequence}. The example in [1] shows that there exists an entire function of infinite order whose Julia set has only one limiting direction. In this note we shall prove

Theorem 1. Let f be a transcendental entire function of lower order λ < ∞. Then there exists a closed interval I ∈ R such that all θ ∈ I are the common limiting directions of J(f^{(n)}), n = 0, ±1, ±2, ⋯, and mes I ≥ π/ max(\frac{1}{2}, λ). Here f^{(n)} denotes the n-th derivative or the n-th integral primitive of f for n ≥ 0 or n ≤ 0 respectively.

We know that Mittag-Leffler’s function

E_{α}(z) = ∑_{n=0}^{∞} \frac{z^{n}}{Γ(1 + αn)}, 0 < α < 2,

is a transcendental entire function of order 1/α. Put Ω(−θ, θ) = {z ∈ C | −θ < arg < θ}. By the discussion used in [3] it is easy to verify that for any E_{α}(z), 0 < α < 2, there exists a constant k > 0, such that

f_{αk}(C \ \overline{Ω}(−\frac{α}{2π} , \frac{α}{2π} )) ⊂ C \ \overline{Ω}(−\frac{α}{2π} , \frac{α}{2π} ),

2000 Mathematics Subject Classification: Primary 30D35, 37F10.

Project supported by the National Natural Science Foundation of China.

here f_{αk} = E_{α}(z) − k. Hence J(f_{αk}) ⊂ \overline{Ω}(−α/2π, α/2π). This shows that the estimate of the length of the closed interval I in Theorem 1 is sharp.

Liverpool ([5]) proved that: if J(f) lies in the half-plane {z ∈ C | Re z ≥ 0} for a transcendental entire function f of order ≤ 1, then there exists a positive constant c such that for any horizontal strip region S with width c, J(f) ∩ S is unbounded; here a strip region means a region between two parallel straight lines. It is easy to see from Theorem 1 that f is of lower order ≥ 1 provided J(f) lies in a half-plane. Therefore, Liverpool’s result is valid only for entire functions with order and lower order one. We shall prove

Theorem 2. Let f be a transcendental entire function of lower order 1, J(f) lie in the half plane {z ∈ C | Re z ≥ 0}. Then there exists a positive constant c such that all J(f^{(n)}) ∩ S, n = 0, ±1, ±2, ⋯, are unbounded for any non-vertical strip region S with width c.

Liverpool ([5]) has pointed out that

\{z ∈ C | (2n + \frac{1}{2})π < Im z < (2n + \frac{3}{2})π\} ∩ J(e^{z} − 1) = ∅, n = 0, ±1, ±2, ….

But we shall prove this kind of “gap strips” will disappear for a class of entire functions of order > 1.

Theorem 3. Let f be a transcendental entire function of order ϱ > 1, and all limiting directions of J(f) belong to (−π/ϱ, π/ϱ). Then all J(f^{(n)}) ∩ S, n = 0, ±1, ±2, ⋯, are unbounded for an arbitrary horizontal strip region S.

Theorem 4. Let f be a transcendental entire function of order ϱ > 1, and lower order λ > \frac{1}{2} . If all limiting directions of J(f) belong to [−π/2λ, π/2λ], then all J(f^{(n)}) ∩ S, n = 0, ±1, ±2, …, are unbounded for an arbitrary strip region S which is parallel to θ ∈ (−π/2λ, π/2λ).

2. Some lemmas

In order to prove the above results, we investigate the growth of f on its Fatou set. The following two lemmas are the improvements of the main results in [2] and [5] respectively. For z_{0} ∈ C and θ, δ ∈ R, put

Ω(z_{0}, θ, δ) = {z ∈ C | | arg(z − z_{0}) − θ| < δ}.

We have

Lemma 1. Let f be a transcendental entire function, and Ω(z_{0}, θ, δ) ⊂ N(f). Then

|f(z)| = O(|z|)^{π/δ}, z ∈ Ω(z_{0}, θ, δ′)

for arbitrary δ′ ∈ (0, δ).

Proof. Since Ω(z_{0}, θ, δ) ⊂ N(f), there is an unbounded component G_{0} of N(f) such that Ω(z_{0}, θ, δ) ⊂ G_{0} . By [1] we know that every component of N(f) is a simply connected hyperbolic domain. Let f(G_{0}) belong to some component G of N(f). It is easy to verify that the mapping

w = h(z) = \frac{(e^{−iθ}z − e^{−iθ}z_{0})^{π/2δ} − 1}{(e^{−iθ}z − e^{−iθ}z_{0})^{π/2δ} + 1}

maps Ω(z_{0}, θ, δ) conformally onto the unit disk {|w| < 1}. Put h^{−1}(0) = a ∈ Ω(z_{0}, θ, δ). By the Riemann theorem, there is a conformal mapping w = g(z): G → {|w| < 1} satisfying g(f(a)) = 0 and g′(f(a)) > 0. Hence F (w) = g ∘ f ∘ h^{−1}(w) is an analytic mapping from the unit disk to itself. By the Schwarz lemma,

(1)  |F (w)| ≤ |w|,  |w| < 1.

Since g^{−1} is univalent on {|w| < 1}, by Koebe’s distortion theorem we have

(2)  |(g^{−1}(w) − f(a))g′(f(a))| ≤ \frac{|w|}{(1 − |w|)^{2}} ,  |w| < 1.

Since f = g^{−1} ∘ F ∘ h, it follows from (1) and (2) that

(3)  |f(z)| ≤ |f(a)| + \frac{1}{|g′(f(a))|(1 − |h(z)|^{2})},  z ∈ Ω(z_{0}, θ, δ).

For arbitrary z ∈ Ω(z_{0}, θ, δ′), put

η = z − z_{0} = re^{iα},  σ = \frac{π}{2δ} ,  λ = sin \frac{δ′π}{2δ} > 0.

Then

|h(z)|^{2} = |\frac{1 − (cos σ(α − θ))/r^{α} + i(sin σ(α − θ))/r^{α} + o(1/r^{α})}{1 + (cos σ(α − θ))/r^{α} − i(sin σ(α − θ))/r^{α} + o(1/r^{α})}|^{2}

= \frac{1 − 2(cos σ(α − θ))/r^{α} + o(1/r^{α})}{1 + 2(cos σ(α − θ))/r^{α} + o(1/r^{α})}.

Thus

\begin{aligned}1 − |h(z)| & > \frac{1 − |h(z)|^{2}}{2} = \frac{4(cos σ(α − θ))/r^{α} + o(1/r^{α})}{1 + 2(cos σ(α − θ))/r^{α} + o(1/r^{α})} \\ & ≥ \frac{4λ/r^{α} + o(1/r^{α})}{1 + 2/r^{α} + o(1/r^{α})}. \end{aligned}

By the above inequality and (3) we can easily deduce the result of Lemma 1. The proof of Lemma 1 is complete.

For any real numbers a > 0 and A > 0, put

H(a, A) = {z ∈ C | Re z > a, | Im z| < A}.

We have

Lemma 2. Let f be a transcendental entire function, and H(a, A) ⊂ N(f), then

|f(z)| = O(exp \frac{π}{A}|z|), z ∈ H(a, A′)

for arbitrary A′ ∈ (0, A).

Proof. Let G_{0} , G be two components of N(f) such that H(a, A) ⊂ G_{0} , f(G_{0}) ⊂ G. It is easy to verify that

w = h_{1}(z) = exp(\frac{π}{2A}z − \frac{απ}{2A})

maps H(a, A) conformally onto {Re w > 0}\{|w| < 1}, and w = h_{2}(z) = (z−2)/z maps {Re z > 1} conformally onto {|w| < 1}. By the Riemann theorem, there exists an univalent analytic function g(z) which maps G onto {|w| < 1}. Hence F (w) = g ∘ f ∘ h_{1}^{−1} ∘ h_{2}^{−1}(w) is an analytic mapping from the unit disk to itself. As in the proof of Lemma 1 we obtain

(4)  |g^{−1} ∘ F (w)| = O(\frac{|w|}{(1 − |w|)^{2}} ),  |w| < 1.

Obviously, f = g^{−1} ∘ F ∘ h_{2} ∘ h_{1} and h_{1}(z) ∈ {Re w > 1} for z ∈ H(a, A′) and sufficiently large |z|. By similar calculations as in the proof of Lemma 1, we can deduce the result of Lemma 2. The proof of Lemma 2 is complete.

Below we shall use the fundamental concepts and basic notations of Nevanlinna’s theory ([4]).

Lemma 3 ([7]). Let f be a transcendental entire function satisfying

lim_{\overline{r→∞}} \frac{T (r, f)}{r^{m}} = 0

for some fixed natural number m. Then for arbitrary α ∈ [0, 2π), the set

J(f) ∩ [ ⋃_{k=1}^{m}\{z ∈ C | \frac{2k − 1}{m}π + α < arg z < \frac{2k}{m} π + α\}]

is unbounded.

Lemma 4 ([4]). Let f be a transcendental entire function. Then

m(r, \frac{f′}{f} ) = O(log rT (r, f)), r → ∞,

at most with an exceptional set of r whose linear measure is finite.

3. The proofs of the theorems

The proof of Theorem 1. We distinguish the following two cases:

(A) Suppose f is of lower order λ < \frac{1}{2} . We shall prove that, all θ ∈ [0, 2π) are the limiting directions of all J(f^{(n)}), n = 0, ±1, ±2, · · · . Assume this statement is not true, then there exist θ ∈ [0, 2π) and an integer n_{0} such that θ is not a limiting direction of J(f^{n_{0}}). Therefore J(f^{n_{0}}) ∩ Ω(0, θ, δ) is bounded for some constant δ > 0. By Lemma 1,

(5)  |f^{(n_{0})}(z)| = O(|z|^{k}),  arg z = θ;

here k is a positive constant. Since the lower order of f^{n_{0}}(z) is less than \frac{1}{2} , by (5) and Wiman’s theorem on minimum modulus (see [4]) we get a contradiction.

(B) Suppose f is of lower order λ ≥ \frac{1}{2} , put

E_{n} = {e^{iθ} | θ is a limiting direction of J(f^{(n)})}.

Obviously, E_{n} is a closed set on the unit circle Γ. Denote E = ⋂_{n∈Z} E_{n}, here Z is the set of integers. It is easy to see that the arguments of the points in E are the common limiting directions of all J(f^{(n)}), and the components of E are closed arcs on Γ. Put

γ = \{α | α is an open arc on Γ with length < \frac{π}{λ} , and its endpoints are not in E\}.

Assume the maximum component of E is of length < π/λ, then the set γ covers Γ. So there exist finitely many α_{1}, α_{2}, …, α_{p} ∈ γ such that ⋃_{j=1}^{p} α_{j} ⊃ Γ. Denote the arguments of two endpoints of α_{j} by θ_{j_{1}}, θ_{j_{2}} , θ_{j_{1}} < θ_{j_{2}} , respectively, and suppose θ_{j_{1}} is not the limiting direction of J(f^{(n_{j_{1}} )}), θ_{j_{2}} is not the limiting direction of J(f^{(n_{j_{2}} )}). By Lemma 1,

(6)  |f^{(n_{j_{1}} )}(z)| = O(|z|^{k_{j_{1}}} ),  arg z = θ_{j_{1}},
(7)  |f^{(n_{j_{2}} )}(z)| = O(|z|^{k_{j_{2}}} ),  arg z = θ_{j_{2}}.

Here k_{j_{1}} , k_{j_{2}} are two positive constants.

Put m = min_{1≤j≤p}(n_{j_{1}}, n_{j_{2}}). Note

f^{(n_{j_{1}} −1)}(z) = ∫_{0}^{z} f^{(n_{j_{1}} )}(η) dη + c,

where c is a constant, and the above integral path is the segment of a straight line from 0 to z. From the above equality and (10) we deduce

|f^{(n_{j_{1}} −1)}(z)| = O(|z|^{k_{j_{1}} +1}), arg z = θ_{j_{1}}.

Repeating the above discussion, we can obtain

|f^{(m)}(z)| = O(|z|^{k_{1}}), arg z = θ_{j_{1}},

where k_{1} is a positive constant. By the same method we have

|f^{(m)}(z)| = O(|z|^{k_{2}}), arg z = θ_{j_{2}},

where k_{2} is a positive constant. Note that θ_{j_{2}} − θ_{j_{1}} < π/λ. By the Phragmén– Lindelöf principle we have

|f^{(m)}(z)| = O(|z|^{k}), θ_{j_{1}} ≤ arg z ≤ θ_{j_{2}},

where k = max(k_{1}, k_{2}). Since α_{1}, α_{2}, …, α_{p} cover Γ, it follows that f is a polynomial. This contradicts the transcendence of f. The proof of Theorem 1 is complete.

The proof of Theorem 2. Assume the conclusion of this theorem is not true; then there exists a sequence of non-vertical strip regions S_{j} with width c_{j} → ∞, and a sequence of J(f^{(n_{j})}) such that J(f^{(n_{j})}) ∩ S_{j} is bounded. Let S_{j} be parallel to the ray arg z = θ_{j} ∈ (−\frac{1}{2}π, \frac{1}{2}π). Choose points z_{j} ∈ S_{j}, j = 1, 2, …. Then the ray L_{j} : z = z_{j} + te^{iθ_{j}}, t > 0, lies on S_{j}. By Lemma 2,

(8)  |f^{(n_{j})}(z)| = O(exp \frac{2π}{c_{j}} |z|),  z ∈ L_{j}.

We distinguish two cases:

(A) Suppose there are infinitely many n_{j} > 0 such that (8) holds. Integrating f^{(n_{j})}(z), by (8) we easily obtain

|f^{(n_{j}−1)}(z)| = O(|z| exp \frac{2π}{c_{j}} |z|), z ∈ L_{j}.

Repeating this procedure we can get

(9)  |f(z)| = O(|z|^{n_{j}} exp \frac{2π}{c_{j}} |z|),  z ∈ L_{j}.

Since L_{j} is not vertical, we can draw two rays:

L′_{j} : z = z_{j} + te^{iα_{j}}, t > 0, L′′_{j} : z = z_{j} + te^{iβ_{j}}, t > 0,

satisfying α_{j} < β_{j}, α_{j}, β_{j} ∈ (\frac{1}{2}π, \frac{3}{2}π). The angle from L_{j} to L′_{j} and the angle from L′′_{j} to L_{j} are both less than π. Since there are no points of J(f) in the left half-plane, by Lemma 1

(10)  |f(z)| = O(|z|^{2}),  z ∈ L′_{j} or L′′_{j}.

By (9), (10) and the Phragmén–Lindelöf principle we have

|f(z)| = O(|z|^{n_{j}} exp \frac{2π}{c_{j}} |z|), z ∈ C.

We thus obtain

\overline{lim_{r→∞}} \frac{T (r, f)}{r} ≤ \frac{2π}{c_{j}} .

Letting c_{j} → ∞ we get

lim_{r→∞} \frac{T (r, f)}{r} = 0.

By Lemma 3, we deduce a contradiction.

(B) Suppose there are infinitely many n_{j} ≤ 0 such that (8) holds. As in (A), we can draw the ray L′_{j} and the ray L′′_{j}; hence (10) follows. Using (10) to estimate the integrand, we can deduce

|f^{(n_{j})}(z)| = O(|z|^{2+n_{j}}), z ∈ L′_{j} or L′′_{j}.

It follows from this equality, (8) and the Phragmén–Lindelöf principle that

(11)  \overline{lim_{r→∞}} \frac{T (r, f^{(n_{j})})}{r} ≤ \frac{2π}{c_{j}} .

On the other hand, by Lemma 4,

T (r, f^{(n_{j}+1)}) ≤ T (r, f^{(n_{j})}) + m(r, \frac{(f^{(n_{j})})′}{f^{(n_{j})}} ) ≤ (1 + o(1))T (r, f^{(n_{j})}) + k_{j} log r

for sufficiently large r, r ∉ E_{j}^{1} , mes E_{j}^{1} < ∞. Here k_{j} is a positive constant. Note that n_{j} ≤ 0. Repeating the above estimation, we obtain

(12)  T (r, f) ≤ (1 + o(1))T (r, f^{(n_{j})}) + K_{j} log r

for r ∉ E_{j}^{1} , mes E_{j}^{1} < ∞. Here K_{j} is a positive constant. By (11) and (12), we have

lim_{\overline{r→∞}} \frac{T (r, f)}{r} ≤ \frac{2π}{c_{j}} .

Furthermore, by the same method as used in (A), we can deduce a contradiction. The proof of Theorem 2 is complete.

The proof of Theorem 3. Assume there exist a horizontal strip region S and an integer n such that J(f^{(n)}) ∩ S is bounded. Denote the width of S by c. Choose a point z_{0} ∈ S, and draw the ray L : z = z_{0} + t, t > 0. By Lemma 2,

(13)  |f^{(n)}(z)| = O(exp \frac{2π}{c} |z|),  z ∈ L.

Since all limiting directions of J(f) belong to (−π/ϱ, π/ϱ), we can draw two rays: L′ : z = z_{0} + te^{iθ}, t > 0, and L′′ : z = z_{0} + te^{−iθ}, t > 0, such that (−θ, θ) ⊂ (−π/ϱ, π/ϱ) and all limiting directions of J(f) belong to (−θ, θ). By Lemma 1,

(14)  |f(z)| = O(|z|^{k}),  z ∈ L′ or L′′;

here k is a positive constant. Put m = min(n, 0). Using (13) and (14) to estimate the integrand, we can obtain

(15)  |f^{(m)}(z)| = O(|z|^{k_{1}} exp \frac{2π}{c} |z|),  z ∈ L,
(16)  |f^{(m)}(z)| = O(|z|^{k_{2}}),  z ∈ L′ or L′′;

here k_{1} , k_{2} are two positive constants. It follows from (15), (16) and the Phragmén–Lindelöf principle that

(17)  |f^{(m)}(z)| = O(|z|^{k_{1}} exp \frac{2π}{c} |z|),  −θ ≤ arg(z − z_{0}) ≤ θ.

Since all limiting directions of J(f) belong to (−π/ϱ, π/ϱ), by Lemma 1, there exists a positive constant k such that (14) holds for θ ≤ arg(z − z_{0}) ≤ 2π − θ. This and (17) imply that f^{(m)} is of order ≤ 1. This is a contradiction. The proof of Theorem 3 is complete.

The proof of Theorem 4. Assume there exist a strip region S which parallels θ ∈ (−π/2λ, π/2λ), and some J(f^{(n)}) such that J(f^{(n)}) ∩ S is bounded. Denote the width of S by c. Choose a point z_{0} ∈ S, by Lemma 2,

|f^{(n)}(z)| = O(exp \frac{2π}{c} |z|), z ∈ L : z = z_{0} + te^{iθ}, t > 0.

Obviously, we can draw two rays

L′ : z = z_{0} + te^{iθ_{1}}, t > 0, L′′ : z = z_{0} + te^{iθ_{2}}, t > 0,

such that (θ_{1}, θ_{2}) ⊃ [−π/2λ, π/2λ], θ_{2} − θ < π/λ and θ − θ_{1} < π/λ. Using the same method as in the proof of Theorem 3, we can deduce

|f^{(n)}(z)| = O(exp \frac{2π}{c} |z|),  θ_{1} ≤ arg(z − z_{0}) ≤ θ_{2},
|f^{(n)}(z)| = O(|z|^{2}),  θ_{2} ≤ arg(z − z_{0}) ≤ θ_{1} + 2π.

It follows that f is of order ≤ 1. This is a contradiction. The proof of Theorem 4 is thus complete.

On limiting directions of Julia sets 399

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Received 2 February 2000