#### Abstract

We study the problem of normal families of meromorphic functions concerning polynomials and shared values. We prove that a family of meromorphic functions in a domain D is normal if, for each function , , where is a polynomial with the origin as zero, is a positive integer, and , are two finite constants.

#### 1. Introduction and Main Results

Let D be a domain in and let be a family of meromorphic functions defined in D. is said to be normal in D, in the sense of Montel, if, for any sequence , there exists a subsequence such that converges spherically locally uniformly in D to a meromorphic function or (see [1]).

Let and be meromorphic functions in D, and let and be complex numbers. If whenever , we write If and , we write If , we say that and share on D.

In 1992, Schwick [2] firstly found a connection between normality criteria and shared values as follows.

Theorem A. *Let be a family of meromorphic functions in D, and let be distinct complex numbers. If, for each , and share , then is normal in D.*

In recent years, more results about normality criteria concerning shared values have been found (see [3–5]). The next theorem was proved by Fang and Zalcman in [4], which was an important normality criteria concerning shared values.

Theorem B. *Let be a family of meromorphic functions in D; let be a positive integer; let and be two finite values. If, for every , all zeros of f have multiplicity at least and , then is normal in D.*

Based on the idea in [3], we will improve and extend Theorem B in the following results.

Theorem 1. *Let be a family of meromorphic functions in D; let be a positive integer; let and be two finite values. Let P be a polynomial with the origin as zero. If, for every , all zeros of f have multiplicity at least and , then is normal in D.*

*Example 2. *Let , , . Then, all zeros of have multiplicity at least 2. For each , we have , . Thus, . , but is not normal in D. This shows that the condition is necessary.

Theorem 3. *Let be a family of meromorphic functions in D; let be a positive integer; let and be two finite values. Let P be a polynomial with the origin as zero and , which has at least one multiple zero. If, for every , all zeros of have multiplicity at least and , then is normal in D.*

#### 2. Some Lemmas

In order to prove our results, we need the following lemmas.

Lemma 1 (see [6]). *Let be a family of meromorphic functions on the unit disc satisfying all zeros of functions in which have multiplicity ≥ p and all poles of functions in . Let be a real number satisfying . Then, is not normal at a point if and only if there exist*(i)

*points , ;*(ii)

*positive numbers , ;*(iii)

*functions such that spherically uniformly on each compact subset of , where is a nonconstant meromorphic function satisfying the zeros of which are of multiplicities and the poles of which are of multiplicities . Moreover, the order of is not greater than 2.*

Here, is the spherical derivative.

Lemma 2 (see [7]). *Let be a family of meromorphic functions on the complex plane . If the spherical derivative of is bounded on , then the order of is at most two. Especially, when is an entire function on the complex plane , then the order of is at most one.*

Lemma 3 (see [8]). *Let be a transcendental meromorphic function on the complex plane , and let be a positive integer. Then assumes every nonzero finite complex value infinitely often.*

Lemma 4 (see [9]). *Let and be two positive integers, and let be a transcendental meromorphic function. Then assumes every nonzero finite complex value infinitely often.*

Lemma 5 (see [5]). *Let be a nonzero finite complex number, let be two positive integers, and let be a nonconstant rational function; all of whose zeros have multiplicities at least . For the polynomial case, if or , then has at least two distinct zeros; for the nonpolynomial rational case, the conclusion holds if .*

#### 3. Proof of Theorems

We only prove Theorem 1, and the proof of Theorem 3 is similar to the second case in the proof of Theorem 1.

Next, we consider two cases.

*Case 1. *Suppose the origin is a simple zero of . We assume that , where . Without loss of generality, we may assume that . Since normality is a local property, it is enough to show that is normal at each point of D. Suppose, on the contrary, that is not normal at . By Lemma 1 (with ), there exist functions , points , and positive numbers such that
converges spherically uniformly on compact subsets of , where is a nonconstant meromorphic function on and all of whose zeros have multiplicity at least . Moreover, the order of is at most 2. Therefore, we have
uniformly on compact subsets of disjoint from the poles of .

We claim (i) ; (ii) .(i) Suppose that , then . If , is an entire function and . By Lemma 2, the order of is at most 1. Assume , where and are two constants. But , which is a contradiction. Therefore, we have . According to Hurwitz’s theorem, there exists a sequence , such that for sufficiently large. It follows that , so that
Thus
which contradicts . This proves (i).(ii) Suppose that , then . If , is a polynomial whose degree is less than , which contradicts that all zeros of have multiplicity at least . So that . Since , by Hurwitz’s theorem, there exists a sequence , such that for sufficiently large. It follows that , so that
which contradicts . This proves (ii).

Next, we assume that is a transcendental meromorphic function. When , by Lemma 3, assumes every nonzero finite complex value infinitely often, a contradiction. When , by Nevanlinna's second fundamental theorem,
where .

By Nevanlinna’s first fundamental theorem, we have
Combining (8) with (9), we have
According to logarithmic derivative theorem,
we have
It follows that
Since and the zeros of have multiplicity exactly , we have by (10) and (13)
Thus , a contradiction. Hence is a rational function.

Now, suppose that is a nonpolynomial rational function. Let
where are constants such that and and are coprime polynomials with and . If ,
where and are coprime polynomials and, by induction, . Then, by (16), has solutions, a contradiction. Hence . Thus, we have
Since , is a nonzero constant . Since , must have a pole in the finite plane. It follows that has a finite pole of order at least ; thus, ,
Then, the equation has solutions, which contradict .

Finally, suppose that is a polynomial. Since and the zeros of are of multiplicity , we have , where . Thus, has solutions, a contradiction.

*Case 2. *Suppose the origin is a multiple zero of . Without loss of generality, we assume that . Let , where is a positive integer and are constants. Suppose is not normal at . By Lemma 1 (with ), there exist functions , points , and positive numbers such that
converges spherically uniformly on compact subsets of , where is a nonconstant meromorphic function on , and all of whose zeros have multiplicity at least . Moreover, the order of is at most 2. We have
uniformly on compact subsets of disjoint from the poles of . If , has no zeros and poles. That is to say, is an entire function and . Since the spherical derivative of is bounded, by Lemma 2, we have . Hence suppose , where and are constants. Thus, , a contradiction. By Lemmas 4 and 5, has zeros. Suppose . Since , by Hurwitz’s theorem, there exists a sequence , such that (for sufficiently large)
It follows that , so that
Thus,
which contradicts . This completes the proof.

#### Acknowledgment

This paper is supported by the National Natural Science Foundation of China (no. 61074016).