Bounded Fatou Components of Composite Transcendental Entire Functions with Gaps

Let f be a transcendental entire function. We write f = f, and f = f ∘f for n ≥ 2 for the nth iterate of f. The Fatou set or set of normality F(f) of f consists of all z in the complex planeCwhich has a neighborhoodU such that the family {f | U : n ≥ 1} is a normal family. The Julia set J(f) of f is J(f) = C \ F(f). For the fundamental results in the iteration theory of rational and entire functions, we refer to the original papers of Fatou [1–3], and Julia [4] and the books of Beardon [5], Carleson and Gamelin [6], Milnor [7], and Ren [8]. Let U be a connected component of F(f). Then f(U) ⊆


Introduction
Let  be a transcendental entire function.We write  1 = , and   =  ∘  −1 for  ≥ 2 for the th iterate of .The Fatou set or set of normality () of  consists of all  in the complex plane C which has a neighborhood  such that the family {  |  :  ≥ 1} is a normal family.The Julia set () of  is () = C \ ().For the fundamental results in the iteration theory of rational and entire functions, we refer to the original papers of Fatou [1][2][3], and Julia [4] and the books of Beardon [5], Carleson and Gamelin [6], Milnor [7], and Ren [8].
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  = 1, then  is called invariant.If for some integer  ≥ 1,   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 [9] proposed whether every component of () is bounded if the growth of  is sufficiently small.The appropriate growth condition would appear to be of order 1/2, minimal type at most.In [9], Baker observed that this condition is best possible in the following sense: for any sufficiently large positive , the function is of order 1/2, mean type, and has an unbounded component  of () containing a segment [ 0 , ∞) of the positive real axis, such that   () → ∞ as  → ∞, locally uniformly in .
It is conjectured that if the order of  is less than 1/2, minimal type, then every component of () is bounded.
Suppose that () = ∑ ∞ =0     is an entire function with gaps; that is, some of the   are zero, in a certain sense.Then the function has the form () = ∑ ∞ =0      .We say that () has Fabry gaps if   / → ∞ as  → ∞, and () has Fejér gaps if ∑ ∞ =1 (1/  ) < ∞.Wang [18] 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 [18] obtained the following result.Theorem 1.Let () = ∑ ∞ =0      be an entire function with 0 <  ≤  < ∞.If () has Fabry gaps, then every component of () is bounded.
Cao and Wang [20] proved the following result.
Singh [21] proved the following result.Theorem 4. Let L be the set of all entire functions  such that, for given  > 0, holds for all  outside a set of logarithmic density 0. Let F = ⋃ ≥1 F  where F  is the set of all transcendental entire functions  such that

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 C. The logarithmic measure of a set  is defined by ∫  (/).If  ⊂ [1, ∞), (, ) 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 6 (see [23]).Let  be an entire function of finite order with Fabry gaps.Then for given  > 0, holds for all r outside a set of logarithmic density 0.
Lemma 9 (see [26]).For an entire function () with Fejér gaps and  > 0, Lemma 10 (see [9]).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 || ≥ 1 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  > 1 and  > 0 such that, for all  ≥ , there exists  satisfying  ≤  ≤   and (, ) = (, ).
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  > 1 and  > 0 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 𝑓 be a transcendental entire function of finite order with
for some  > 2. Then there exist  > 1 and  > 0 such that, for all  ≥ , there exists  satisfying  ≤  ≤   and (, ) = (, ).

Main Results
In 2012, the authors proved some results on the bounded Fatou components of transcendental entire functions with gaps; see [27].In this paper, we investigate the iteration of the composite entire functions with gaps and obtain the following results. then So Which gives a contradiction if we take any  > ((1 + ) /).
Remark 16.If  satisfies (, ) ≥ exp(exp(log )  ), 0 <  < 1, then  ≥ 0, 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.