In the paper, we take up a new method to prove a result of value distribution of meromorphic functions: let f be a meromorphic function in
, and let 
, where P is a polynomial. Suppose that all zeros of f have multiplicity at least 
, except possibly finite many, and 
as 
. Then 
has infinitely many zeros.
The value distribution theory of meromorphic functions occupies one of the central places in Complex Analysis which now has been applied to complex dynanics, complex differential and functional equations, Diophantine equations and others.
In his excellent paper [
In this paper, we study the value distribution of transcendental meromorphic functions, all but finitely many of whose zeros have multiplicity at least
, where 
is a positive integer.
In 2008, Liu et al. [
Theorem A Let 
be an integer, let 
be a meromorphic function of infinite order 
in
, and let
, where 
is a polynomial. Suppose that 1) all zeros of 
have multiplicity at least
, except possibly finitely many, and 2) all poles of 
are multiple, except possibly finitely many.
Then 
has infinitely many zeros.
Theorem B Let 
be an integer, let 
be a meromorphic function of finite order 
in
, and let
, where 
is a polynomial. Suppose that 1) all zeros of 
have multiplicity at least
, except possibly finitely many, and 2)
.
Then 
has infinitely many zeros.
In the present paper, we prove the following result, which is a significant improvement of Theorem 1.
Theorem 1 Let 
be an integer, let 
be a meromorphic function of order 
in
, and let
, where 
is a polynomial. Suppose that all zeros of 
have multiplicity at least
, except possibly finitely many. Then 
has infinitely many zeros.
Theorem 1 and Theorem 2 taken together imply the following result.
Theorem 2 Let 
be an integer, let 
be a meromorphic function in
, and let
, where 
is a polynomial. Suppose that 1) all zeros of 
have multiplicity at least
, except possibly finitely many, and 2) 
as
.
Then 
has infinitely many zeros.
We use the following notation. Let 
be complex plane and 
be a domain in
. For 
and
, 
and
. We write 
in 
to indicate that the sequence 
converges to 
in the spherical metric uniformly on compact subsets of 
and 
in 
if the convergence is in the Euclidean metric.
Let 
be a meromorphic function in
. Set
The Ahlfors-Shimizu characteristic is defined by
Remark Let 
denote the usual Nevanlinna characteristic function. Since 
is bounded as a function of
, we can replace 
with 
in the paper.
The order 
of the meromorphic function 
is defined as
Lemma 1 [
a sequence of holomorphic functions in 
such that 
locally uniformly in
, where 
is univalent in
. Let 
be a sequence of functions meromorphic in 
such that for each
1) all zeros of 
have multiplicity at least
; and 2)
.
Then 
is quasinormal of order 1 in
. If, moreover, no subsequence of 
is normal at
, then
locally uniformly in 
and there exists 
such that 
for all
.
Remark Since Lemma 1 is not stated explicitly in [
is quasinormal of order 1 is essentially identical to that of Theorem 
of [
. It then follows from Lemma 7 that 
locally uniformly on
. The bound on 
follows from Lemma 9 of [
Lemma 2 [5, Lemma 2] Let 
be a family of functions meromorphic in
, all of whose zeros have multiplicity at least
, and suppose that there exists 
such that 
whenever
. Then if 
is not normal at
, there exist, for each
1) points
,
;
2) functions
; and 3) positive numbers 
such that 
in
, where 
is a nonconstant meromorphic function in
all of whose zeros have multiplicity at least
, such that
.
Lemma 3 Let 
be a meromorphic function of order 
in
, then there exist 
and 
such that
Proof We claim that there exist 
and 
such that
Otherwise there would exist 
and 
such that
for all
. From this follows
and hence
Now we have 
which contradicts the hypothesis that
.
Observing that 
hence there exists a sequence 
such that 
and 
as
. Let
. Obviously, 
and 
, and hence 
as
.
Lemma 4 Let 
and
. Let 
be a transcendental meromorphic function, all of whose zeros have multiplicity at least
. Set
. Suppose that
. Then there exists a sequence
and 
such that
as
.
Proof Since 
and
, we have
. By Lemma 3, there exist 
and 
such that
Set
. Clearly,
. Thus 
is not normal at 0. Obviously, all zeros of 
have multiplicity at least 
in
, and hence all zeros of 
have multiplicity at least 
in 
for sufficiently large
. Using Lemma 2 for
, there exist points
, and positive numbers 
and a subsequence of 
(that we continue to call
) such that
in
, where 
is a nonconstant meromorphic function in
, all of whose zeros have multiplicity at least
.
We claim that
, where 
is a constant. Otherwise, 
, where 
and 
are constants. Then, either 
is a constant function, or all zeros of 
have multiplicity at most
. A contradiction.
Let 
be not a zero or pole of
, and let
. Now we have
where
. Since 
and 
is not a zero or pole of
, we have
, 
and 
as
, where
.
Set 
and
, where
. Clearly,
where 
satisfying 
as
.
Now, we have 
and
Set
. Obviously, 
and
, and hence 
as
.
Proof We assume that 
has at most finitely many zeros and derive a contradiction. Let 
as
, where 
and
.
Set
. By Lemma 4, there exists a sequence 
and 
such that
and
Set
. By (1.4),
Hence, no subsequence of 
is normal at
.
Since 
has at most finitely many zeros, we have for sufficiently large
,
Observing that
in
. It follows from Lemma 1 (applied to 
in
), and there exists 
such that for all 
Set
. Then
and hence
Using the simple inequality
for
, we have
The second term on the right of (1.7) is
Putting (1.7), (1.8), and (1.9) together, we have for 
and sufficiently large
,
It follows from (1.1), (1.6), and (1.10),
Thus,
which contradicts (1.3).
This work was supported by National Natural Science Foundation of China (No.11001081, No.11226095).
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