The Hausdorff dimension of Julia sets of meromorphic functions in the Speiser class

Abstract

We show that for each $d\in (0,2]$ there exists a meromorphic function f such that the inverse function of f has three singularities and the Julia set of f has Hausdorff dimension d.

1 Introduction and main results

We will be concerned with the iteration of transcendental meromorphic functions $f:\mathbb {C}\rightarrow \widehat{{\mathbb {C}}}:=\mathbb {C}\cup \{\infty \}$. The main objects studied here are the Fatou set F(f), consisting of all $z\in {\mathbb {C}}$ for which the iterates $f^k$ of f are defined and form a normal family in some neighborhood of z, and the Julia set $J(f):=\widehat{{\mathbb {C}}}{\setminus } F(f)$. For an introduction to the dynamics of transcendental meromorphic functions we refer to [10].

The Speiser class $\mathcal {S}$ consists of all transcendental meromorphic functions f for which the set ${\text {sing}}(f^{-1})$ of singularities of the inverse, i.e., the set of critical and asymptotic values of f, is finite. More precisely, if q denotes the cardinality of ${\text {sing}}(f^{-1})$, then we write $f\in \mathcal {S}_q$. Equivalently, $f\in \mathcal {S}$ if there exist finitely many points $a_1,\dots ,a_q\in \widehat{{\mathbb {C}}}$ such that

$$\begin{aligned} f:\mathbb {C}{\setminus } f^{-1}(\{a_1,\dots ,a_q\})\rightarrow \widehat{{\mathbb {C}}}{\setminus }\{a_1,\dots ,a_q\} \end{aligned}$$

is a covering map. And if q is the minimal number with this property, then $f\in \mathcal {S}_q$. The monodromy theorem implies that we always have $q\ge 2$.

The Eremenko–Lyubich class $\mathcal {B}$ consists of all transcendental meromorphic functions f for which ${\text {sing}}(f^{-1}){\setminus } \{\infty \}$ is a bounded subset of $\mathbb {C}$. Thus

$$\begin{aligned} \mathcal {S}=\bigcup _{q=2}^\infty \mathcal {S}_q \subset \mathcal {B}. \end{aligned}$$

Both the Speiser and the Eremenko–Lyubich class play an important role in transcendental dynamics. The similarities and differences between these classes are addressed in a number of recent papers [2, 12,13,14, 21]. A survey of some of these as well as many other results concerning the dynamics of functions in $\mathcal {S}$ and $\mathcal {B}$ is given in [41].

Considerable attention has been paid to the Hausdorff dimension of Julia sets; see [30] and [45] for surveys. We will denote the Hausdorff dimension of a subset A of $\mathbb {C}$ by ${\text {dim}}A$.

Baker [4] showed that if f is a transcendental entire function, then J(f) contains continua so that ${\text {dim}}J(f)\ge 1$. Stallard [44, Theorem 1.1] showed that for all $d\in (1,2)$ there exists a transcendental entire function f with ${\text {dim}}J(f)=d$ while Bishop [15] constructed an example with ${\text {dim}}J(f)=1$. Stallard’s examples are actually in the Eremenko–Lyubich class $\mathcal {B}$. Previously she had shown that ${\text {dim}}J(f)>1$ for entire $f\in \mathcal {B}$. In particular, this is the case for entire functions in $\mathcal {S}$.

Albrecht and Bishop [1] showed that given $\delta >0$ there exists an entire function $f\in \mathcal {S}$ such that ${\text {dim}}J(f)<1+\delta$. In fact, these were the first examples of entire functions in $\mathcal {S}$ for which the Julia set has dimension strictly less than 2. In their examples the inverse has three finite singularities. Since every non-constant entire function has the asymptotic value $\infty$ by Iversen’s theorem, and since we include $\infty$ in ${\text {sing}}(f^{-1})$, their examples are in $\mathcal {S}_4$.

For transcendental meromorphic functions f we have ${\text {dim}}J(f)>0$ by a result of Stallard [42]. On the other hand, she showed [43, Theorem 5] that for all $d\in (0,1)$ there exists a transcendental meromorphic function f such that ${\text {dim}}J(f)=d$. Again her examples are in $\mathcal {B}$. Together with her result covering the interval (1, 2) mentioned above, and since $J(\exp z)=\mathbb {C}$ by a result of Misiurewicz [38] and $J(\tan z)=\mathbb {R}$, it follows that for all $d\in (0,2]$ there exists $f\in \mathcal {B}$ such that ${\text {dim}}J(f)=d$.

We shall show that such examples also exist in the Speiser class $\mathcal {S}$ and in fact in $\mathcal {S}_3$.

Theorem 1.1

Let $d\in (0,2]$. Then there exists a function $f\in \mathcal {S}_3$ such that ${\text {dim}}J(f)=d$.

We note that if $f\in \mathcal {S}_2$, then ${\text {dim}}J(f)>1/2$; see [8, Theorem 3.11] and [34, Remark 3.2 and Section 4.3]. On the other hand, Barański [8, Section 4] showed that for $f_\lambda (z)=\lambda \tan z\in \mathcal {S}_2$ the function $\lambda \mapsto {\text {dim}}J(f_\lambda )$ maps (0, 1] monotonically and continuously onto (1/2, 1]. It seems likely that for $d\in (1,2]$ there also exists $f\in \mathcal {S}_2$ such that ${\text {dim}}J(f)=d$. The main interest of Theorem 1.1 thus lies in the case where $0<d\le 1/2$.

2 Preliminary results

The escaping set I(f) of a meromorphic function f is defined as the set of all $z\in \mathbb {C}$ for which $f^n(z) \rightarrow \infty$ as $n\rightarrow \infty$. We always have $I(f)\ne \emptyset$ and $J(f)=\partial I(f)$. This was shown by Eremenko [22] for transcendental entire f and by Domínguez [17] for transcendental meromorphic f.

For $f\in \mathcal {B}$ we have $I(f)\subset J(f)$. This is due to Eremenko and Lyubich [24, Theorem 1] for entire f and to Rippon and Stallard [40, Theorem A] for meromorphic f.

Theorem 1.1 complements the result of [2] where it was shown that for all $d\in [0,2]$ there exists a meromorphic function $f\in \mathcal {S}$ such that ${\text {dim}}I(f)=d$.

An important concept in the theory of meromorphic functions is the order; see, e.g., [28]. The following result was proved in [11, Theorem 1.1]. Here and in the following $\mathbb {N}=\{1,2,3,\dots \}$.

Lemma 2.1

Let $f\in \mathcal {B}$ be of finite order $\rho$. Suppose that $\infty$ is not an asymptotic value of f and that there exists $M\in \mathbb {N}$ such that all but finitely many poles of f have multiplicity at most M. Then

$$\begin{aligned} {\text {dim}}I(f)\le \frac{2M\rho }{2+M\rho }. \end{aligned}$$

(2.1)

We will not actually use this lemma. Instead we will use the following closely related result, which can be proved by the same method.

Lemma 2.2

Let f be as in the Lemma 2.1. Suppose, furthermore, that $f(0)\ne 0$. For $\lambda \in \mathbb {C}{\setminus }\{0\}$ define $f_\lambda (z)=f(\lambda z)$. Then

$$\begin{aligned} \limsup _{\lambda \rightarrow 0} {\text {dim}}J(f_\lambda ) \le \frac{2M\rho }{2+M\rho }. \end{aligned}$$

Proof

It is easy to see that $f_\lambda$ also has order $\rho$, for all $\lambda \in \mathbb {C}{\setminus }\{0\}$. We proceed as in [11]. Let $(a_j)$ be the sequence of poles of f and let $m_j$ be the multiplicity of $a_j$. Let $b_j\in \mathbb {C}{\setminus }\{0\}$ be such that

$$\begin{aligned} f(z)\sim \left( \frac{b_j}{z-a_j}\right) ^{m_j} \quad \text {as}\ z\rightarrow a_j . \end{aligned}$$

Let

$$\begin{aligned} t> \frac{2M \rho }{2+ M \rho }. \end{aligned}$$

By [11, Lemma 3.1] we have

$$\begin{aligned} \sum _{j=1}^\infty \left( \frac{|b_j|}{|a_j|^{1+1/M}}\right) ^t < \infty . \end{aligned}$$

(2.2)

Let $a_j^\lambda$ and $b_j^\lambda$ be the corresponding values for $f_\lambda$. Then $a_j^\lambda =a_j/\lambda$ and $b_j^\lambda =b_j/\lambda$ so that

$$\begin{aligned} \sum _{j=1}^\infty \left( \frac{|b_j^\lambda |}{|a_j^\lambda |^{1+1/M}}\right) ^t =|\lambda |^{t/M}\sum _{j=1}^\infty \left( \frac{|b_j|}{|a_j|^{1+1/M}}\right) ^t . \end{aligned}$$

(2.3)

As in [11] we choose $R_0>|f(0)|$ such that ${\text {sing}}(f_\lambda ^{-1})={\text {sing}}(f^{-1})\subset D(0,R_0)$ and choose $R\ge (16R_0)^M$. Here and in the following D(a, r) denotes the open disk of radius r around a point $a\in \mathbb {C}$. Choosing $\lambda$ small we can achieve that $f_\lambda (D(0,3R))\subset D(0,R)$ so that $D(0,3R)\subset F(f_\lambda )$.

We put $B(R):=\{z:|z|>R\}\cup \{\infty \}$ and, as in [11, pp. 5376f.], consider the collection $E_l$ of all components V of $f_\lambda ^{-l}(B(R))$ for which $f_\lambda ^k(V)\subset B(3R)$ for $0\le k\le l-1$. Then $E_l$ is a cover of $J(f_\lambda )$ and we have

$$\begin{aligned} \sum _{V\in E_l}\!\left( {\text {diam}}_{\chi }(V)\right) ^t = \frac{1}{M}\!\left( \frac{32}{(2R)^{1/M}24}\right) ^{\! t} \left( M (2^{1/M} 24)^{t} \sum _{j=1}^\infty \!\left( \frac{|b_j^\lambda |}{|a_j^\lambda |^{1+1/M}}\right) ^{\! t} \right) ^{\! l}. \end{aligned}$$

Here ${\text {diam}}_{\chi }(V)$ denotes the spherical diameter of V. Together with (2.2) and (2.3) the last estimate yields that if $\lambda$ is sufficiently small, then

$$\begin{aligned} \sum _{V\in E_l}\left( {\text {diam}}_{\chi }(V)\right) ^t\rightarrow 0 \quad \text {as}\ l\rightarrow \infty . \end{aligned}$$

Hence ${\text {dim}}J(f_\lambda )\le t$ for small $\lambda$. $\square$

It follows from a recent result of Mayer and Urbański [35] that Lemma 2.2 and (2.1) can be sharpened to

$$\begin{aligned} \lim _{\lambda \rightarrow 0} {\text {dim}}J(f_\lambda )={\text {dim}}I(f). \end{aligned}$$

In fact, their result says that ${\text {dim}}I(f)$ is the infimum of the set of all $t>0$ for which (2.2) holds.

For entire functions, the following result can be found in [24, Section 3] and [21, Proposition 2.3].

Lemma 2.3

Let $f,g\in \mathcal {S}_3$ and suppose that there exist homeomorphisms $\psi :\widehat{{\mathbb {C}}}\rightarrow \widehat{{\mathbb {C}}}$ and $\phi :\mathbb {C}\rightarrow \mathbb {C}$ such that $\psi \circ f=g\circ \phi$. Then there exist a fractional linear transformation $\alpha :\widehat{{\mathbb {C}}}\rightarrow \widehat{{\mathbb {C}}}$ and an affine map $\beta :\mathbb {C}\rightarrow \mathbb {C}$ such that $\alpha \circ f=g\circ \beta$.

Proof

Since $f\in \mathcal {S}_3$ it follows from [21, Observation 1.10] that there exists a fractional linear transformation $\alpha :\widehat{{\mathbb {C}}}\rightarrow \widehat{{\mathbb {C}}}$ which is isotopic to $\psi$ relative to ${\text {sing}}(f^{-1})$. We follow the argument in the proof of [21, Proposition 2.3(a)]. Let $(\psi _t)_{t\in [0,1]}$ be the isotopy between $\psi _0=\psi$ and $\psi _1=\alpha$. By the isotopy lifting property, there exists a unique isotopy $(\phi _t)_{t\in [0,1]}$ in

$$\begin{aligned} U:=f^{-1}\!\left( \widehat{{\mathbb {C}}}{\setminus }{\text {sing}}(f^{-1})\right) \end{aligned}$$

such that $\phi _0=\phi$ and $\psi _t\circ f=g\circ \phi _t$ in U for all $t\in [0,1]$. It remains to show that $\phi _t$ extends continuously to the preimages of the singular values of f and coincides with $\phi$ there for all t.

Let $z_0\in \mathbb {C}$ be a preimage of $v_0\in {\text {sing}}(f^{-1})$. We have to show that $\phi _t(z)\rightarrow \phi (z_0)$ as $z\rightarrow z_0$. We take a small neighborhood D of $z_0$ such that $f:D\rightarrow f(D)$ is a proper map, with no critical points except possibly at $z_0$. We may also assume that $f(z)\ne v_0$ for $z\in {\overline{D}}{\setminus } \{z_0\}$. It then suffices to show that $\phi _t(z)\in \phi (D)$ if z is sufficiently close to $z_0$. By the continuity of f and the properties of an isotopy, we have

$$\begin{aligned} \psi _t(f(z))\rightarrow \psi _t(f(z_0))=\psi (v_0)\quad \text {as}\ z\rightarrow z_0, \end{aligned}$$

uniformly in $t\in [0,1]$. It follows that

$$\begin{aligned} g(\phi _t(z))=\psi _t(f(z))\in \psi (f(D))=g(\phi (D)) \end{aligned}$$

if z is sufficiently close to $z_0$. Thus $\phi _t(z)$ can be obtained by analytic continuation of $g^{-1}$ along the curve $t\mapsto \psi _t(f(z))$. It follows that $\phi _t(z)\in \phi (D)$ if z is sufficiently close to $z_0$.

With $\beta :=\phi _1$ we have $\alpha \circ f=g\circ \beta$. If z is not a critical point of f, then $\beta$ is of the form $g^{-1}\circ \alpha \circ f$ near z for some branch of the inverse of g. Hence $\beta$ is holomorphic. Since $\beta :\mathbb {C}\rightarrow \mathbb {C}$ is a homeomorphism, this implies that $\beta$ is affine. $\square$

The following result is due to Kotus and Urbański [29].

Lemma 2.4

Let f be an elliptic function. Let q be the maximal multiplicity of the poles of f. Then

$$\begin{aligned} {\text {dim}}J(f)> \frac{2q}{q+1}. \end{aligned}$$

We shall need a variation of this result.

Lemma 2.5

Let f be as in Lemma 2.4. If $(f_n)$ is a sequence of meromorphic functions which converges locally uniformly to f with respect to the spherical metric, then

$$\begin{aligned} {\text {dim}}J(f_n)> \frac{2q}{q+1} \end{aligned}$$

for all large n.

Lemma 2.5 can be deduced from the work of Kotus and Urbański; see Remark 2.1 below. But for the convenience of the reader we will include a proof of Lemma 2.5, thereby reproving Lemma 2.4. Here we will use the following result [25, Proposition 9.7].

Lemma 2.6

Let $S_1,\dots ,S_m$ be contractions on a closed subset K of $\mathbb {R}^d$ such that there exists $b_1,\dots ,b_m\in (0,1)$ with

$$\begin{aligned} b_k|u-v|\le |S_k(u)-S_k(v)| \quad \text {for } u,v\in K \end{aligned}$$

(2.4)

and $1\le k\le m$. Suppose that $K_0$ is a non-empty compact subset of K with

$$\begin{aligned} K_0=\bigcup _{k=1}^m S_k(K_0) \end{aligned}$$

and $S_j(K_0)\cap S_k(K_0)=\emptyset$ for $j\ne k$. Let $t>0$ with

$$\begin{aligned} \sum _{k=1}^m b_k^t=1. \end{aligned}$$

Then ${\text {dim}}K_0\ge t$.

Proof of Lemmas 2.4 and 2.5

Let $(a_j)$ be the sequence of poles of multiplicity q. For sufficiently large R there is a neighborhood $U_j$ of $a_j$ such that

$$\begin{aligned} f:U_j{\setminus }\{a_j\}\rightarrow \{z\in \mathbb {C}:|z|>R\} \end{aligned}$$

is a covering map of degree q. There exists $r_1>0$ such that $D(a_j,r_1)\subset U_j$ for all j. Let $0<r_0<r_1$. Then there exists $M\in \mathbb {N}$ such that

$$\begin{aligned} \overline{D(a_k,r_0)}\subset \overline{U_k}\subset f(D(a_j,r_0)) \end{aligned}$$

for all $j,k\ge M$. Thus for $j,k\ge M$ there exists $V_{j,k}\subset D(a_j,r_0)$ such that $f:V_{j,k}\rightarrow D(a_k,r_0)$ is biholomorphic.

Let $W_k:=f^{-1}(V_{k,M})\cap D(a_M,r_0)$. Then

$$\begin{aligned} f^2:W_{k}\rightarrow D(a_M,r_0) \end{aligned}$$

is biholomorphic. Moreover, $f^2$ extends to a bijective map from $\overline{W_k}$ to $K:=\overline{D(a_M,r_0)}$. Let $S_k:K\rightarrow \overline{W_{k}}$ be the inverse function of $f^2:\overline{W_{k}}\rightarrow K$. Then $S_k$ extends to an injective map $S_k:D(a_M,r_1)\rightarrow \mathbb {C}$. Choosing $r_0\le (2-\sqrt{3})r_1$ we conclude that $W_k=S_k(D(a_M,r_0))$ is convex; see [19, Theorem 2.13].

In order to apply Lemma 2.6 we note that since f has a pole of multiplicity q at $a_j$, there exists $c_1>0$ such that

$$\begin{aligned} |f'(z)|\le c_1 |f(z)|^{(q+1)/q} \quad \text {for}\ z\in D(a_j,r_0). \end{aligned}$$

This implies that there exists $c_2>0$ such that

$$\begin{aligned} |f'(z)|\le c_2 |a_k|^{(q+1)/q} \quad \text {if}\ z\in D(a_j,r_0) \ \text {and}\ f(z)\in D(a_k,r_0). \end{aligned}$$

With $c_3:= c_2^2 |a_M|^{(q+1)/q}$ this yields that

$$\begin{aligned} |(f^2)'(z)|=|f'(f(z))f'(z)|\le c_3 |a_k|^{(q+1)/q} \quad \text {for}\ z\in W_k. \end{aligned}$$

Since $W_k$ is convex this yields that

$$\begin{aligned} |u-v|=|f^2(S_k(u))-f^2(S_k(v))|\le c_3 |a_k|^{(q+1)/q}|S_k(u)-S_k(v)| \quad \text {for}\ u,v\in K. \end{aligned}$$

It follows that (2.4) holds with

$$\begin{aligned} b_k:=\frac{1}{c_3 |a_k|^{(q+1)/q}} . \end{aligned}$$

Let now $N\ge M$ and define $t>0$ by

$$\begin{aligned} \sum _{k=M}^N b_k^t=1. \end{aligned}$$

It follows from Lemma 2.6 that the limit set of the iterated function system $\{S_k:M\le k\le N\}$ has Hausdorff dimension at least t. It is easily seen that this limit set is contained in the Julia set. Thus ${\text {dim}}J(f)\ge t$. Since

$$\begin{aligned} \sum _{k=M}^\infty \frac{1}{|a_k|^2}=\infty \end{aligned}$$

we have $t>2q/(q+1)$ if N is large enough. This yields Lemma 2.4.

To prove Lemma 2.5 we note that for large n there are domains $W_k^n$ and $U_M^n$ close to $W_k$ and $U_M$ such that

$$\begin{aligned} f_n^2:W_k^n\rightarrow U_M^n \end{aligned}$$

is biholomorphic, and the inverse function $S_k^n$ satisfies (2.4) with a constant $b_k^n$ instead of $b_k$, with $b_k^n\rightarrow b_k$ as $n\rightarrow \infty$. The conclusion then follows again from Lemma 2.6. $\square$

Remark 2.1

The argument of Kotus and Urbański [29] is similar to the one used above, but they use an infinite iterated function system and apply results of Mauldin and Urbański [33] concerning such systems. However, it would suffice to consider a sufficiently large finite subsystem. This would yield Lemma 2.5 as above.

The proof actually yields that the hyperbolic dimension of f and $f_n$, and not only the Hausdorff dimension of their Julia sets, have the given lower bound; see [9, 39] for a discussion of the hyperbolic dimension of meromorphic functions.

Lemma 2.7

Let $f\in \mathcal {S}$. Suppose that f has an attracting fixed point whose immediate attracting basin contains all finite singularities of $f^{-1}$. Suppose also that $\infty$ is not an asymptotic value of f and that there exists a uniform bound on the multiplicities of the poles of f. Then J(f) is totally disconnected and F(f) is connected.

There are several closely related results in the literature, see [5, Theorem G], [18, Theorem A and Corollary  3.2] and [47, Theorem 2.7]. However, none of these results seems to apply exactly to the situation we have.

Zheng [47, Theorem 2.7] showed that the conclusion of Lemma 2.7 holds if $\infty \notin {\text {sing}}(f^{-1})$. Thus his result would apply if the poles are assumed to be simple, while we allow multiple poles. Hawkins and Koss ([27, Theorem 3.12]; see also [26, Theorem 3.2]) do not require that the poles are simple, but they restrict to elliptic functions.

Our proof of Lemma 2.7 will use some ideas from the papers mentioned. A difference to the methods employed there, however, is that we will use the following consequence of the Grötzsch inequality; see [16, Section 5.2] and [36, Corollary A.7]. Here and in the following ${\text {mod}}(A)$ denotes the modulus of an annulus A.

Lemma 2.8

Let $(G_k)$ be a sequence of simply connected domains in $\mathbb {C}$ such that $A_k:=G_k{\setminus } \overline{G_{k+1}}$ is an annulus for all $k\in \mathbb {N}$. Suppose that

$$\begin{aligned} \sum _{k=1}^\infty {\text {mod}}(A_k) =\infty . \end{aligned}$$

Then $\bigcap _{k=1}^\infty G_k$ consists of a single point.

Proof of Lemma 2.7

Let $\xi$ be the attracting fixed point whose attracting basin W contains all finite singularities of $f^{-1}$. Then the postsingular set

$$\begin{aligned} P(f):=\overline{\bigcup _{n=0}^\infty f^n\!\left( {\text {sing}}(f^{-1}){\setminus }\{\infty \}\right) } \end{aligned}$$

is a compact subset of W. It is not difficult to see that there exist Jordan domains U and V such that

$$\begin{aligned} \{\xi \}\cup P(f)\subset U\subset {\overline{U}}\subset V \subset {\overline{V}} \subset W. \end{aligned}$$

Then $A:=V{\setminus }{\overline{U}}$ is an annulus. Clearly, $A\subset W$.

Let $(a_j)$ be the sequence of poles of f and let $m_j$ denote the multiplicity of $a_j$. By hypothesis, there exists $M\in \mathbb {N}$ such that $m_j\le M$ for all j. Let $Y_j$ be the component of the preimage of $\widehat{{\mathbb {C}}}{\setminus } {\overline{U}}$ that contains $a_j$. Then $f:Y_j{\setminus }\{a_j\}\rightarrow \mathbb {C}{\setminus } {\overline{U}}$ is a covering of degree $m_j$. Putting $B_j:=f^{-1}(A)\cap Y_j$ we find that $B_j$ is an annulus with

$$\begin{aligned} {\text {mod}}(B_j)=\frac{1}{m_j}{\text {mod}}(A)\ge \frac{1}{M}{\text {mod}}(A). \end{aligned}$$

To prove that J(f) is totally disconnected, let $z\in J(f)$. We want to show that the component of J(f) containing z consists of the point z only. First we note that for all $n\in \mathbb {N}$ there exists $j(n)\in \mathbb {N}$ such that $f^n(z)\in Y_{j(n)}$. Let $X_n$ be the component of $f^{-n}(Y_{j(n)})$ containing z. Note that $Y_{j(n)}$ and $X_n$ are Jordan domains. Since $\partial U$ and hence $\partial X_n$ are contained in F(f), the component of J(f) containing z is contained in the intersection of the $X_n$. It thus suffices to prove that this intersection consists of only one point.

In order to do so we note that since $P(f)\subset U$, the map $f^{n}:X_n\rightarrow Y_{j(n)}$ is biholomorphic. This implies that

$$\begin{aligned} C_n:=f^{-n}(B_{j(n)})\cap X_n= f^{-n-1}(A)\cap X_n \end{aligned}$$

(2.5)

is an annulus satisfying

$$\begin{aligned} {\text {mod}}(C_n)={\text {mod}}(B_{j(n)})\ge \frac{1}{M}{\text {mod}}(A). \end{aligned}$$

Moreover, $X_n$ is equal to the union of $C_n$ and the component of $\mathbb {C}{\setminus } C_n$ that contains z.

Since the closure of A is a compact subset of the attracting basin W of $\xi$, there exists $p\in \mathbb {N}$ such that $f^p(A)\subset U$. In particular,

$$\begin{aligned} f^p(A)\cap A=\emptyset . \end{aligned}$$

(2.6)

This implies that $C_{n+p}\cap C_n=\emptyset$ for all $n\in \mathbb {N}$. In fact, if $w\in C_{n+p}\cap C_n$, then $f^{n+p+1}(w)\in f^p(A)\cap A$ by (2.5), contradicting (2.6). It follows that

$$\begin{aligned} X_{n+p}\subset X_n{\setminus } C_n, \end{aligned}$$

and hence that

$$\begin{aligned} {\text {mod}}\!\left( X_n{\setminus }\overline{X_{n+p}}\right) \ge {\text {mod}}(C_n)\ge \frac{1}{M}{\text {mod}}(A). \end{aligned}$$

As already mentioned above it now follows from Lemma 2.8, applied with $G_k=X_{1+pk}$, that J(f) is totally disconnected. Of course, this yields that F(f) is connected.

3 Proof of Theorem 1.1

Let G be the conformal map from the triangle with vertices 0, $\pi /2$ and $i\pi /2$ onto the lower half-plane such that

$$\begin{aligned} G(\pi /2)=0, \quad G(0)=1 \quad \text {and}\quad G(i\pi /2)=\infty . \end{aligned}$$

Extending this to the whole plane by reflections we obtain an elliptic function G. The critical values of G are 0, 1 and $\infty$ so that $G\in \mathcal {S}_3$. The zeros and poles of G have multiplicity 4 while the 1-points have multiplicity 2. We also note that $G(\mathbb {R})=[0,1]$.

We may express G in terms of the Weierstrass $\wp$-function with periods $\pi$ and $\pi i$. The critical values of $\wp$ are $e_1$, $e_2$, $e_3$ and $\infty$, with

$$\begin{aligned} e_2=\wp ((\pi +i \pi )/2)=0 \quad \text {and}\quad e_1=\wp (\pi /2)=-e_3=-\wp (i\pi /2) . \end{aligned}$$

It follows from this that

$$\begin{aligned} G(z)=\left( \frac{\wp (z+i\pi /2)}{e_1}\right) ^2 . \end{aligned}$$

First we construct an example of a function $f\in \mathcal {S}_3$ where the Julia set is the whole sphere and thus has Hausdorff dimension 2. To this end we consider the function

$$\begin{aligned} f(z):=i G(\pi z/2). \end{aligned}$$

Then $f\in \mathcal {S}_3$. The critical values of f are 0, i and $\infty$ and we have $f(0)=i$ and $f(i)=\infty$. To prove that $J(f)=\widehat{{\mathbb {C}}}$ we note that f has no wandering domains [6] and no Baker domains [40, Corollary to Theorem A]. All other types of components of F(f) are related to the singularities of $f^{-1}$; see [10, Theorem 7] for the exact statement. Since the points in ${\text {sing}}(f^{-1})\cap \mathbb {C}=\{0,i\}$ are mapped to $\infty$ by f or $f^2$ this yields that $F(f)=\emptyset$ and hence $J(f)=\widehat{{\mathbb {C}}}$ as claimed.

To construct functions in $\mathcal {S}_3$ for which the Julia set has Hausdorff dimension $d\in (0,2)$, we consider, for $p\in \mathbb {N}$ and small $\eta \in (0,\pi /2)$, the function H defined by $H(z):=\eta G(z)^p$. The critical values of H are 0, $\eta$ and $\infty$. Thus $H\in \mathcal {S}_3$. We have $H(0)=\eta$ and $H(\pi /2)=0$. Also, H decreases in the interval $(0,\pi /2)$. Choosing $\eta$ sufficiently small we can achieve that H has an attracting fixed point $\xi \in (0,\pi /2)$ whose attracting basin contains $[0,\eta ]$ and thus, since $H(\mathbb {R})=[0,\eta ]$, also contains $\mathbb {R}$. Lemma 2.7 implies that this attracting basin coincides with F(H) and that J(H) is totally disconnected.

Since $H((0,\pi /2))=(0,\eta )$ we actually have $\xi \in (0,\eta )$. Choosing $\eta$ small we can also achieve that $H''(z)\ne 0$ for $0<|z|\le \eta$. In fact, $H''(x)<0$ for $0<x\le \eta$.

Let now m be a (large) odd integer. Then

$$\begin{aligned} h_m(z)=H(m\arcsin (z/m)) . \end{aligned}$$

defines a meromorphic function $h_m\in \mathcal {S}_3$. Similar examples were already considered by Teichmüller [46, p. 734], and later by Bank and Kaufman [7, Section 5], Langley [31, Section 2] and Eremenko [23].

The elliptic function $H(z)=h_m(m\sin (z/m))$ has order 2. A result of Edrei and Fuchs [20, Corollary  1.2] thus yields that $h_m$ has order 0. In fact, as in the papers cited above we find that there exists a constant c such that the Nevanlinna characteristic satisfies $T(r,h_m)\sim c( \log r)^2$ as $r\rightarrow \infty$.

For large m the function $h_m$ has an attracting fixed point $\xi _m$, with $\xi _m\rightarrow \xi$ as $m\rightarrow \infty$, such that the attracting basin of $\xi _m$ contains the interval $[0,\eta ]$ and hence $\mathbb {R}$. Lemma 2.7 implies that this attracting basin is connected and coincides with $F(h_m)$. Choosing m large we can also achieve that $h_m$ decreases in the interval $[0,\eta ]$ and that $h_m''(x)<0$ for $0<x\le \eta$. This implies that $h_m'$ decreases in the interval $[0,\eta ]$.

The poles of H and $h_m$ have multiplicity 4p. Except for the zeros at $\pm m$, which have multiplicity 2p, the zeros of $h_m$ also have multiplicity 4p. The $\eta$-points on the real axis have multiplicity 2, but if $p\ge 2$, then H and $h_m$ also have simple $\eta$-points (corresponding to the points where G takes the p-th roots of unity).

As the poles of H have multiplicity 4p, Lemma 2.4 yields that

$$\begin{aligned} {\text {dim}}J(H) > \frac{8p}{4p+1} . \end{aligned}$$

Moreover, it follows from Lemma 2.5 that if m is sufficiently large, then

$$\begin{aligned} {\text {dim}}J(h_m)> \frac{8p}{4p+1}. \end{aligned}$$

(3.1)

We fix such a value of m and, for $\lambda \in (0,1]$, we put $f_\lambda (z):=h_m(\lambda z)$ so that $f_1=h_m$.

Since $h_m$ has order 0, Lemma 2.2 yields that

$$\begin{aligned} \lim _{\lambda \rightarrow 0}{\text {dim}}J(f_\lambda ) = 0. \end{aligned}$$

(3.2)

We will show that the function $\lambda \mapsto {\text {dim}}J(f_\lambda )$ is continuous in the interval (0, 1]. Since $f_1=h_m$ it then follows from (3.1) and (3.2) that for all $d\in (0,8p/(4p+1)]$ there exists $\lambda \in (0,1]$ such that ${\text {dim}}J(f_\lambda )=d$. Since p can be chosen arbitrarily large, this yields the conclusion.

It remains to prove that $\lambda \mapsto {\text {dim}}J(f_\lambda )$ is continuous. Recalling that $h_m$ and $h_m'$ are decreasing in the interval $[0,\eta ]$ we can deduce that $f_\lambda$ has an attracting fixed point $\zeta _\lambda \in (0,\eta )$ and that the multiplier

$$\begin{aligned} m_\lambda :=f_\lambda '(\zeta _\lambda )=\lambda h_m'(\lambda \zeta _\lambda ) \end{aligned}$$

is a decreasing function of $\lambda$ in the interval (0, 1]. Note here that $\zeta _1=\xi _m$ and $m_1=h_m'(\xi _m)<0$. As before it follows from Lemma 2.7 that the attracting basin of $\zeta _\lambda$ is connected and coincides with the Fatou set of $f_\lambda$.

Let $\lambda \in (0,1]$. Kœnigs’ theorem [37, Theorem 8.2] yields that there exists a function g holomorphic and injective in some neighborhood U of $\zeta _\lambda$ such that $g(\zeta _\lambda )=0$, $g'(\zeta _\lambda )=1$ and

$$\begin{aligned} g(f_\lambda (g^{-1}(z)))=m_\lambda z \end{aligned}$$

for all $z\in g(U)$. For $\kappa \in (0,1]$ we put

$$\begin{aligned} \gamma =\frac{\log |m_\kappa |}{\log |m_\lambda |} -1 \end{aligned}$$

and define $h:\mathbb {C}\rightarrow \mathbb {C}$, $h(z)= z|z|^\gamma$. Then

$$\begin{aligned} h(m_\lambda h^{-1}(z))=m_\kappa z. \end{aligned}$$

With $\phi =h\circ g:U\rightarrow \mathbb {C}$ we then have

$$\begin{aligned} \phi (f_\lambda (\phi ^{-1}(z)))=m_\kappa z \end{aligned}$$

(3.3)

for $z\in \phi (U)$. The maps h and $\phi$ are K-quasiconformal with

$$\begin{aligned} K= \max \left\{ \frac{\log |m_\kappa |}{\log |m_\lambda |}, \frac{\log |m_\lambda |}{\log |m_\kappa |}\right\} . \end{aligned}$$

(3.4)

For a detailed account of quasiconformal mappings, we refer to [32]. The complex dilatation $\mu (z):=\phi _{{\overline{z}}}(z)/\phi _z(z)$ satisfies

$$\begin{aligned} \mu (f_\lambda (z))= \mu (z) \frac{f_\lambda '(z)}{\;\overline{f_\lambda '(z)}\;} \end{aligned}$$

(3.5)

if $z,f_\lambda (z)\in U$. We may use (3.5) to extend $\mu$ to $\mathbb {C}$. More precisely, we put $\mu (z)=0$ if $z\in J(f_\lambda )$ or if $z\in F(f_\lambda )$ and $(f_\lambda ^n)'(z)=0$ for some $n\in \mathbb {N}$. For the remaining points $z\in F(f_\lambda )$ we define

$$\begin{aligned} \mu (z) = \mu (f_\lambda ^n(z)) \frac{\;\overline{(f_\lambda ^n)'(z)}\;}{(f_\lambda ^n)'(z)}, \end{aligned}$$

(3.6)

where n is chosen so large that $f_\lambda ^n(z)\in U$. Using (3.5) it is easily seen that $\mu$ is well-defined, i.e., the definition does not depend on the value of n chosen in (3.6). We find that (3.5) holds for all z.

Let $\psi :\mathbb {C}\rightarrow \mathbb {C}$ be the solution of the Beltrami equation

$$\begin{aligned} \mu (z)=\frac{\psi _{{\overline{z}}}(z)}{\psi _z(z)}, \end{aligned}$$

normalized by $\psi (0)=0$ and $\psi (\eta )=\eta$. It follows from (3.5) that

$$\begin{aligned} k:=\psi \circ f_\lambda \circ \psi ^{-1} \end{aligned}$$

(3.7)

is meromorphic. Since $f_\lambda$ is symmetric with respect to the real axis, the same applies to g, $\phi$, $\mu$, $\psi$ and k. By definition, $f_\lambda$ is even. In order to show that k is also even, we first note that since $f_\lambda$ is even, it follows from (3.5) that $\mu$ is even. This implies that $\psi (z)$ and $\psi (-z)$ have the same complex dilatation. Hence there exists an affine map L such that $\psi (z)=L(\psi (-z))$. Since $\psi (0)=0$ we have $L(0)=0$ so that L has the form $L(z)=az$ for some $a\in \mathbb {C}{\setminus }\{0\}$. We also see that L is real on the real axis so that $a\in \mathbb {R}{\setminus }\{0\}$. Since $a\psi (i)=L(\psi (i))=\psi (-i)=\overline{\psi (i)}$ we find that $|a|=1$. As $\psi$ is injective this implies that $a=-1$ so that $\psi$ is odd. Hence k is even.

Since the complex dilatations of $\phi$ and $\psi$ agree in U, we have $\phi =\tau \circ \psi$ for some function $\tau$ holomorphic and injective on $\psi (U)$. Together with (3.3) this implies that

$$\begin{aligned} k(z)= \psi (f_\lambda (\psi ^{-1}(z)))=\tau ^{-1}(m_\kappa \tau (z)) . \end{aligned}$$

Thus $\tau ^{-1}(0)=\psi (\zeta _\lambda )$ is a fixed point of k of multiplier $m_\kappa$.

Another function with a fixed point of multiplier $m_\kappa$ is $f_\kappa$. We will show that $k=f_\kappa$. In order to do so we note that both k and $f_\kappa$ are in $\mathcal {S}_3$, with critical values 0, $\eta$ and $\infty$. It follows from Lemma 2.3 and (3.7) that there exist a fractional linear transformation $\alpha$ and an affine map $\beta$ such that $\alpha \circ k=f_\lambda \circ \beta$. Since all poles of k and $f_\kappa$ have multiplicity 4p, all but two zeros of both functions have multiplicity 4p, and all $\eta$-points of both functions have multiplicity 2 or 1, we find that $\alpha (0)=0$, $\alpha (\eta )=\eta$ and $\alpha (\infty )=\infty$. Hence $\alpha (z)\equiv z$ so that $k=f_\lambda \circ \beta$.

As $\beta$ is affine we have $-\beta (-z)=\beta (z)-2\beta (0)$. Noting that k and $f_\lambda$ are even we deduce that

$$\begin{aligned} f_\lambda (\beta (z)-2\beta (0))= f_\lambda (-\beta (-z)) = f_\lambda (\beta (-z))=k(-z)=k(z)=f_\lambda (\beta (z)). \end{aligned}$$

Since periodic functions have order at least 1 while $f_\lambda$ has order 0, this implies that $\beta (0)=0$ so that $\beta$ has the form $\beta (z)=cz$ for some constant c. Thus $k(z)=f_\lambda (cz)=h_m(\lambda cz)$. As k has an attracting fixed point of multiplier $m_\kappa$ this yields that $c=\kappa /\lambda$ so that $k(z)=h_m(\kappa z)=f_\kappa (z)$.

Inserting $k=f_\kappa$ in (3.7), we obtain

$$\begin{aligned} f_\kappa =\psi \circ f_\lambda \circ \psi ^{-1}. \end{aligned}$$

This implies that

$$\begin{aligned} J(f_\kappa )= \psi (J(f_\lambda )). \end{aligned}$$

As $\psi$ is K-quasiconformal, and thus Hölder continuous with exponent 1/K, we deduce that

$$\begin{aligned} \frac{1}{K} {\text {dim}}J(f_\lambda )\le {\text {dim}}J(f_\kappa )\le K{\text {dim}}J(f_\lambda ). \end{aligned}$$

(3.8)

It follows from (3.4) that $K\rightarrow 1$ as $\kappa \rightarrow \lambda$. Thus we deduce from (3.8) that ${\text {dim}}J(f_\kappa )\rightarrow {\text {dim}}J(f_\lambda )$ as $\kappa \rightarrow \lambda$. Hence $\lambda \mapsto {\text {dim}}J(f_\lambda )$ is continuous. $\square$

Remark 3.1

A celebrated result of Astala [3, Corollary 1.3] says that (3.8) can be improved to

$$\begin{aligned} \frac{1}{K} \left( \frac{1}{{\text {dim}}J(f_\lambda )}-\frac{1}{2}\right) \le \frac{1}{{\text {dim}}J(f_\kappa )}-\frac{1}{2} \le K \left( \frac{1}{{\text {dim}}J(f_\lambda )}-\frac{1}{2}\right) . \end{aligned}$$

For our purposes, however, the weaker and simpler estimate (3.8) suffices.