Some properties of Fatou and Julia sets of transcendental meromorphic functions

Theorem 1.1. Let f{z) be a transcendental meromorphic function in C with li = fi(f) < oo and 6 = 6(oo, /) > 0. If \i = 0, then E = [0,2n); If fi > 0 and Jf has an unbounded component, then $$\operatorname{mes} E\geqslant \min\biggl\{2\pi,\frac{4}{\mu}\arcsin\sqrt{\frac{\delta}{2}}\biggr\}.$$ We make some remarks on Theorem 1.1. - (1) If Jf has only bounded components, we do not know if Theorem 1.1 holds. In this case, Ff has at most an unbounded component. If Ff has no unbounded components, it is obvious that E = [0,2n). If Ff has only an unbounded component U, and if U is wandering or periodic of period at least two, then / is bounded in U, from the proof of Theorem 1.1 it follows that Theorem 1.1 holds. Then we are left with the case when U is invariant. In this special case, if for some a S Jf, CF{(CL) > 0 (please see the statement before Lemma 2.2 for its definition), then Theorem 1.1 still follows from Lemma 2.2 and the proof of Theorem 1.1. - (2) The condition that 6(oo, f) > 0 is necessary. Observe f(z) = A tan 2, Agfi, the real axis. It is easy to get /\(/) = 1 and {(00, /) — 0. It was proved in [3] that when A > 1, Jf = 5ft, then E =* {0, TT}, and mes.E = 0. When 0 < A < 1, the Julia set of f(z) = A tans is a Cantor set and the Fatou set consists of one unbounded component, but since f(z) has only two singularity values, it was proved in [11] that for any a 6 Jf, Cpf(a) > 0. (3) Baker [2] investigated the radial distribution of the Julia set of a transcendental entire function and constructed an entire function with infinite lower order whose Julia set lies in a horizontal strip. It is well known that an entire function / may only have unbounded simply connnected components of the Fatou set and <5(oo, /) = 1. Therefore, the condition that / has a finite lower order is necessary in Theorem 1.1. A further discussion on this subject of entire functions with finite lower order was-made in [9] after Baker [2]. Their methods are not available for the case of meromorphic functions. Next we consider when mes E = 2TT. If mes E < 2n, then Ff must contain unbounded angle domains. Now [12, Theorem 3] says that Ff contains no unbounded angle domains, if for arbitrary positive integer m, the following holds (1) $$\limsup_{r \to \infty} \frac{L(r, f)}{r^m} = \infty,$$ where L{r,f) := min{|/(z)|}. Thus we have

Theorem 1.2. Let f(z) be a transcendental meromorphic function in C satisfying (I). Then E= [O,2TT). REMARK. (1) above suggests a further discussion of non-existence of the unbounded periodic components of Fj, which was investigated in Zheng [11, 12].

Theorem 1.3. Let fj (j = 1,2,.. .,N) be transcendental meromorphic functions. Assume that there exists a sequence {rn} of positive numbers which tends to infinity such that (2) $$\lim_{n\to\infty}\frac{L(r_n,f_1)}{r_n}=\infty,$$ and for each j and sufficiently large n, there is a Rj>n ^ rn, such that (3) $$L(R_{j,n}, f_j) > r_n, \quad j = 2, \dots, N.$$ Define g(z) = fxo ••• o /AT(2). Let D be a hyperbolic domain in C such that for p > 0, gp (z) : D -» D is analytic. If for some a € D, gnp{a) —t b € 3D, assume, in addition, that b is not an essential singularity point ofg(z). Then D is bounded. We make remarks on Theorem 1.3. - (i) Theorem 1.3 is a generalisation of results in [12]. For example it was proved in [12] that a transcendental meromorphic function has no unbounded (pre)periodic Fatou components if it satisfies (2). - (ii) If / is a transcendental meromorphic function of order A = A(/) < 1/2 and 6(oo, /) > 1 — cos 7rA, then for arbitrarily large r > 0, we have a R ^ r such that L(R,f) > r. In fact, we can take A(/) < a < 1/2 such that 6(oo,f) > 1 - cos7ra. From [6], the set $$F := \left\{ r > 1 : \log L(r, f) > \frac{\pi \alpha}{\sin \pi \alpha} \left( \cos \pi \alpha + \delta(\infty, f) - 1 \right) T(r, f) \right\}$$ has lower logarithmic density at least 1 - (A(/))/ a > 0. Therefore, for all sufficiently large r > 0, there exists a R e (r 1/d , r) such that $$L(R, f) > e^{\beta T(R, f)} > R^d \geqslant r,$$ where ft — [(?ra)/(sin 7ra)] (COSTTQ; + <5(oo,/) - l) > 0 and (a/A(/)) < d < +oo. This paper was mainly completed by the first author.

Lemma 2.1. Let f(z) be transcendental and meromorphic in C with finite positive lower order \i — fi(f) and such that S — <5(oo, /) > 0. Define for r > 0 (4) $$D(r) := \left\{ \theta \in [0, 2\pi) : \log^+ \left| f(re^{i\theta}) \right| > \frac{1}{\log r} T(r, f) \right\}.$$ Then there exists an unbounded sequence {r;} ofr such that for sufficiently small e > 0 we have a jo such that when j Js jo, (5) $$\operatorname{mes} D(r_j) \geqslant \min \left\{ 2\pi, \frac{4}{\mu} \arcsin \sqrt{\frac{\delta}{2}} \right\} - \varepsilon.$$ An open set is called hyperbolic if it has at least three boundary points in C. We define the hyperbolic metric on an open set by the hyperbolic metrics of its components. Let Wbe a hyperbolic open set in C. For an a € C\W, define $$C_W(a) = \inf \{ \lambda_W(z) | z - a| : \forall z \in W \},\$$ where \w{z) is the hyperbolic density on W. It is well-known that \w(z)S\v(z) ^ 1, z € W, where 5w{z) is the Euclidean distance of z to dW and if every component of W is simply connected, then Cw{a) ^ 1/2- For r > 0 and 8i,92 6 [0,2ir),8i < 62, define $$\Omega(r;\theta_1,\theta_2) := \{z : \arg z \in (\theta_1,\theta_2), |z| > r\}.$$

Lemma 2.2. Let f(z) be analytic in ft(ro;#i,02), U a hyperbolic domain and $$f: \Omega(r_0; \theta_1, \theta_2) \to U.$$ If there exists a point a € dU\{oo}, such that Cy(a) > 0, then there exists a constant d > 0 such that for sufficiently small e > 0, we have (6) $$|f(z)| = O(|z|^d), z \to \infty, z \in \Omega(r_0; \theta_1 + \varepsilon, \theta_2 - \varepsilon).$$ PROOF: Write Q = £l(ro;9i,02)- Since f(Q,) C U, from the Schwarz-Pick Lemma we have (7) $$\lambda_U(f(z))|f'(z)| \leq \lambda_{\Omega}(z), \ z \in \Omega.$$ From the definition of Cu(d), we have (8) $$\lambda_{U}(f(z))|f'(z)| \geqslant C_{U}(a)\frac{|f'(z)|}{|f(z)-a|}, \ z \in \Omega.$$ On the other hand, since for e > 0 and z € QQ — Q(ro;6i + e,62 - e), 6Q(Z) ^ |z|sine, we have (9) $$\lambda_{\Omega}(z) \leq [|z| \sin \varepsilon]^{-1}, \ z \in \Omega_{0}.$$ Combining (7), (8) and (9) gives (10) $$c\frac{\left|f'(z)\right|\left|dz\right|}{\left|f(z)-a\right|} \leqslant \frac{1}{|z|}|dz|, \ z \in \Omega_0,$$ where $c = C_U(a) \sin \varepsilon$ . We draw a curve $\gamma$ in $\Omega_0$ from a fixed point b to z by connecting b and $|b|e^{i\theta}$ , $\theta = \arg z$ along the circle $\{w : |w| = |b|\}$ and $|b|e^{i\theta}$ and z along the radial $\{w : \arg w = \theta\}$ . Then integrating the both sides of (10) along $\gamma$ implies the desired inequality (6). PROOF OF THEOREM 1.1: If $\mu(f) = 0$ , from the remark (ii) on Theorem 1.3 we know that f satisfies (1) and therefore from Theorem 1.2 we have $E = [0, 2\pi)$ . Now we assume $\mu > 0$ . For the sake of convenience, put $$\sigma = \min \bigg\{ 2\pi, \frac{4}{\mu} \arcsin \sqrt{\frac{\delta}{2}} \bigg\}.$$ Now we conversely suppose that $\operatorname{mes} E < \sigma$ . Take a t > 0 such that $\sigma - \operatorname{mes} E > t > 0$ . Since E is closed, $S = [0, 2\pi) \setminus E$ consists of (at most countablely many) open intervals I from which we can then find finitely many open intervals $I_i$ (i = 1, 2, ..., m) such that $\operatorname{mes} \left( S \setminus \bigcup_{i=1}^m I_i \right) < K/2$ , where $K = \sigma - \operatorname{mes} E - t > 0$ . Under the assumption of Theorem 1.1, an application of Lemma 2.1 implies that there exists a sequence $\{r_j\}$ of positive numbers such that $\operatorname{mes} D(r_j) > \sigma - t > 0$ , where $D(r_j)$ is defined as in (4). Obviously $$\operatorname{mes}(D(r_i) \cap S) = \operatorname{mes}(D(r_i)) \setminus (E \cap D(r_i)) \ge \operatorname{mes} D(r_i) - \operatorname{mes} E \ge K > 0.$$ Thus for each j we have $$\operatorname{mes}\left(\left(\bigcup_{i=1}^{m} I_{i}\right) \cap D(r_{j})\right) = \operatorname{mes}\left(S \cap D(r_{j})\right) - \operatorname{mes}\left(\left(S \setminus \bigcup_{i=1}^{m} I_{i}\right) \cap D(r_{j})\right)$$ $$> K - \frac{K}{2} = \frac{K}{2},$$ so there exists an open interval $I = I_{i_0} \subset S$ such that for infinitely many j, (11) $$\operatorname{mes}(D(r_j) \cap I) > \frac{K}{2m} > 0.$$ It is easy to see that we can assume that for each j, (11) holds. Write $I = (\alpha, \beta)$ . Take a positive number $\varepsilon$ such that (12) $$\operatorname{mes}(D(r_j) \cap I_{2\varepsilon}) > \frac{K}{3m} > 0 \ (j = 1, 2, ...),$$ where we denote by $I_d$ the interval $(\alpha + d, \beta - d)$ for $0 < 2d < \beta - \alpha$ . It is easy to see from $I \cap E = \emptyset$ that there exists a positive R such that $\Omega(R; I_{\varepsilon}) = \{z \in C : |z| \ge R \text{ and } \arg z \in I_{\varepsilon}\} \subset F_f$ . Now by applying Lemma 2.2 to f in $\Omega(R; I_{\varepsilon})$ , we have $$|f(z)| \leqslant C_0 |z|^p, \ z \in \Omega(R; I_{2\varepsilon}),$$ where Co and p are both positive constants. Then (14) $$\int_{I_{2\epsilon} \cap D(r_j)} \log^+ |f(r_j e^{i\theta})| d\theta \leqslant \int_I p \log r_j d\theta + O(1) = O(\log r_j).$$ On the other hand, applying Lemma 2.1 to / gives (15) $$\int_{I_{2\epsilon} \cap D(r_j)} \log^+ \left| f(r_j e^{i\theta}) \right| d\theta \geqslant \operatorname{mes} \left( I_{2\epsilon} \cap D(r_j) \right) \frac{T(r_j, f)}{\log r_j} > \frac{K}{3m} \frac{T(r_j, f)}{\log r_j}.$$ Combining (14) with (15) gives $$T(r_j, f) = O((\log r_j)^2).$$ Then fj.(f) = 0. We get a contradiction. . D Theorem 1.1 follows.

Lemma 3.1. Let D be a domain with at least three boundary points and f be meromorphic in C except possibly at most countably many essential sigularity points such that f(D) C D. Then one of the following mutually exclusive possibilities can occur. - (1) There is a subsequence of {fn {z)} which converges to z in D. - (2) fn (z) -\b&T),~D is the closure ofD.* Lemma 3.1 can be proved from the arguments of Heins [8]. We also need the following result, which is due to Zheng [12] and of independent significance.

Lemma 3.2. Let U be an unbounded hyperbolic domain and f : U —> U analytic. If $$f^n(z)|_U \to \infty, \quad n \to \infty,$$ then there exists a curve 7 tending to infinity and a constant £ > 1, such that f(j)Cj and $$\frac{|z|}{\mathcal{L}} \leqslant |f(z)| \leqslant \mathcal{L}|z|, \ \forall z \in \gamma$$ PROOF OF THEOREM 1.3: Suppose that D is unbounded. Take a point a e D and a positive number M such that (16) $$|a| < M \text{ and } |f_j \circ \cdots \circ f_N(g^i(a))| < M, \text{ for } j = 1, 2, \dots, N \text{ and } i = 0, 1, \dots, p.$$ From (2) and (3) for arbitrarily large K > 3 we can take a sufficiently large R> M such that $$(17) L(R, f_1) > KR$$ and for each j there exists a $M \leq R_j \leq R$ such that $$(18) L(R_i, f_i) > R.$$ Now we draw in D a curve $\gamma$ connecting a and a point in $\{|z|=R\}$ such that $\gamma \in \{|z| \le R\}$ . It is easy to see that $\{|z|=R_N\} \cap \gamma \neq \emptyset$ and from (16), $f_N(\gamma) \cap \{|z| < M\} \neq \emptyset$ , and then from (18) when j=N we have $$(19) f_N(\gamma) \cap \{|z| = R\} \neq \emptyset.$$ Obviously from (16), $f_{N-1} \circ f_N(\gamma) \cap \{|z| < M\} \neq \emptyset$ and from (19), $f_N(\gamma) \cap \{|z| = R_{N-1}\} \neq \emptyset$ . Again applying (18) when j = N-1 gives that $f_{N-1} \circ f_N(\gamma) \cap \{|z| = R\} \neq \emptyset$ . Inductively we get $$(20) f_2 \circ \cdots \circ f_{N-1} \circ f_N(\gamma) \cap \{|z| = R\} \neq \emptyset.$$ Then there exists a point $z_1 \in \gamma$ such that $\left| f_2 \circ \cdots \circ f_{N-1} \circ f_N(z_1) \right| = R$ , and from (17) we have $\left| g(z_1) \right| > KR$ . This implies that $g(\gamma) \cap \{|z| = R\} \neq \emptyset$ . Take a segment of $g(\gamma)$ , denoted by $\gamma_1$ , from g(a) to $\{|z| = R\}$ such that $\gamma_1 \subset \{|z| \leqslant R\}$ . Repeating the above process to $\gamma_1$ gives that $g(\gamma_1) \cap \{|z| = R\} \neq \emptyset$ , so $g^2(\gamma) \cap \{|z| = R\} \neq \emptyset$ . Inductively we obtain $g^{p-1}(\gamma) \cap \{|z| = R\} \neq \emptyset$ . Now take a curve $\gamma_{p-1}$ from $g^{p-1}(\gamma)$ connecting $g^{p-1}(a)$ and a point in $\{|z| = R\}$ such that $\gamma_{p-1} \subset \{|z| \leqslant R\}$ . As we did for $\gamma$ in the first step, we have (20) for $\gamma_{p-1}$ , so that we have (20) for $g^{p-1}(\gamma)$ . Then there exists a point $z^ \in \gamma$ such that $\left| f_2 \circ \cdots \circ f_{N-1} \circ f_N(g^{p-1}(z^)) \right| = R$ , and from (17) we have $\left| g^p(z^*) \right| > KR$ . If $\left\{g^n(a)\right\}$ is unbounded, then for some $0\leqslant m\leqslant p-1$ , $g^{np+m}|_D\to\infty(n\to\infty)$ . It is easy to see from unboundedness of D and (2) and (3) that $g^m(D)$ is also unbounded. Since $g^p(g^m(D))\subset g^m(D)$ , applying Lemma 3.2 to $g^p$ in $g^m(D)$ gives the existence of $\Gamma$ tending to $\infty$ from $g^m(a)$ in $g^m(D)$ such that $\left|g^p(z)\right|\leqslant L|z|,\ z\in\Gamma$ , where L is a positive constant. In the above discussion, we replace $\gamma$ by a segment curve of $\Gamma$ in $\left\{|z|\leqslant R\right\}$ , D by $g^m(D)$ and K by 2L, then we have a point $z^\in\Gamma$ such that $\left|g^p(z^)\right|>2LR$ . On the other hand, $\left|g^p(z^)\right|\leqslant L|z^|\leqslant LR$ , so that $LR\geqslant 2LR$ . We get a contradiction. Then $\left\{g^n(a)\right\}$ is bounded. If $\{g^n|_D\}$ has only constant limit functions, then from Lemma 3.1 $g^{np+m}(a) \to p_m \ (n \to \infty), 0 \le m \le p-1$ . If $p_0 \in D$ , then $p_m \in g^m(D), 0 \le m \le p-1$ so that we can require that (16) also holds for i > p. If $p_0 \in \partial D$ , we can also do this under the assumption of Theorem 1.3 about this case. Thus it is clear that $g^{np}(\gamma)$ cannot tend to $p_0$ . A contradiction we get gives that $\{g^n|_D\}$ has a non-constant limit function. Then all the limit points of $\{g^n(a)\}$ must be analytic points of $f_j \circ \cdots \circ f_N(z)$ , $(j=1,2,\ldots,p)$ so that we can make (16) hold for i > p. Applying Lemma 3.1 implies that there is a subsequence $\{g^{nkp}\}$ of $\{g^{np}\}$ such that $$g^{n_k p}(z) \to z, (k \to \infty)$$ in D. When njt is sufficiently large, we have (21) $$|g^{n_k p}(z)| < (K-1)|z|, z \in \gamma.$$ On the other hand, from (21) and the discussion in the first paragraph, there is a znk s 7 such that $$KR < |g^{n_k p}(z_{n_k})| < (K-1)|z_{n_k}| \leq (K-1)R.$$ This is impossible. D Theorem 1.3 follows.