Exponential polynomials with Fatou and non-escaping sets of finite Lebesgue measure

We give conditions ensuring that the Fatou set and the complement of the fast escaping set of an exponential polynomial $f$ have finite Lebesgue measure. Essentially, these conditions are designed such that $|f(z)|\ge\exp(|z|^\alpha)$ for some $\alpha>0$ and all $z$ outside a set of finite Lebesgue measure.


Introduction and results
Let f be a transcendental entire function, and let f n denote the n-th iterate of f . The Fatou set F (f ) of f is the set of all z ∈ C such that the iterates (f n ) n∈N 0 form a normal family in a neighborhood of z, and the Julia set J(f ) is the complement of F (f ). These sets play an important role in complex dynamics. Clearly, F (f ) is open, and J(f ) is closed. Moreover, J(f ) is always non-empty, and either J(f ) = C, or J(f ) has empty interior. An introduction to the dynamics of transcendental entire functions can be found in [2]. The escaping set of f is defined by I(f ) := {z : f n (z) → ∞ as n → ∞}.
Eremenko [7] showed that I(f ) is always non-empty, and that J(f ) = ∂I(f ). For r > 0, let M (r, f ) := max |z|=r |f (z)| denote the maximum modulus of f , and let M n (r, f ) be its n-th iterate with respect to r. The fast escaping set A(f ) is a subset of the escaping set. It was introduced by Bergweiler and Hinkkanen [5], and is defined by A(f ) := {z : there exists l ∈ N such that |f n (z)| ≥ M n−l (R, f ) for n > l}, where R is chosen such that M (r, f ) > r for r ≥ R.
We are interested in the Lebesgue measure of the sets defined above. McMullen [9] showed that the Julia set of f (z) = sin(az + b), a = 0, has positive Lebesgue measure. In fact, it can be seen from the proof that also J(f ) ∩ A(f ) has positive measure. Sixsmith [13] proved that if f (z) = q j=1 a j exp ω j q z , where q ≥ 2, a j ∈ C \ {0}, and ω q = exp(2πi/q), then J(f ) ∩ A(f ) has positive measure. Sixsmith remarked without proof that his result remains true for where q ≥ 3, a j , b j ∈ C\{0}, arg(b j ) < arg(b j+1 ) < arg(b j )+π for j ∈ {1, ..., q −1}, and arg(b q ) > arg(b 1 ) + π, with the argument chosen in [0, 2π). Bergweiler and Chyzhykov [4] gave conditions ensuring that the Julia set and the escaping set of a transcendental entire function of completely regular growth have positive measure. These conditions are satisfied for the functions (1). In fact, they are also satisfied if one allows arg(b j+1 ) = arg(b j ) + π for some j ∈ {1, ..., q − 1} or arg(b q ) = arg(b 1 ) + π. Further criteria for Julia sets and (fast) escaping sets to have positive measure are given in [1,3].
For certain functions it is possible to obtain stronger results in the sense that one can bound the size of the complement of the Julia set or (fast) escaping set. Schubert [12] used McMullen's methods to prove that if f (z) = sinh(z), then the Lebesgue measure of F (f ) and C \ I(f ) is finite in any horizontal strip of width 2π. In fact, the proof shows that one may replace I(f ) by the fast escaping set A(f ) here. Schubert's result was generalized by Zhang and Yang [14] to functions of the form f (z) = P (e z )/e z , where P is a polynomial of degree at least 2 satisfying P (0) = 0.
There seem to be no papers whose main aim is to show that the Lebesgue measure of the Fatou set or the complement of the (fast) escaping set of certain transcendental entire functions is finite. However, there are some results occurring in papers mainly treating a different subject. We mention two of them. Hemke [8,Theorem 5.1] showed where P, Q 1 , Q 2 are polynomials with Q 1 , Q 2 ≡ 0 and deg(P ) ≥ 3, then the Lebesgue measure of C \ I(f ) is finite. One example for such a function is f (z) = sin(z 3 ). A result of Bock [6,Example 2] says that if f (z) = sin(πz), then F (f ) = ∅, and (f n (z)) tends to infinity for almost all z ∈ C. This is different for f (z) = sin(z), which is conjugate to the function sinh(z) considered by Schubert. From f (0) = 0 and f (0) = 1 it follows that F (f ) = ∅, and that there exists a component of F (f ) where (f n (z)) tends to zero. Since F (f ) is open and 2πperiodic, the Lebesgue measure of F (f ) and C \ I(f ) is infinite. Also, the Fatou set and non-escaping set of f (z) = sin(z 2 ) have infinite measure. To see this, note that f has a superattracting fixed point at zero. Let ε > 0 such that D(0, ε) is contained in the attractive basin of zero. There exists δ ∈ (0, π/2) such that sin(D(πk, δ)) ⊂ D(0, ε) for all k ∈ Z. Let D k := D(πk, δ) and p(z) = z 2 . Then p −1 (D k ) is contained in the attractive basin of zero of the function f . For a measurable set A ⊂ C let meas(A) denote the Lebesgue measure of A. We get .
Summing up over all k yields that the attractive basin of zero has infinite measure. See Figure 1 for an illustration of the non-escaping sets of sin(z), sin(z 2 ), and sin(z 3 ). In this paper, we consider exponential polynomials of the form where Q j and P j are polynomials with deg(P j ) < d. We give conditions ensuring that the Lebesgue measure of the complement of J(f ) ∩ A(f ) is finite.
where d ∈ N with d ≥ 3, P j and Q j are polynomials with Q j ≡ 0 and deg(P j ) < d, Note that the conditions on the b j imply that N ≥ 3. Recall that Sixsmith's result for the functions (1) remains true if arg(b j+1 ) = arg(b j ) + π for some j ∈ {1, ..., q − 1} or arg(b q ) = arg(b 1 ) + π. This is not true in general for Theorem 1.1, as the following example shows.
Then h has a superattracting fixed point at zero, and the attractive basin of zero has infinite Lebesgue measure. In particular, the Lebesgue measure of C \ (A(h) ∩ J(h)) is infinite.
We will verify this in Section 5. However, under certain additional conditions on the polynomials P j , the statement of Theorem 1.1 remains true if arg(b j+1 ) = arg(b j ) for some j ∈ {1, ..., N − 1} or arg(b N ) = arg(b 1 ) + π. This is the following result.
If there exists j ∈ {1, ..., N − 1} such that arg(b j+1 ) = arg(b j ) + π, or if arg(b N ) = arg(b 1 ) + π, in addition suppose that there are k, l ∈ {1, ..., N } with arg(b k ) = arg(b j ) and arg(b l ) = arg(b j+1 ), or arg(b k ) = arg(b 1 ) and arg(b l ) = arg(b N ), respectively, such that the polynomials P k , P l can be written in the form with polynomials g, g k , g l satisfying deg(g) Note that the conditions on the b j imply that N ≥ 2. Theorem 1.1 is a special case of Theorem 1.3. Also, the functions (2) considered by Hemke satisfy the assumptions of Theorem 1.3.
Throughout the rest of the paper, let f be an entire function satisfying the assumptions of Theorem 1.3. In Section 2, we will show that f can be approximated by simpler functions in large parts of the complex plane, and use this to prove that |f (z)| is large outside a set of finite measure. Then, in Section 3, we show that f is injective in certain small disks. We finish the proof of Theorem 1.3 in Section 4, using a construction similar to one that occurred in McMullen's paper [9], and has since then been used by various authors. Finally, in Section 5, we verify the properties of Example 1. 2   2 The behaviour of f In this section, we prove several properties of the function f . We first introduce some and define the sets Moreover, define "exceptional sets" for l ∈ {1, 2} (see Figure 2).
Lemma 2.1. The Lebesgue measure of E 1 and E 2 is finite.
We will later prove that the function f behaves "nicely" outside E 1 . For z 0 ∈ C and r > 0, we denote by D(z 0 , r) := {z : |z − z 0 | < r} the open disk of radius r around z 0 . Lemma 2.1 follows directly from Lemma 2.2. Let P be a polynomial of degree d, with d as before. Then the Lebesgue measure of P −1 (U 2 ) is finite.
The next Lemma yields that if R 0 > 0 is large, then in each component of By continuity, m depends only on the connected component of C \ (E 1 ∪ D(0, R 0 )) containing z, and not on z itself. We get if R 0 and hence |z| is sufficiently large. This is the result for f . Moreover, The result for f now follows from similar estimates as above and the fact that Analogously, the result for f follows from Remark. In order to prove Lemma 2.3, we did not need any assumptions on the arguments of the b j . In particular, the statement remains true without the additional condition (3) in the case that arg(b j+1 ) = arg(b j ) + π for some j ∈ {1, ..., N − 1} or This is different for the next result.
Remark. Without the additional condition (3) in the case that arg(b j+1 ) = arg(b j )+π for some j ∈ {1, ..., N } or arg(b N ) = arg(b 1 ) + π, the statement of Lemma 2.4 is not true in general. We will prove in Section 5 that the function h(z) = 1 2 exp(z 3 + iz) − if |z| is sufficiently large. Thus, it suffices to show that there exists j ∈ {1, ..., N } with We first consider the case that f satisfies the assumptions of Theorem 1.1, that is, Then there is a constant C > 0, such that for all z ∈ C there exists j ∈ {1, ..., N } with if |z| is sufficiently large. Now suppose that f does not satisfy the assumptions of Theorem 1.1. Then the assumptions of Theorem 1.3 imply that there are j, k ∈ {1, ..., N } satisfying | arg(b j ) − arg(b k )| = π, and polynomials g, g j , g k with deg(g) ≤ d−1 and max{deg(g j ), deg(g k )} ≤ d − 3, such that P j = b j g + g j and P k = b k g + g k .
With β j := arg(b j ) we have b j = |b j |e iβ j and b k = −|b k |e iβ j .

Injectivity
The aim of this section is to prove that f is injective in certain disks contained in C\E 1 . We start with a basic injectivity criterion (see, e.g., [10, Proposition 1.10]). We also require the following criterion.
Lemma 3.2. Let z 0 ∈ C and r > 0. Let h be holomorphic in D(z 0 , r). Suppose that h (ζ) = 0 for all ζ ∈ D(z 0 , r) and Then h is injective in D(z 0 , r).
This follows directly from Becker's univalence criterion (see, e.g., [11,Theorem 6.7]). However, Lemma 3.2 may also be proved by much more elementary arguments using Lemma 3.1. We sketch the proof here.
We now state the main result of this section.
For z ∈ C and A ⊂ C let dist(z, A) := inf{|z − a| : a ∈ A} denote the Euclidean distance of z and A.
The following Corollary is an immediate consequence of Lemma 3.4 and Lemma 3.3.

Proof of Theorem 1.3
In this section, we prove Theorem 1.3. First, we collect several results that we require. For α > 0 consider the function We will use the following result [3, Lemma 2.1].
Lemma 4.5. Let z 0 ∈ C, r > 0, and ρ ∈ (0, 1). Suppose that h : D(z 0 , r) → C is holomorphic and injective. Then, for all z ∈ D(z 0 , ρr), Moreover, max |z|≤ρr |h (z)| min |z|≤ρr |h (z)| Proof of Theorem 1.3. Let B 0 be a large open square centred at zero with sides parallel to the real and imaginary axis. LetS be a collection of closed squares in C with sides parallel to the real and imaginary axis such that • for all S ∈S, the side length s of S satisfies with σ as in Lemma 3.3.
If the side length of B 0 is sufficiently large, this can be achieved as follows. First, divide C \ B 0 into squares of a fixed size so that the side length of all squares satisfies the lower bound. If the side length s of a square does not satisfy the upper bound, divide it into four squares of side length s/2, and then continue this procedure until the side length of the squares satisfies the upper bound.
Let S be the collection of all S ∈S such that dist(S, E 1 ) > 2σ min z∈S |z| d−1 . By Lemma 3.4 and the definition of S, if B 0 is sufficiently large. Next, we construct a subset of A(f ) ∩ J(f ) as an intersection of nested sets. Fix a square S 0 ∈ S. Let K 0 := {S 0 } and, for n ∈ N, let K n := {T n ⊂ S 0 : f n (T n ) ∈ S and T n ⊂ T n−1 for some T n−1 ∈ K n−1 }.
We first show that To do so, let z ∈ T . Then f n (z) ∈ C \ E 1 for all n ∈ N 0 . Let α ∈ (0, ν) and β > d.
if |z| is large. By denote the density of A in B. We will show that for any square S ∈ S, where, as before, 0 < α < ν. See Figure 3 for an illustration of pack(f (S)).
Altogether, we get provided the square B 0 is sufficiently large. Thus, if B 0 is large. Let n ∈ N 0 and T n ∈ K n . Then f n (T n ) ∈ S. By (6) applied to S = f n (T n ) and Lemma 2.4, We use this to prove that the set T n \ T n+1 ∈K n+1 T n+1 has small density in T n . For all k ∈ {1, ..., n}, there is a square S k ∈ S such that f k (T n ) ⊂ S k . In particular, f n+1 is injective in T n . Thus, To estimate max z∈Tn |(f n+1 ) (z)| min z∈Tn |(f n+1 ) (z)| , let w 0 ∈ f k (T n ). Then Induction yields diam(f k (T n )) < 1 2 n−k diam(f n (T n )) ≤ In particular, for z k ∈ f k (T n ), f k (T n ) ⊂ D z k , 2 k−n σ max z∈f k (S 0 ) |z| d−1 ⊂ D z k , 2 k−n σ|z k | −(d−1) .