Finite Type Transcendental Entire Functions Whose Buried Points Set Contains Unbounded Positive 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.