The Normality Criteria of Meromorphic Functions Concerning Shared Fixed-Points
Abstract
We study the normality criteria of meromorphic functions concerning shared fixed-points; we obtain the following: Let ℱ be a family of meromorphic functions defined in a domain D and k ≥ 2 a positive integer. For every f ∈ 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 ℱ is normal.
1. Introduction and Main Results
Let ℱ be a family of meromorphic functions defined in the domain D⊆ℂ. If any sequence {fn} ⊂ ℱ contains a subsequence that converges spherically locally uniformly in D, to a meromorphic function or ∞, we say that ℱ is normal in D (see [1, 2]).
Let f(z) be a meromorphic function in a domain D, and z ∈ ℂ. 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 ∈ ℱ, all zeros of f are of multiplicity at least k + 2, and ff(k) ≠ a, then ℱ is normal in D.
In 2011, Hu and Meng [4] extended Theorem 2 as follows.
Theorem 3. Let ℱ 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 ∈ ℱ, ff(k) and gg(k) share a in D, and, for every pair of functions f, g ∈ ℱ, all zeros of f are of multiplicity at least k + 1, then ℱ 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 ℱ be a family of meromorphic functions defined in a domain D. Let k be a positive integer. If, for every f ∈ ℱ such that f ≠ 0, ff(k) and gg(k) share z in D for every pair of function f, g ∈ ℱ, then ℱ is normal in D.
Theorem 5 (main theorem). Let ℱ be a family of meromorphic functions defined in a domain D. Let k ≥ 2 be a positive integer. For every f ∈ 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 ∈ ℱ, then ℱ is normal in D.
Theorem 6 (main theorem). Let ℱ be a family of meromorphic functions defined in a domain D. Let k be a positive integer. For every f ∈ 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 ∈ ℱ, then ℱ is normal in D.
2. Some Lemmas
In order to prove our theorems, we require the following results.
Lemma 7 (see [5], [6].)Let ℱ be a family of meromorphic functions in a domain D, and let k be a positive integer, such that each function f ∈ 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 ∈ ℱ. If ℱ is not normal at z0 ∈ D, then, for each 0 ≤ α ≤ k, there exist a sequence of points zn ∈ D, zn → z0, a sequence of positive numbers ρn → 0+, and a subsequence of functions fn ∈ ℱ such that
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
For simplicity, we denote N : = n1 + n2 + ⋯+nt. Obviously, N ≥ t.
From (2), we have
Differentiating (3), we get
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
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
Case 2. If f is a rational but not a polynomial. We see
For simplicity, we denote
Differentiating (13), we have
Now we distinguish two subcases.
Case 2.1. Supposing that ff(k) − z has exactly one zero z0, from (13), we obtain
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
Case 2.1.2. l = 2N + kt − 1. Next we distinguish two subcases: M > N and M ≤ N.
When M > N, similar to the Case 2.1.1, it follows that 2M < 2N; from (13) and (16), we get a contradiction.
If M ≤ N, by combining (15) and (18), we may give the following inequality:
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
Case 2. If f is a rational but not a polynomial, we set
For simplicity, we denote
Differentiating (29), we have
Now we distinguish two subcases.
Case 2.1. Supposing that ff(k) − z has exactly one zero z0, from (13), we obtain
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
- (i)
If k ≤ 4, we have
()
- (ii)
If k ≥ 5, then
()
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 ℱ1 = {Fj : Fj(z) = fj(z)/z1/2∣fj ∈ ℱ}. If ℱ1 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
Here we distinguish two cases.
Case 1.1. Supposing that zj/ρj → c, c is a finite complex number. Then
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 D1∩D2 = ∅ and such that H(ξ)H(k)(ξ) − ξ has no other zeros in D1 ∪ D2 except for ξ0 and , where
By Hurwitz’s Theorem, for sufficiently large j, there exist points ξj ∈ D1, such that
By the assumption in Theorem 5 that and share z, it follows that
Case 1.2. We may suppose that zj/ρj → ∞. We have
Thus, we have
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 D1∩D2 = ∅ and such that g(ξ)g(k)(ξ) − 1 has no other zeros in D1 ∪ D2 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, ℱ1 is normal at 0.
From Cases 1.1 and 1.2, we know that ℱ1 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) ∈ ℱ1, there exists δ > 0 such that if F(z) ∈ ℱ1, 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 ∈ ℱ, when |z | = δ/2, we have
Case ii. There exists a subsequence of fj, and we may still denote it as fj such that fj(0) = 0. Since f ∈ 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 ℱ is normal at z = 0.
By Cases i and ii, ℱ 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
From (61) we get
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 D1∩D2 = ∅ and such that gg(k) − z0 has no other zeros in D1 ∪ D2 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.