Transcendental Meromorphic Functions Whose First Order Derivatives Have Finitely Many Zeros ()
1. Introduction and Main Result
Let f be a meromorphic function in
, define
. W. Bergweiler [1] gave a conjecture in 2001 as follow: Conjecture 1: Let f be a transcendental meromorphic function in
. If
for all
, then
is unbounded.
Bergweiler pointed that let
, Conjecture 1 is equivalent to the following one.
Conjecture 2: Let g be a transcendental meromorphic function in
. Suppose that
does not have zeros. Then there exist a sequence
of fixed points of g such that
.
Bergweiler [1] has separately proved Conjecture 1 is affirmative for finite order meromorphic functions and entire functions; Jianming Chang [2] has confirmed the conjecture for infinite order meromorphic functions for the first time, which is based on theory of normal and quasinormal families.
For the conjecture,
and
are essentially equivalent. In fact, if
, then
and
is also transcendental meromorphic function; the zeros of f and the zeros of
are the same and
is unbounded if and only if
is unbounded.
Considering the discussion above, it’s natural to research the problem that whether the conclusion is true if
but not
, the problem is radically different to the conjecture and gives a important supplement. We can give a example to show that the problem is significant. Let
, it’s obvious that
and
is unbounded. In details, we have
Theorem 1. Let f be meromorphic in
and the order is greater than 2. If
has finitely many zeros and f takes a finite non-zero value finitely many times, then
is unbounded.
Theorem 2. Let f be entire in
and the order is greater than 1. If
has finitely many zeros and f takes a finite non-zero value finitely many times, then
is unbounded.
2. Preliminary Lemmas
Lemma 1. Let f be a meromorphic function. If the spherical derivative
of
is bounded. Then the order of
is at most 2.
For details of lemma 1, can see [3]
Remark: Let
be a sequence of meromorphic functions,
means
locally uniformly convergence to meromorphic function
; for a meromorphic function f in
, let
.
Lemma 2. Let
be a family of functions meromorphic in a domain D. Suppose that there exist
such that
for all
. For any given
satisfying
, if
is not normal, then there exist a sequence
in
, a sequence
in D, a sequence
of positive real numbers and a non-constant finite order function f which is meromorphic in
such that
for some
and
Moreover, the spherical derivative
of f satisfies
for all
.
For details of lemma 2, can see [4]
In the case no hypothesis on
is required, the case
is due to Zalcman [5], and the case
is due to Pang [6] [7]. In the hypothesis of Lemma 2, if all
have no zero in D, then
and the conclusion is still right according to the proof of Lemma 2 in which K can take 0.
Lemma 3. Let f be a meromorphic function, D be an bounded domain and c be constant in
, if
has
zeros in D and
has
zeros in D which are all the zeros of
. Then each discriminating zero of
and
in D is the same one.
Proof. Let
with
be a meromorphic function which have no zero in D satisfying
and
in which
.
Because
has
zeros in D which are all the zeros of
, there exist
points in
be zeros of
and without loss of generality we may assume
.
As
, we can deduce that
from the above it follows that
and the proof of Lemma 3 is complete.
Lemma 4. Let f be a holomorphic function, if the spherical derivative of f is bounded. Then the order of f is at most 1.
For details of lemma 4, can see [8].
3. Proof of Theorem 1
Proof. we apply Lemma 1 to obtain a sequence
such that
.
, let
, it’s easy to apply Marty’s theorem to know
is not normal at 0. Suppose
only have finitely many zeros (
), there exist a subsequence of
we still suppose it’s
such that
have no zero in
, thus according to Lemma 2, there exist a sequence
, a sequence
of positive real numbers and a non-constant finite order function
such that when
,
and
in
and
satisfies
for all
.
, let
, there exist entire functions
and
such that
and
have no common non-trivial divisor and
, then
(1)
For
and
, the derivative of (1) is
(2)
here we divide two cases:
Case 1:
have no zero in
.
Because
is bounded, then we apply Lemma 4 to have that the order of
and
are at most 1. On the other hand,
, we can deduce
or constant
and
or
.
here we first proof that
, if
, we have
then we have that
(3)
(4)
From (4) it can be deduced that because
have no zero,
have no zero in any bounded domain when n is large enough,. Then so are
unite (3). What’ more,
has no zero and pole and cannot take
in any bounded domain, which contradict with Picard’s Theorem, therefore
.
From (1) we have
Because
have zero in
, from above it can be deduced that
have to have zeros which convergence to the zeros of
. From (2) when
then
or
and
have to be unbounded.
Case 2:
have zero in
. We will proof the case is impossible.
Take a finite zero c of
and it’s multiple is
then there exist some sufficiently small neighborhood
of c such that
only have one zero of
. (2) can be expressed as
Because
take
finitely many times and from (2)
have no zero in
then when n is large enough,
have k zeros in
, which are all the zeros of
due to that
only has finitely many zeros; therefore, c is the zero of
with
multiple and
have
zeros in
.
(1) can be expressed as
From (1) we have that
(5)
Here we divide two cases for (5):
Subcase 2.1: If
is normal in
.
From (5) we have
(6)
when n is large enough, notice that
have no zero in
, therefore
have no zero in
and according to (6),
have no zero in
contradict with
have
zeros in
.
Subcase 2.2: If
is not normal in
. Let
Then
is not normal and have no zero in
, we apply Lemma 2 to obtain
and
of positive real numbers and a non-constant finite order function
such that
, and
here we will prove that
has no simple pole if it exist; let
be the pole of
, for
cannot always be
, there exist closed disc
such that
and
are holomorphic in
and
uniformly in
and so are
.
Notice that
cannot be constant, there exist
such that
We firstly show that the discriminating zeros of
in
are all the zeros of
in
when n is large enough. In fact, we have that the k zeros of
as same as
, which are all belong to the
zeros of
and
in
, then Lemma 3 can be used to prove the conclusion and we further have
which means
has to have multiple pole if it exist.
Notice
only has finitely many zeros, then
have no multiple zero in
when n is large enough.
Considering
, since
have no zero in
when n is large enough,
are analytic in
and according to Hurwitz’s Theorem,
has no multiple pole. With the assert above,
have no pole and be entire.
Notice that
and Lemma 4, the order of
is at most 1. Since
have no zero in
when n is large enough, then
have no zero in any bounded domain, from the above it follows that
and
.
is
(7)
Let
then the derivative of (7) is
(8)
(2) can be expressed as
, let
be the k order derivative of
, then the k order derivative of (2) is
with
have no zero in
. Let
then we have
(9)
The k order derivative of (8) is
(10)
(9) + (10) is
It shows that
have to have zeros in
when n is large enough, however, from (10) and Hurwitz’s theorem, it is impossible; this gives a contradiction and the proof of Theorem 1 is complete.
4. Remarks
It follows from the proof of Theorem 1 that the hypothesis for order can be replaced by greater than 1 for entire functions. In fact, from Lemma 4, we can obtain a sequence
such that
. Then using the start point of proof of Theorem 1,
, let
, it’s easy to apply Marty’s theorem to know
is not normal at 0. Suppose
only has finitely many zeros (
), there exist a subsequence of
. We still suppose it’s
such that
has no zero in
. Thus according to Lemma 2, there exist a sequence
, a sequence
of positive real numbers and a non-constant finite order function
such that when
,
and
in
and
satisfies
for all
. For
and
we apply Lemma 6 to have the order of g at most 1, and
and we have
and the first order derivative is
If
, then
and
is unbounded. Then the proof of Theorem 2 is complete.
The requirement for order in Theorem 2 is sharp, let
, then
.
By the equivalence between the conjecture 1 and 2, we can have two corollaries from Theorem 1 and 2.
Corollary 1. Let g be meromorphic in
and the order is greater than 2. If
has finitely many zeros and
takes a finite non-zero value finitely many times, then g has a sequence
of fixed points such that
.
Corollary 2. Let g be entire in
and the order is greater than 1. If
has finitely many zeros and
takes a finite non-zero value finitely many times, then g has a sequence
of fixed points such that
.
Acknowledgements
I thank the Editor and the referee for their comments. Research of F. Guo is funded by the Yunnan province Science Foundation grant 2016FD015. This support is greatly appreciated.