Volume 2014, Issue 1 654294
Research Article
Open Access

The Normality Criteria of Meromorphic Functions Concerning Shared Fixed-Points

Wei Chen

Wei Chen

School of Mathematics Science, Xinjiang Normal University, Urumqi, Xinjiang 830054, China xjnu.edu.cn

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Qi Yang

Qi Yang

School of Mathematics Science, Xinjiang Normal University, Urumqi, Xinjiang 830054, China xjnu.edu.cn

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Wen-jun Yuan

Wen-jun Yuan

School of Mathematics and Information Sciences, Guangzhou University, Guangzhou, Guangdong 510006, China gzhu.edu.cn

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Hong-gen Tian

Corresponding Author

Hong-gen Tian

School of Mathematics Science, Xinjiang Normal University, Urumqi, Xinjiang 830054, China xjnu.edu.cn

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First published: 06 April 2014
Citations: 1
Academic Editor: Chuanxi Qian

Abstract

We study the normality criteria of meromorphic functions concerning shared fixed-points; we obtain the following: Let be a family of meromorphic functions defined in a domain D and k ≥ 2 a positive integer. For every f, all zeros of f are of multiplicity at least k + 2 and all poles of f are multiple. If ff(k) and gg(k) share z in D for each pair of functions f and g, then is normal.

1. Introduction and Main Results

Let be a family of meromorphic functions defined in the domain D. If any sequence {fn} ⊂ contains a subsequence that converges spherically locally uniformly in D, to a meromorphic function or , we say that is normal in D (see [1, 2]).

Let f(z) be a meromorphic function in a domain D, and z. We say z is a fixed-point of f(z) when f(z) = z. Let f(z) and g(z) be two meromorphic functions in D; if f(z) − z and g(z) − z have the same zeros (ignoring multiplicity), then we say f(z) and g(z) share fixed-points.

In 2009, Lu and Gu [3] proved the following results.

Theorem 1. Let k be a positive integer and f a transcendental. If all zeros of f have multiplicity at least k + 2, then ff(k) assumes every finite nonzero value infinitely often.

Theorem 2. Let f be a family of meromorphic functions defined in a domain D. Suppose that k is a positive integer and a ≠ 0 is a finite complex number. If, for each f, all zeros of f are of multiplicity at least k + 2, and ff(k)a, then is normal in D.

In 2011, Hu and Meng [4] extended Theorem 2 as follows.

Theorem 3. Let be a family of meromorphic functions defined in a domain D. Let k be a positive integer and a ≠ 0 be a finite complex number. If, for every f, ff(k) and gg(k) share a in D, and, for every pair of functions f, g, all zeros of f are of multiplicity at least k + 1, then is normal in D.

A natural question is the following: what can be said if the finite complex number a in Theorem 3 is replaced by the fixed-point z? In this paper, we answer this question by proving the following theorems.

Theorem 4 (main theorem). Let be a family of meromorphic functions defined in a domain D. Let k be a positive integer. If, for every f such that f ≠ 0, ff(k) and gg(k) share z in D for every pair of function f, g, then is normal in D.

Theorem 5 (main theorem). Let be a family of meromorphic functions defined in a domain D. Let k ≥ 2 be a positive integer. For every f, all zeros of f have multiplicity at least k + 2 and all poles of f are multiple. If ff(k) and gg(k) share z in D for every pair of function f, g, then is normal in D.

Theorem 6 (main theorem). Let be a family of meromorphic functions defined in a domain D. Let k be a positive integer. For every f, f has only zeros with multiplicity at least k + 1 and f has poles at most [(k − 1)/2]. If ff(k) and gg(k) share z in D for every pair of function f, g, then is normal in D.

2. Some Lemmas

In order to prove our theorems, we require the following results.

Lemma 7 (see [5], [6].)Let be a family of meromorphic functions in a domain D, and let k be a positive integer, such that each function f has only zeros of multiplicity at least k, and suppose that there exists A ≥ 1 such that |f(k)(z)| ≤ A whenever f(z) = 0, f. If is not normal at z0D, then, for each 0 ≤ αk, there exist a sequence of points znD, znz0, a sequence of positive numbers ρn → 0+, and a subsequence of functions fn such that

()
locally uniformly with respect to the spherical metric in , where g is a nonconstant meromorphic function, whose zeros all have multiplicity at least k, such that g#(ζ) ≤ g#(0) = kA + 1. Moreover, g has order at most 2.

Here as usual, g#(ζ) = |g(ζ)|/(1+|g(ζ)|2) is the spherical derivative.

Lemma 8. Let k be a integer and f a nonconstant rational meromorphic function such that f ≠ 0; then ff(k)z has at least two distinct zeros.

Proof. Since f ≠ 0, set

()
where A is a nonzero constant and nj  (j = 1,2, …, t) are positive integers.

For simplicity, we denote N : = n1 + n2 + ⋯+nt. Obviously, Nt.

From (2), we have

()
where p(z)≢0 is a polynomial and deg⁡⁡(P) = k(t − 1).

Differentiating (3), we get

()
where P1(z)≢0 is a polynomial and deg⁡⁡(P1) = (k + 2)(t − 1).

Next, we assume, to the contrary, that ff(k)z has at most one zero. We distinguish two cases.

Case 1. If ff(k)z has exactly one zero z0. From (3), we obtain

()
where B is a nonconstant. Differentiating (5), we have
()
where P2(z)≢0 is a polynomial and deg⁡(P2) ≤ 2t.

From (4) and (6), we obtain

()
then −k ≥ 1; this is a contradiction.

Case 2. If ff(k)z has no zero. Then we have 2N + kt + 1 = 0 from (5), which is a contradiction.

The proof is complete.

Lemma 9. Let k ≥ 2 be an integer and f a nonconstant rational meromorphic function. If f has only zeros with multiplicity at least k + 2 and all poles of f are multiple, then ff(k)z has at least two distinct zeros.

Proof. Assume, to the contrary, that ff(k)z has at most one zero.

Case 1. If f is a polynomial, since all zeros of f are of multiplicity at least k + 2, then we can know that ff(k) has at least one zero with multiplicity k + 4. So ff(k)z has at least one zero and (ff(k))  has zeros with multiplicity at least k + 3. According to the assumption, we obtain that ff(k)z has only a zero z0; then there exists a nonzero constant A and an integer l ≥ 2 such that

()
So we have
()
which, however, has only simple zeros. This is a contraction.

Case 2. If f is a rational but not a polynomial. We see

()
where A is a nonzero constant and mj ≥ (k + 2)  (j = 1,2, …, s).

For simplicity, we denote

()
From (10), we have
()
()
where g(z)≢0 is a polynomial and deg⁡⁡(g) ≤ k(s + t − 1).

Differentiating (13), we have

()
()
where g1(z)≢0,  g2(z)≢0 are polynomials and deg⁡⁡(g1)≤(k + 1)(s + t − 1), deg⁡⁡(g2)≤(k + 2)(s + t − 1).

Now we distinguish two subcases.

Case 2.1. Supposing that ff(k)z has exactly one zero z0, from (13), we obtain

()
Differentiating (16), we have
()
()
where ,   are polynomials, , and . From (14) and (17), we have z0αj  (j = 1,2, …, s).

Further, we distinguish two subcases.

Case 2.1.1. l ≠ 2N + kt − 1. From (16), it is easily obtained that deg⁡⁡(P) ≥ deg⁡⁡(Q). Thus, (13) implies

()
So MN + k/2; that is, M > N. From (13) and (16), noting that z0αj(j = 1,2, …, s), we have
()
It follows that 2M ≤ (k + 2)s + 2t; combining this inequality with (11), we obtain
()
which is impossible.

Case 2.1.2. l = 2N + kt − 1. Next we distinguish two subcases: M > N and MN.

When M > N, similar to the Case 2.1.1, it follows that 2M < 2N; from (13) and (16), we get a contradiction.

If MN, by combining (15) and (18), we may give the following inequality:

()
and hence
()
This is a contradiction.

Case 2.2. Suppose that ff(k)z has no zero; then l = 0 for (16). Similarly with the proof of Case 2.1, we also obtain a contradiction.

Lemma 10. Let k be an integer and f a nonconstant rational meromorphic function. If f has only zeros with multiplicity at least k + 1 and f has poles at most [(k − 1)/2], then ff(k)z has at least two distinct zeros.

Proof. Assume, to the contrary, that ff(k)z has at most one zero.

Case 1. If f is a polynomial, since all zeros of f are of multiplicity at least k + 1, then we can know that ff(k) has at least one zero with multiplicity k + 2. So we see that ff(k)z has at least one zero and (ff(k))  has zeros with multiplicity at least k + 1. According to the assumption, we obtain that ff(k)z has only a zero z0; then there exists a nonzero constant A and an integer l ≥ 2 such that

()
So we have
()
which, however, has only simple zeros. This is a contraction.

Case 2. If f is a rational but not a polynomial, we set

()
where A is a nonzero constant and mj ≥ (k + 2)(j = 1,2, …, s).

For simplicity, we denote

()
From (26), we have
()
()
where g(z)≢0 is a polynomial and deg⁡⁡(g) ≤ k(s + t − 1).

Differentiating (29), we have

()
()
where g1(z)≢0, g2(z)≢0 are polynomials and deg⁡⁡(g1) ≤ (k + 1)(s + t − 1) and deg⁡⁡(g2) ≤ (k + 2)(s + t − 1).

Now we distinguish two subcases.

Case 2.1. Supposing that ff(k)z has exactly one zero z0, from (13), we obtain

()
Differentiating (32), we have
()
()
where , are polynomials, , and . From (30) and (33), we have z0αj  (j = 1,2, …, s). From (31) and (34), we see
()
So M ≤ ((k + 2)s + 2t)/2. Since M ≥ (k + 1)s, we have (k + 1)sM ≤ ((k + 2)s + 2t)/2, so s ≤ 2t/k ≤ [k − 1]/k < 1.

Therefore, f has no zeros. According to Lemma 8, we have a contradiction.

Case 2.2. Suppose that ff(k)z has no zero; then l = 0 for (32). Similarly with the proof of Case 2.1, we also obtain a contradiction.

Lemma 11 (see [7].)Let f be a transcendental meromorphic function, and let φ(z) be a small function such that T(r, φ) = S(r, f); then

()

Lemma 12. Let f be a transcendental meromorphic function; let k be a positive integer; let P(z)(≢0) be a polynomial. If all zeros of f have multiplicity at least k + 1, then f(z)f(k)(z)  −  P(z) has infinitely many zeros.

Proof. We denote

()
then
()
Since p(z) is a polynomial, we see T(r, 1/P(z)) = s(r, f). Now we distinguish two cases as follows.
  • (i)

    If k ≤ 4, we have

    ()

By Lemma 11, we have
()
Thus,
()
  • (ii)

    If k ≥ 5, then

    ()

By Lemma 11, we have
()

So

()

From (41) and (44), we can deduce that (1/P)ff(k) − 1 has infinitely many zeros; thus, ff(k)P has infinitely many zeros.

Lemma 13 (see [4].)Take a positive integer k and a nonzero complex number a. If f is a nonconstant meromorphic function such that f has only zeros of multiplicity at least k + 1, then ff(k)a has at least two distinct zeros.

3. Proof of Theorems

Proof of Theorem 5. Consider the following.

Case 1. For z0 = 0, let 1 = {Fj : Fj(z) = fj(z)/z1/2fj}. If 1 is not normal at 0, by Lemma 7, there exist a sequence {zj} of complex numbers with zj → 0 and a sequence {ρj} of positive numbers with ρj → 0 such that

()
locally uniformly on compact subsets of , where g(ξ) is a nonconstant meromorphic function in .

Here we distinguish two cases.

Case 1.1. Supposing that zj/ρjc, c is a finite complex number. Then

()
locally uniformly on compact subsets of disjoint from the poles of g, where H(ξ) is a nonconstant meromorphic function in , all of whose zeros have multiplicity at least k + 2 and all poles of which have multiplicity at least 2. Hence
()
spherically locally uniformly in C disjoint from the poles of g.

If H(ξ)H(k)(ξ) ≡ ξ, since H(ξ) has zeros with multiplicity at least k + 2, obviously there is a contradiction. Hence, H(ξ)H(k)(ξ)≢ξ. Since the multiplicity of all zeros of H(ξ) is at least k + 2, so by Lemmas 9 and 12, H(ξ)H(k)(ξ) − ξ has at least two distinct zeros.

Suppose that ξ0, are two distinct zeros of H(ξ)H(k)(ξ) − ξ. We choose a positive number δ small enough such that D1D2 = and such that H(ξ)H(k)(ξ) − ξ has no other zeros in D1D2 except for ξ0 and , where

()

By Hurwitz’s Theorem, for sufficiently large j, there exist points ξjD1, such that

()

By the assumption in Theorem 5 that and share z, it follows that

()
Fix m, take j, and note ρjξj → 0, ; we obtain
()
Since the zeros of has no accumulation point, for sufficiently large j, we have
()
Therefore, when j is large enough, . This contradicts with the facts ξjD1, , D1D2 = . Thus 1 is normal at 0.

Case 1.2. We may suppose that zj/ρj. We have

()
where . Thus we have
()
Since , l = 0,1, …, k, we have
()
On the other hand, for l = 1,2, …, k, we have
()

Thus, we have

()
spherically locally uniformly in disjoint from the poles of g.

If g(ξ)g(k)(ξ) ≡ 1, then g has no zeros. Of course, g also has no poles. Since g is a nonconstant meromorphic function of order at most 2, then there exists constant ci  (i = 1,2), (c1, c2)≠(0,0), and ; obviously, this is contrary to the case g(ξ)g(k)(ξ) ≡ 1. Hence, g(ξ)g(k)(ξ)≢1.

Since the multiplicity of all zeros of g is at least k + 1 and all poles of which have multiplicity at least 2, thus, by Lemma 13, g(ξ)g(k)(ξ) − 1 has at least two distinct zeros.

Suppose that ξ1, are two distinct zeros of g(ξ)g(k)(ξ) − 1. We choose a positive number δ small enough such that D1D2 = and such that g(ξ)g(k)(ξ) − 1 has no other zeros in D1D2 except for ξ1 and , where

()

By Hurwitz’s Theorem, for sufficiently large j there exist points , such that

()

Similar to the proof of Case 1.1, we get a contradiction. Then, 1 is normal at 0.

From Cases 1.1 and 1.2, we know that 1 is normal at 0; there exists Δ = {z:|z | < ρ} and a subsequence of Fj, and we may still denote it as Fj, such that Fj converges spherically locally uniformly to a meromorphic function F(z) or in Δ.

Here we distinguish two cases.

Case i. When j is large enough, fj(0) ≠ 0. Then F(0) = . Thus, for each Fj(z) ∈ 1, there exists δ > 0 such that if F(z) ∈ 1, then |F(z)| > 1 for all z ∈ Δδ = {z:|z | < δ}. Thus, for sufficiently large j, |Fj(z)| ≥ 1, 1/fj is holomorphic in Δδ. Therefore, for all fj, when |z | = δ/2, we have

()
By maximum Principle and Montel’s Theorem, is normal at z = 0.

Case ii. There exists a subsequence of fj, and we may still denote it as fj such that fj(0) = 0. Since f, the multiplicity of all zeros of f is at least k + 2; then Fj(0) = 0. Thus, there exists 0 < r < ρ such that Fj(z) is holomorphic in Δr = {z:|z | < r} and has a unique zero z = 0 in Δr. Then Fj converges spherically locally uniformly to a holomorphic function F(z) in Δr; fj converges spherically locally uniformly to a holomorphic function F(z)z1/2 in Δr. Hence is normal at z = 0.

By Cases i and ii, is normal at z = 0.

Case 2. For z0 ≠ 0, suppose, to the contrary, that F is not normal in D. Then there exists at least one point z0 such that F is not normal at the point z0. Then by Lemma 7, there exist a sequence of complex numbers with and a sequence {ρn} of positive numbers with ρn → 0 such that

()
locally uniformly on compact subsets of C, where g(ξ) is a nonconstant meromorphic function in C, whose zeros all have multiplicity at least k + 2. Moreover, g(ξ) has order at most 2.

From (61) we get

()
also locally uniformly with respect to the spherical metric besides the poles of g(ξ).

If gg(k)z0, then g has no zeros. Of course, g also has no poles. Since g is a nonconstant meromorphic function of order at most 2, then there exists constant ci  (i = 1,2) ≠ 0, and . Obviously, this is contrary to the case gg(k)z0. Hence gg(k)z0.

By Lemmas 12 and 13, we deduce that gg(k)z0 has at least two distinct zeros. Next we show that it is impossible. Let ξ0 and be two distinct zeros of gg(k)z0. We choose a positive number δ small enough such that D1D2 = and such that gg(k)z0 has no other zeros in D1D2 expect for ξ0 and , where

()

Similar to the proof of Case 1, we get a contradiction. This finally completes the proof of Theorem 5.

Similar to the proof of Theorem 5, combining with Lemmas 8 and 10 and Lemma 12, we can prove Theorems 4 and 6 easily; here we omit the proof.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

This work was supported by the Visiting Scholar Program of Chern Institute of Mathematics at Nankai University. The third author would like to express his hearty thanks to Chern Institute of Mathematics that provided very comfortable research environments to him while working as visiting scholar. The authors thank the referees for reading the paper very carefully and making a number of valuable suggestions to improve the readability of the paper. Foundation item: Nature Science Foundation of China (11271090); Nature Science Foundation of Guangdong Province (S2012010010121); Graduate Research and Innovation Projects (XJGRI2013131) of Xinjiang Province.

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