Advances in Pure Mathematics
Vol.09 No.11(2019), Article ID:96360,9 pages
10.4236/apm.2019.911045
Transcendental Meromorphic Functions Whose First Order Derivatives Have Finitely Many Zeros
Feng Guo
Department of Mathematics, Yunnan Normal University, Kunming, China
Copyright © 2019 by author(s) and Scientific Research Publishing Inc.
This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).
http://creativecommons.org/licenses/by/4.0/
Received: October 14, 2019; Accepted: November 11, 2019; Published: November 14, 2019
ABSTRACT
Let f be a meromorphic function in . If the order of f is greater than 2, has finitely many zeros and f takes a non-zero finite value finitely times, and then is unbounded.
Keywords:
Transcendental Meromorphic Functions, Derivatives, 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.
Conflicts of Interest
The author declares no conflicts of interest regarding the publication of this paper.
Cite this paper
Guo, F. (2019) Transcendental Meromorphic Functions Whose First Order Derivatives Have Finitely Many Zeros. Advances in Pure Mathematics, 9, 925-933. https://doi.org/10.4236/apm.2019.911045
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