#### Abstract

We prove that composite transcendental entire functions with certain gaps have no unbounded Fatou component.

#### 1. Introduction

Let be a transcendental entire function. We write , and for for the th iterate of . The Fatou set or set of normality of consists of all in the complex plane which has a neighborhood such that the family is a normal family. The Julia set of is . For the fundamental results in the iteration theory of rational and entire functions, we refer to the original papers of Fatou , and Julia  and the books of Beardon , Carleson and Gamelin , Milnor , and Ren .

Let be a connected component of . Then , where is a component of . If there is a smallest positive integer such that , then is periodic of period . In particular, if , then is called invariant. If for some integer , is periodic, while is not periodic then is called preperiodic. If is periodic and then is called a Baker domain. If all the are disjoint, that is, for , then is called a wandering domain. Let be a transcendental entire function. In 1981, Baker  proposed whether every component of is bounded if the growth of is sufficiently small. The appropriate growth condition would appear to be of order , minimal type at most. In , Baker observed that this condition is best possible in the following sense: for any sufficiently large positive , the function is of order , mean type, and has an unbounded component of containing a segment of the positive real axis, such that as , locally uniformly in .

It is conjectured that if the order of is less than , minimal type, then every component of is bounded. It is still open for wandering domains, although there are several remarkable results for wandering domains under the assumptions that the growth satisfies, in addition, some regular conditions; see .

Suppose that is an entire function with gaps; that is, some of the are zero, in a certain sense. Then the function has the form . We say that has Fabry gaps if as , and has Fejér gaps if .

Wang  proved that every component of the Fatou set of an entire function with certain gaps is bounded, by using the properties of the entire functions with such gaps. Wang  obtained the following result.

Theorem 1. Let be an entire function with . If has Fabry gaps, then every component of is bounded.

For Fejér gap, Wang  proposed the following problem.

Let be an entire function with Fejér gaps, that is, Is every component of bounded?

For composite of entire function, Qiao  proved the following result.

Theorem 2. Let be a transcendental entire function, where are entire functions with order . Then every nonwandering component is bounded.

Cao and Wang  proved the following result.

Theorem 3. Let , where are nonconstant holomorphic maps, each having order less than 1/2. If there is a number such that the lower order of is greater than 0, then every component of is bounded.

Singh  proved the following result.

Theorem 4. Let be the set of all entire functions such that, for given , holds for all outside a set of logarithmic density 0. Let where is the set of all transcendental entire functions such that If , where , then every component of is bounded.

#### 2. Preliminaries

We use the standard notations for the maximum modulus , minimum modulus , order of growth , and lower order of growth of a function ; namely, Briefly, we also denote maximum modulus and minimum modulus by and .

Let be a set in . The logarithmic measure of a set is defined by . If , denote the part of in the interval , that is, , then the upper logarithmic density of the set is defined by the lower logarithmic density of the set is defined by If the upper and lower logarithmic density are equal, their common value is called the logarithmic density of .

Lemma 5 (see ). Let be a transcendental entire function. Then there exists such that, for all and all ,

Lemma 6 (see ). Let be an entire function of finite order with Fabry gaps. Then for given , holds for all r outside a set of logarithmic density 0.

Lemma 7 (see ). Let be an entire function with for some . Then for given , holds for all outside a set of logarithmic density 0.

Lemma 8 (see ). If satisfies the gap-condition , then for given , holds for all outside a set of finite logarithmic measure.

Lemma 9 (see ). For an entire function with Fejér gaps and ,

Lemma 10 (see ). Let be a domain and a compact subset of . Let be the family of all holomorphic functions g on which omit the values 0, 1 and satisfy the condition that on . Then there exist constants and such that , for every and every .

Lemma 11. Let be a transcendental entire function of finite order with Fabry gaps. Then there exist and such that, for all , there exists satisfying and .

Proof. By Lemma 6, given any , we have , for all outside a set of logarithmic density 0. Let . Then there exists such that, for all , there exists such that . If not, then there exists a sequence such that , for every . Thus . Socontradicting .
Set . There exists such that . By Lemma 6, Since , there exists , where such that . This completes the proof of Lemma 11.

By Lemma 9 and the same method in the proof of Lemma 11, we can prove the following result.

Lemma 12. Let be a transcendental entire function of finite order with Fejér gaps. Then there exist and such that, for all , there exists satisfying and .

By Lemma 7 and the same method in the proof of Lemma 11, we can prove the following result.

Lemma 13. Let be a transcendental entire function of finite order with for some . Then there exist and such that, for all , there exists satisfying and .

#### 3. Main Results

In 2012, the authors proved some results on the bounded Fatou components of transcendental entire functions with gaps; see . In this paper, we investigate the iteration of the composite entire functions with gaps and obtain the following results.

Theorem 14. Let be a transcendental entire function, where have Fabry gaps of finite order. If then has no unbounded component.

Proof. The proof follows the idea of Theorems 3 and 4. Since is a transcendental entire function of finite order with Fabry gaps, by Lemma 11, there exist and sufficiently large such that, for , there exists such that andSince is a transcendental entire function of finite order, must be a transcendental entire function. It follows that there exists a number such that for all , and there exists a number such that for all sufficiently large. In fact, if there is a sequence which tend to such that then So Which gives a contradiction if we take any .
Given , define inductively It is easy to see that, for all , Thus and as for all .
Take number sufficiently large, for , by Lemma 11, there exists such that for . By the hypotheses of Theorem 14, (19), and Lemma 11, we havefor ; andSuppose on contrary that has an unbounded component . Without loss of generality we may assume 0, 1 belong to . Hence each map omits the values 0, 1 in . It follows from the unboundedness and connectivity of that there exists such that meets the circles for all ,  .
We choose a value such that and note that must contain a path joining a point to a point . It is clear that contains two subsets , such that joins to and contains and joins to . We know that and so . Also and so . Hence contains an arc joining a point to a point . Similarly contains an arc joining a point to a point .
Repeating the process inductively we obtain that contains an arc joining to a point and contains an arc joining a point to a point .
Since and are two subsets of , it follows that must contain an arc joining to the point . By induction it now follows that contains an arc joining a point to the point .
Thus is a component of containing , and, on , takes a value of modulus at least and as . Thus we conclude that locally uniformly in . Hence there exists such that, for all , for all . Thus the family satisfies the conditions of Lemma 10 on , and so there exist constants , such that for all and for all . Choose with such that and ; we have for all which contradicts the fact that is a transcendental entire function and as . This completes the proof of Theorem 14.

Corollary 15 (see ). Let be a transcendental entire function of finite order with Fabry gaps. If then has no unbounded component.

Remark 16. If satisfies , , then , so Corollary 15 is an extension of Theorem 1.
By Lemma 12 and the same method used in the proof of Theorem 14, we can show the following result.

Theorem 17. Let be a transcendental entire function, where have Fejér gaps. If then has no unbounded component.

By Theorem 17, we have the following.

Corollary 18 (see ). Let be a transcendental entire function with Fejér gaps. If then has no unbounded component.

Remark 19. Corollary 18 partly answers the problem of Wang on the Fejér gaps.
By Lemma 13 and the same method of the proof of Theorem 14, we can show Theorem 20.

Theorem 20. Let be a transcendental entire function, where are entire functions with the gap-conditions If then has no unbounded component.

By Theorem 20, we have the following.

Corollary 21. Let be a transcendental entire function, where are entire functions with the gap-conditions for some . If then has no unbounded component.

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

This work was supported by the National Natural Science Foundation of China (Grant no. 11261002), the Natural Science Foundation of Yunnan Province of China (Grant no. 2012FZ167), and the Educational Commission of Yunnan Province of China (Grant no. 2012z121).