ABSTRACT

Let with being real coefficient polynomial and it's leading coefficient be positive, , when and satisfy two certain conditions, buried point set of contains unbounded positive real interval.

Keywords:Transcendental Entire Functions, Julia Set, Buried Points Set, Real Interval

1. Introduction and Main Result

Let be an entire function on the complex plane. Define the iterated sequence of as

can be divided into two sets:

is called Fatou set, which is open and contains at most countably many components. is called Julia set, and it’s closed and perfect. The fundamental theory of complex dynamical system can refer to [1] -[4] .

For an entire function, let, with sing be the set of singular value. If is not a smooth covering map over any neighborhood of, then is a singular value. If, we call is finite type entire function. The basic properties of the type entire function can refer to [5] .

I.N. Baker has first structured the transcendental entire function whose Julia set is (See [6] ):

Theorem A:

For a certain real value, has the whole complex plane for its Julia set.

Notice the set. Baker’s result show that the set is nonempty; Jang, C.M. proved that this set contains infinitely many elements in [7] ; Qiao, J. proved that the set is unbounded in [8] . What’s more, Qiao has researched the buried sets in [9] , which contains unbounded positive real interval:

Theorem B:

If for and, then belongs to the set of buried points.

Here we study the function with is real coefficient polynomial and it's leading coefficient is positive, expend the function in Theorem B:

Theorem 1 Let, is real coefficient polynomial and it’s leading coefficient is positive, the zeros of are which are real, unbounded positive real interval with. and satisfy:

(1);

(2) with is real zeros of.

Then belongs to the set of buried points set.

Remark: Qiao has given the example that satisfy the condition of Theorem B in [9] , then the example show that the function satisfy conditions in Theorem 1 is nonempty.

2. Proof of Theorem 1

Lemma 1 Let be an entire funcion of finite type. Then each Fatou component is eventually periodic, and has only finitely many periodic compinents. They are attractive domains, superattractive domains, parabolic domains or Siegel discs.

Lemma 2 Let be a transcendental entire function, be a component of. If is an attractive, a superattractive or a parabolic periodic domain, then the cycle of D contains at least one singularity of; if D is a Siegel disc, then the forward orbits of the singularities of are dense on.

Lemma 3 Let be transcendental entire function of finite type. Then for.

Proof of Theorem 1: The singularities of are 0 and, then

if, so from Lemma 2 have no Siegel disc and from Lemma 1,the periodic component of only be attractive, superattractive or parabolic. and from we can have, then 0 is a repelling fixed point, from Lemma 1 and 2, has at most cycles of periodic components

such that and there exist such that

*

Here we first proof belong to. If there exist and, then is contained in a component of and from above we have that there exist, a cycle of component with and such that and are in the same domain, from * we have that. However, means when, when, then is montone increasing sequence, by the relation

we have which give a contradiction to Lemma 3.

Then we will proof that belongs to the set of buried points. If there exist a point and is on the boundary of a component of, from the discussion above, we know there exist some, a cycle of component with such that when,

and there exist and some, such that .

Let, it’s easy to have that and,. Without the loss of generality, we can let,

Here we prove are all in the same connected component of.If not, there exist and two different component of called and such that

We can make curve, such that and belong to different components of, then and belong to different components of, that gives a contradiction to.

Let be an bounded continuum containing and, we will prove that

. If not, and can form a bounded domain and have no interior point, therefore have to have a bounded domain, notice that, let, due to

, then, notice that and when therefore

, from we have. That means is a unbounded component, however any component of have to turn into cycle from Lemma 1, therefore the components of

are all unbounded, it’s contradiction. What’s more, we can have that

that means is the common boundary of with. The common boundary is at most of two domains, therefore. Here we divide two cases to discuss:

Case 1:. is. Considering

and * we have

Notice, then, we only need to consider 0 and the real zeros of, from Lemma 2 is an attractive, superattractive, or rational indifferent fixed point, but from conditions (1) and (2) in Theorem 1, 0 and are repelling fixed point, it’s a contradiction.

Case 2:.Without the loss of generality, we take be the component above and be the component under, for with is large enough, take a sequence such that.

Let with, then

We can suppose that, then we have

Notice that and we can easily deduce that and belong to the above half plane when and is large enough, but it contradicts that. The proof is complete.

Acknowledgements

Author thanks professor Jianyong Qiao and Yuhua Li for their discussions and suggestions.

References