- Email: [email protected]
Normal families of meromorphic functions sharing a holomorphic function and the converse of the Bloch principle
Acta Mathematica Scientia 2012,32B(4):1503–1512 http://actams.wipm.ac.cn
NORMAL FAMILIES OF MEROMORPHIC FUNCTIONS SHARING A HOLOMORPHIC FUNCTION AND THE CONVERSE OF THE BLOCH PRINCIPLE∗
Jiang Yunbo (
)
)
Gao Zongsheng (
LMIB and School of Mathematics and Systems Science, Beihang University, Beijing 100191, China E-mail: [email protected]; [email protected]
Abstract In this paper, we investigate normal families of meromorphic functions, prove some theorems of normal families sharing a holomorphic function, and give a counterexample to the converse of the Bloch principle based on the theorems. Key words meromorphic function; holomorphic function; shared function; normal family; Bloch principle 2010 MR Subject Classification
1
30D35; 30D45
Introduction and Main Results
In this paper, we use the standard notations and concepts of the Nevanlinna theory (see [21–25]). Let D be a domain in the complex plane C, and F be a family of meromorphic functions defined in D. F is said to be normal in D, in the sense of Montel, if every sequence fn (z) ∈ F (n = 1, 2, · · ·) has a subsequence fnk (z) (k = 1, 2, · · ·) which converges spherically locally uniformly in D, to a meromorphic functions or ∞ (see [22, 23, 25]). Recently, some results about normal families was gotten (see [12–20, 25]). We use → to stand for convergence, ⇒ to stand for spherical local uniform convergence in C for the convenience of expression. In 1994, Yang and Yang [1] proposed a conjecture: If f is an entire function and k ≥ 2, (k) f f − a (a = 0) has infinitely many zeros. Some work was done to the above conjecture and more general form f (f (k) )n − a (a is a small function of f ), see [2–6]. Zhang and Song [4] proved the following theorem: Theorem A Suppose that f is a transcendental meromorphic function, n, k are two positive integers, then, when n ≥ 2, f (f (k) )n − a(z) has infinitely many zeros, where a(z) ≡ 0 is a small function of f . Alotaibi [5] showed a generalization of Theorem A using a simple proof. ∗ Received
September 16, 2010; revised July 1, 2011. Supported by the NSFC (11171013).
1504
ACTA MATHEMATICA SCIENTIA
Vol.32 Ser.B
There are famous Bloch principle and the converse of Bloch principle in the research of normal families of meromorphic (holomorphic) functions. Bloch principle reads: if a meromorphic (holomorphic) function in the complex plane certainly reduced to a constant when satisfying a property P , then a family of meromorphic (holomorphic) functions in a domain D certainly become a normal family when satisfying the property P . The converse of Bloch principle reads: if a family of meromorphic (holomorphic) functions in a domain D certainly become a normal family when satisfying a property P , then a meromorphic (holomorphic) function in the complex plane certainly reduced to a constant when satisfying the property P . Although Bloch principle is not valid in general, but many normality theorems were got starting form Bloch principle. Influenced by Bloch principle, Hu and Meng [7] proved the following normality theorem corresponding to Theorem A when a is a complex number: Theorem B Take positive integers n and k with n, k ≥ 2 and take a non-zero complex number a. Let F be a family of meromorphic functions in the domain D such that each f ∈ F has only zeros of multiplicity at least k. For each pair (f , g) ∈ F , if f (f (k) )n and g(g (k) )n share a IM, then F is normal in D. Naturally we ask: whether there exist normality theorems corresponding to Theorem A when a is a function. In this paper, we prove the following two theorems. Theorem 1 Let n, k ≥ 2, m ≥ 0 be three integers, and m be divisible by n + 1. Suppose that a(z)(≡ 0) is a holomorphic function with zeros of multiplicity m in a domain D. Let F be a family of holomorphic functions in D, for each f ∈ F , f has only zeros of multiplicity k + m at least. For each pair (f , g) ∈ F , f (f (k) )n and g(g (k) )n share a(z) IM, then F is normal in D. Theorem 2 m, a(z) are supposed as in Theorem 1. Let n ≥ 2m + 2, k ≥ 2 be two positive integers. Let F be a family of meromorphic functions in a domain D, for each f ∈ F , f has only zeros of multiplicity max{k + m,2m + 2} at least. For each pair (f , g) ∈ F , f (f (k) )n and g(g (k) )n share a(z) IM, then F is normal in D. Remark Obviously we can see that Theorem 2 extends Theorem B. Corollary 1 m, a(z) are supposed as in Theorem 1. Let n, k ≥ 2 be two positive integers. Let F be a family of holomorphic functions in a domain D, for each f ∈ F , f has only zeros of multiplicity k + m at least, and f (f (k) )n = a(z)(z ∈ D), then F is normal in D. Corollary 2 m, a(z) are supposed as in Theorem 1. Let n ≥ 2m + 2, k ≥ 2 be two positive integers. Let F be a family of meromorphic functions in a domain D, for each f ∈ F , f has only zeros of multiplicity max{k + m, 2m + 2} at least, and f (z)(f (k) (z))n = a(z)(z ∈ D). Then F is normal in D. In 2005, Lahiri [8] proved the following normality theorem. Theorem C Let F be a family of meromorphic functions in a complex domain D. Let a, b ∈ C and a = 0. Define a =b . Ef = z ∈ D : f (z) + f (z) If there exists a positive constant M such that |f (z)| ≥ M for all f ∈ F whenever z ∈ Ef , then F is a normal family. Recently, Charak and Rieppo [10] proved the following Lahiri type normality theorem.
No.4
Y.B. Jiang & Z.S. Gao: NORMAL FAMILIES OF MEROMORPHIC FUNCTIONS
1505
Theorem D Let F be a family of meromorphic functions in a complex domain D. Let a, b ∈ C and a = 0. Let m1 , m2 , n1 , n2 be nonnegative integers such that m1 n2 − m2 n1 > 0, m1 + m2 ≥ 1, n1 + n2 ≥ 2. Put a = b . Ef = z ∈ D : (f (z))n1 (f (z))m1 + (f (z))n2 (f (z))m2 If there exists a positive constant M such that |f (z)| ≥ M for all f ∈ F whenever z ∈ Ef , then F is a normal family. In this paper, we prove the following Lahiri type normality theorem. Theorem 3 m, a(z) are supposed as in Theorem 1. Let n ≥ 2m + 2, k be two positive integers. Let F be a family of meromorphic functions in a complex domain D, for each f ∈ F , f has only zeros of multiplicity max{k + m, 2m + 2} at least and poles of multiplicity 2 at least. Define Ef = {z ∈ D : f (z)(f (k) (z))n = a(z)}. If there exists a positive constant M such that |f (z)| ≥ M for all f ∈ F whenever z ∈ Ef , then F is a normal family. Theorem 4 m, a(z) are supposed as in Theorem 1. Let n ≥ 2m + 2, k be two positive integers. Let F be a family of meromorphic functions in a complex domain D, for each f ∈ F , f has only zeros of multiplicity max{k + m, 2m + 2} at least and poles of multiplicity 2 at least. Define Ef = {z ∈ D : f (z)(f (k) (z))n = a(z)}. If there exists a complex number b = 0 such that f (z) = b for all f ∈ F whenever z ∈ Ef , then F is a normal family. Lahiri [8] gave a counterexample to the converse of the Bloch principle based on Theorem C, Charak and Rieppo [10] also gave a counterexample to the converse of the Bloch principle based on Theorem D. Li [9] also gave a counterexample to the converse of the Bloch principle based on a theorem he proved in [9]. In this paper, we also give a counterexample to the converse of the Bloch principle based on Theorems 3 and 4, as the following Example 1. Example 1 Let f (z) = e2z , a(z) = e6z , n = 2. Then f (z)(f (z))2 − a(z) = 4e6z − e6z = 3e6z = 0, i.e., there exists a nonconstant meromorphic function f (z) possessing the property described in Theorems 3 and 4.
2
Some Lemmas
Lemma 1 [12] Let F be a family of meromorphic functions on the unit disc Δ, such that all zeros of functions in F have multiplicity greater than or equal to l, and all poles of functions in F have multiplicity greater than or equal to j. Let α be a real number satisfying −l < α < j. Then F is not normal in any neighborhood of z0 ∈ Δ if and only if there exist (i) Points zk ∈ Δ, zk → z0 ; (ii) Positive numbers ρk , ρk → 0 and (iii) Functions fk ∈ F such that ρα k fk (zk + ρk ζ) ⇒ g(ζ), where g(ζ) is a nonconstant meromorphic function, which may be taken to satisfy the normalization g # (z) ≤ g # (0) = 1 (z ∈ C), where g # (z) denotes the spherical derivative of g(z).
1506
ACTA MATHEMATICA SCIENTIA
Vol.32 Ser.B
Lemma 2 Let a(z) be a non-zero polynomial of degree m, and n, k (≥ 2) be two positive integers. If f is a non-constant polynomial with zeros of multiplicity k + m at least, then f (z)(f (k) (z))n − a(z) has at least two distinct zeros, and f (z)(f (k) (z))n − a(z) ≡ 0. Proof Since f is a non-constant polynomial with zeros of multiplicity k +m at least, then the degree of f is k+m at least, thus deg(f (z)(f (k) (z))n ) > deg(a(z)), then f (z)(f (k) (z))n −a(z) has at least one zero, and f (z)(f (k) (z))n − a(z) ≡ 0. If f (z)(f (k) (z))n −a(z) has only one zero. We may assume f (z)(f (k) (z))n −a(z) = λ(z−z0 )l , where λ is a non-zero constant, l is a positive integer. Compare the degrees of a(z) and f (z), we have l = deg(f (z)(f (k) (z))n ) > m + 1, then (f (z)(f (k) (z))n )(m) − λ · l · (l − 1) · · · (l − m + 1)(z − z0 )l−m = a(m) (z) = 0, (f (f (k) )n )(m+1) = λ · l · (l − 1) · · · (l − m)(z − z0 )l−m−1 , thus z0 is the unique zero of (f (f (k) )n )(m+1) . Since f is a non-constant polynomial with zeros of multiplicity k + m at least, then z0 is a zero of f , thus (f (f (k) )n )(m) (z0 ) = 0, it contradicts with (f (f (k) )n )(m) (z0 ) = a(m) (z0 ) = 0. Thus, f (z)(f (k) (z))n − a(z) has at least two distinct zeros. 2 Lemma 3 Let a(z) be a non-zero polynomial of degree m, and n (≥ 2m + 2), k be two positive integers. If f is a non-polynomial rational function, and all the zeros of f have multiplicity 2m + 2 at least, then f (z)(f (k) (z))n − a(z) has two distinct zeros at least, and f (z)(f (k) (z))n − a(z) ≡ 0. Proof Since f is a non-polynomial rational function, then obviously f (z)(f (k) (z))n − a(z) ≡ 0. Let (z − γ1 )p1 (z − γ2 )p2 · · · (z − γs )ps f (z)(f (k) (z))n = C , (2.1) (z − δ1 )q1 (z − δ2 )q2 · · · (z − δt )qt p = p1 + p2 + · · · + ps , q = q1 + q2 + · · · + qt , where C is a non-zero constant, s, t ≥ 1, pi ≥ 2m + 2, qj ≥ n(k + 1) + 1 (i = 1, 2, · · · , s; j = 1, 2, · · · , t) are integers. Differentiate both sides of (2.1), we have (f (f (k) )n ) = C
(z − γ1 )p1 −1 (z − γ2 )p2 −1 · · · (z − γs )ps −1 Q1 (z), (z − δ1 )q1 +1 (z − δ2 )q2 +1 · · · (z − δt )qt +1 (1)
(0)
where Q1 (z) = (p − q)z s+t−1 + · · · + b1 z + b1 (f (f (k) )n )(m+1) = C
(i)
is a polynomial, b1 (i = 0, 1) are constants.
(z − γ1 )p1 −m−1 (z − γ2 )p2 −m−1 · · · (z − γs )ps −m−1 Qm+1 (z), (z − δ1 )q1 +m+1 (z − δ2 )q2 +m+1 · · · (z − δt )qt +m+1 (1)
(2.2) (0)
where Qm+1 (z) = (p − q)(p − q − 1) · · · (p − q − m)z (m+1)(s+t−1) + · · · + bm+1 z + bm+1 is a (i) polynomial, bm+1 (i = 0, 1) are constants. Now we discuss two cases: Case (1) If f (f (k) )n − a(z) has a unique zero, let f (f (k) )n − a(z) =
A(z − z0 )r , (z − δ1 )q1 (z − δ2 )q2 · · · (z − δt )qt
where A is a non-zero constant, r is a positive integer. Subcase (1.1) m ≥ r. Differentiate both sides of (2.3), then (f (f (k) )n ) − a (z) =
(z − δ1
A(z − z0 )r−1 R1 (z) , − δ2 )q2 +1 · · · (z − δt )qt +1
)q1 +1 (z
(2.3)
No.4
1507
Y.B. Jiang & Z.S. Gao: NORMAL FAMILIES OF MEROMORPHIC FUNCTIONS (1)
(0)
where R1 (z) = (r − q)z t + · · · + l1 z + l1 (f (f (k) )n )(r) − a(r) (z) =
(1)
(0)
is a polynomial, l1 , l1
are constants.
A · Rr (z) , (z − δ1 )q1 +r (z − δ2 )q2 +r · · · (z − δt )qt +r (1)
(0)
where Rr (z) = (r − q)(r − q − 1) · · · (r − q − r + 1)z rt + · · · + lr z + lr (0) lr are constants. (f (f (k) )n )(r+1) − a(r+1) (z) =
(1)
is a polynomial, lr ,
A · Rr+1 (z) , (z − δ1 )q1 +r+1 (z − δ2 )q2 +r+1 · · · (z − δt )qt +r+1 (1)
(0)
where Rr+1 (z) = (r − q)(r − q − 1) · · · (r − q − r)z (r+1)t−1 + · · · + lr+1 z + lr+1 is a polynomial, (1) (0) lr+1 , lr+1 are constants. (f (f (k) )n )(m+1) − a(m+1) (z) =
(z − δ1
)q1 +m+1 (z
A · Rm+1 (z) , − δ2 )q2 +m+1 · · · (z − δt )qt +m+1 (1)
(2.4) (0)
where Rm+1 (z) = (r − q)(r − q − 1) · · · (r − q − m)z (m+1)t−(m−r+1) + · · · + lm+1 z + lm+1 is a (1) (0) polynomial, lm+1 , lm+1 are constants. Comparing formula (2.1) with (2.3), we have p > q. From formula (2.2) and (2.4), p − ms− s ≤ (m+ 1)t− (m− r + 1), then r − m ≥ p− ms− s− (m+ 1)t+ 1. Since ms+ s+ (m+ 1)t ≤ p(m+1) q(m+1) p(m+1) p(m+1) 2m+2 + n(k+1) < 2m+2 + n(k+1) < p, thus r − m ≥ 1, which contradicts with m ≥ r. Subcase (1.2) m < r. Differentiate both sides of (2.3), then (f (f (k) )n )(m+1) − a(m+1) (z) =
(z − z0 )r−m−1 ϕ(z) , (z − δ1 )q1 +m+1 (z − δ2 )q2 +m+1 · · · (z − δt )qt +m+1
(2.5)
where ϕ(z) = D(r − q)(r − q − 1) · · · (r − q − m)z (m+1)t + · · · + b1 z + b0 is a polynomial, b1 , b0 are constants. Next, we discuss three subcases: Subcase (1.2.1) when r < q + m. Comparing formula (2.1) with (2.3), we have p > q. From (2.2) and (2.5), we know p − ms − s ≤ (m + 1)t, then p ≤ ms + s + (m + 1)t ≤ p(m+1) q(m+1) 2m+2 + n(k+1) < p, this is a contradiction. Subcase (1.2.2) when r = q + m. If p > q, then, by (2.2) and (2.5), p ≤ ms + q(m+1) s + (m + 1)t ≤ p(m+1) 2m+2 + n(k+1) < p, a contradiction, thus p ≤ q. From (2.2) and (2.5), q(m+1) r − m − 1 ≤ (s + t + 1)(m + 1), then q < r ≤ s(m + 1) + t(m + 1) < p(m+1) 2m+2 + n(k+1) ≤ q, a contradiction. Subcase (1.2.3) when r > q + m. Comparing (2.1) with (2.3) we have p > q. Similar discussion as Subcase (1.2.1) yields a contradiction. Case (2) If f (f (k) )n − a(z) has no zero. Then r = 0 in (2.3), similar discussion to (1) we can get a contradiction. By Cases (1) and (2), we have f (z)(f (k) (z))n − a(z) has two distinct zeros at least. 2 Lemma 4 Let a(z) be a non-zero polynomial of degree m, and n(≥ 2m+2) be an integer. If f is a non-constant rational function with only zeros of multiplicity m+1 at least and poles of multiplicity 2 at least, then f (z)(f (z))n −a(z) has one zero at least, and f (z)(f (z))n −a(z) ≡ 0. Proof If f is a non-constant polynomial, as f has only zeros of multiplicity m + 1 at least, we know deg(f ) ≥ m + 1, deg(f (z)(f (z))n ) ≥ nm + m + 1, then f (z)(f (z))n − a(z) has one zero at least, and f (z)(f (z))n − a(z) ≡ 0.
1508
ACTA MATHEMATICA SCIENTIA
Vol.32 Ser.B
If f is a rational function but not a polynomial, then obviously f (z)(f (z))n − a(z) ≡ 0. Let (z − γ1 )u1 (z − γ2 )u2 · · · (z − γs )us f (z) = B , (z − δ1 )v1 (z − δ2 )v2 · · · (z − δt )vt U = u1 + u2 + · · · + us , V = v1 + v2 + · · · + vt . Then f (z)(f (z))n = B
(z − γ1 )U1 (z − γ2 )U2 · · · (z − γs )Us rn (z) P = , V V V 1 2 t (z − δ1 ) (z − δ2 ) · · · (z − δt ) Q
(2.6)
where r(z) = B(U − V )z s+t−1 + · · · + a1 z + a0 is a polynomial, a1 , a0 are two complex numbers, Ui = (n + 1)ui − n (i = 1, 2, · · · , s), Vj = (n + 1)vj + n (j = 1, 2, · · · , t). Thus, deg(P ) ≤ (n + 1)U + nt − n, deg(Q) = (n + 1)V + nt.
(2.7)
If f (f )n − a(z) has no zero, then f (f )n = a(z) +
P D = , Q Q
(2.8)
where D is a non-zero constant. Then deg(P ) = deg(Q) + m.
(2.9)
Comparing (2.7) with (2.9), we know V < U , so deg(P ) = (n + 1)U + nt − n, then, by (2.9), we get U − V = n+m n+1 . Differentiating (2.6), we get (f (f )n )(m+1) = B
(z − γ1 )U1 −m−1 (z − γ2 )U2 −m−1 · · · (z − γs )Us −m−1 ψ(z) , (z − δ1 )V1 +m+1 (z − δ2 )V2 +m+1 · · · (z − δt )Vt +m+1
(2.10)
where ψ(z) = [(n + m)(s + t − 1) + M − ms − N − mt]z (n+m+1)(s+t−1) + · · · + b1 z + b0 is a polynomial, b1 , b0 are constants, M = U1 + U2 + · · · + Us , N = V1 + V2 + · · · + Vt . Differentiating (2.8), we have (f (f )n )(m+1) =
D(−1)m [N (N + 1) · · · (N + m)]z (m+1)(t−1) + · · · + c1 z + c0 , (z − δ1 )V1 +m+1 (z − δ2 )V2 +m+1 · · · (z − δt )Vt +m+1
(2.11)
where c1 , c0 are constants. Since M − ms − s ≤ (m + 1)(t − 1) and M > (2m + 2)s, so t > s. Thus (3n + 2)t ≤ N = M − n − m + ns + nt ≤ (m + 1)(t − 1) + ms + s − n − m + ns + nt < 2(n + m + 1)t. Hence n < 2m, a contradiction. Then f (z)(f (z))n − a(z) has one zero at least. 2
3
Proofs of Theorems
Proof of Theorem 1 For any point z0 in D, either a(z0 ) = 0 or a(z0 ) = 0. Case (1) When a(z0 ) = 0. We may assume z0 = 0. Thus a(z) = ad z d + ad+1 z d+1 + · · · = d z h(z), where ad (= 0) is a constant, d (≥ 1) is divisible by n + 1. We may assume that d = (n + 1)s, s (≥ 1) being a positive integer.
No.4
1509
Y.B. Jiang & Z.S. Gao: NORMAL FAMILIES OF MEROMORPHIC FUNCTIONS
Let G = {Ht |Ht (z) =
ft (z) z s }.
If G is not normal at 0, then by Lemma 1, there exist points −
nk
still denoted as
zj ρj
zj → 0, positive numbers ρj → 0 and Hj ∈ G such that gj (ζ) = ρj n+1 Hj (zj + ρj ζ) ⇒ g(ζ), where g(ζ) is a non-constant meromorphic function in C, and g # (ζ) ≤ 1. Now, we discuss two cases. Case (1.1) There exists a subsequence of being a finite complex number. Then Fj (ζ) =
fj (ρj ζ) s+ nk ρj n+1
zj ρj ,
zj ρj ))
(ρj ζ)s Hj (zj + ρj (ζ −
=
nk
ρsj ρjn+1
, such that
zj ρj
→ c, c
⇒ ζ s g(ζ − c) = H(ζ),
so (k)
(k)
Fj (ζ)(Fj (ζ))n −
fj (ρj ζ)(fj (ρj ζ))n − a(ρj ζ) a(ρj ζ) = ⇒ H(ζ)(H (k) (ζ))n − ad ζ d . ρdj ρdj
Since for any f ∈ F , the multiplicity of every zero of f is k + m at least, the multiplicity of every zero of H is k + m at least, then from Lemma 2, Lemma 3 and Theorem A, H(ζ)(H (k) (ζ))n − ad ζ d ≡ 0, and H(ζ)(H (k) (ζ))n − ad ζ d has two distinct zeros at least. Assume that ζ1 , ζ2 are two distinct zeros of H(ζ)(H (k) (ζ))n − ad ζ d . We choose σ > 0 properly, such that D(ζ1 , σ) D(ζ2 , σ) = ∅, where D(ζ1 , σ) = {ζ| |ζ − ζ1 | < σ}, D(ζ2 , σ) = {ζ| |ζ − ζ2 | < σ}. (k)
By Hurwitz’s theorem, there exists a subsequence of fj (ρj ζ)(fj (ρj ζ))n − a(ρj ζ), still (k) denoted as fj (ρj ζ)(fj (ρj ζ))n − a(ρj ζ), points ζj → ζ1 , and points ζj → ζ2 , such that, when j is sufficiently large, (k) (k) fj (ρj ζj )(fj (ρj ζj ))n − a(ρj ζj ) = 0, fj (ρj ζj )(fj (ρj ζj ))n − a(ρj ζj ) = 0.
(3.1)
Because, for each pair of functions f and g in F , f (f (k) )n and g(g (k) )n share a(z) in D, thus, for any positive integer r, when j is sufficiently large, fr (ρj ζj )(fr(k) (ρj ζj ))n − a(ρj ζj ) = 0, fr (ρj ζj )(fr(k) (ρj ζj ))n − a(ρj ζj ) = 0.
(3.2)
(k) For fixed r, let j converge to ∞, then ρj ζj → 0, ρj ζj → 0, hence fr (0)(fr (0))n − a(0) = 0. (k) Then, by the isolation property of every zero of fr (ζ)(fr (ζ))n − a(ζ), when j is sufficiently large, ρj ζj = ρj ζj = 0, so when j is sufficiently big, ζ1 = ζ2 = 0, which contradicts with D(ζ1 , σ) D(ζ2 , σ) = ∅. Thus G is normal at 0. zj ρj ,
Case (1.2) There exists a subsequence of Then (k) fj (zj
s
+ ρj ζ) = (zj + ρj ζ)
(k) Hj (zj
+ ρj ζ) +
still denoted as
k
zj ρj
, such that
(k−i)
ci (zj + ρj ζ)s−i Hj
zj ρj
→ ∞.
(zj + ρj ζ)
i=1 nk
= (zj + ρj ζ)s ρjn+1
−k (k) gj (ζ)
+
k
nk
ci (zj + ρj ζ)s−i ρjn+1
i=1
where ci = d(d − 1) · · · (d − i + 1)Ckd when d ≥ i, ci = 0 when d < i.
−(k−i) (k−i) gj (ζ),
1510
ACTA MATHEMATICA SCIENTIA
Since any term in the expansion of (
nk
ρj n+1
−k)(n−l)
(k)
(gj (ζ))n−l (
k
i=1
1 [(zj (zj +ρj ζ)d nk
ci (zj + ρj ζ)s−i ρjn+1
Vol.32 Ser.B n
nk
+ ρj ζ)s ρjn+1 gj (ζ)
−k+i (k−i) gj (ζ))l ]
l=1
Cnl (zj + ρj ζ)s(n−l)
can be stated as
lk l
nk ρjn+1 (k) l n−l s−iν n+1 −k+iν (k−iν ) gj (ζ)Cn (gj (ζ)) (ciν (zj + ρj ζ) ρj gj (ζ)) (zj + ρj ζ)sl ν=1 l
1 (k) (k−iν ) l n−l g = (ζ)C (g (ζ)) (c g (ζ)) ⇒ 0, j iν j n j l
z
( ρjj + ζ)ν=1
iν
ν=1
here l ∈ {1, 2, · · · , n}, all iν ∈ {1, 2, · · · , k}(ν ∈ {1, 2, · · · , l}), then (k)
ad fj (zj + ρj ζ)(fj (zj + ρj ζ))n − ad a(zj + ρj ζ) nk ad 1 (k) gj (ζ)(gj (ζ))n + (zj + ρj ζ)s ρjn+1 gj (ζ) = d h(zj + ρj ζ) (zj + ρj ζ)
l n k nk ( nk −k)(n−l) (k) −k+i (k−i) Cnl (zj + ρj ζ)s(n−l) ρj n+1 (gj (ζ))n−l ci (zj + ρj ζ)s−i ρjn+1 gj (ζ) i=1
l=1
−ad ⇒ g(ζ)(g
(k)
(ζ)) − ad , n
where convergence is in C − {ζ|g(ζ) = ∞}. Then, by Lemma 2, Lemma 3 and Theorem A, we have g(ζ)(g (k) (ζ))n − ad has two distinct zeros at least, and g(ζ)(g (k) (ζ))n − ad ≡ 0. Suppose that ζ3 , ζ4 are two distinct zeros of g(ζ)(g (k) (ζ))n − ad . We choose γ > 0 properly, such that D(ζ3 , γ) D(ζ4 , γ) = ∅, where D(ζ3 , γ) = {ζ| |ζ − ζ3 | < γ}, D(ζ4 , γ) = {ζ| |ζ − ζ4 | < γ}. (k) By Hurwitz’s theorem, there exist a subsequence of fj (zj + ρj ζ)(fj (zj + ρj ζ))n − a(zj + (k)
ρj ζ), still denoted as fj (zj + ρj ζ)(fj (zj + ρj ζ))n − a(zj + ρj ζ), points ζj → ζ3 and points ζj∗ → ζ4 , such that, when j is sufficiently large, (k)
fj (zj + ρj ζj )(fj (zj + ρj ζj ))n − a(zj + ρj ζj ) = 0, (k)
fj (zj + ρj ζj∗ )(fj (zj + ρj ζj∗ ))n − a(zj + ρj ζj∗ ) = 0.
(3.3) (3.4)
Similar to the proof of Case (1.1) we get a contradiction. Then G is normal at 0. From Cases (1.1) and (1.2) we have G is normal at 0, so there exists Δρ = {z : |z| < ρ} and a subsequence Htk of Ht , such that Htk converges spherically locally uniformly to a meromorphic function U (z) or ∞(k → ∞) in Δρ . Now, we discuss two cases (i) and (ii): (i) When k is sufficiently big, ftk (0) = 0. So U (0) = ∞. Then, for arbitrary constant R > 0, ∃σ ∈ (0, ρ), when z ∈ Δσ , we have |U (z)| > R. Hence, for sufficiently large k, |Htk (z)| > R2 . So 1 σ ft is holomorphic in Δσ , and when |z| = 2 , k
2s+1 1 1 = ft (z) Ht (z)z s ≤ Rσ s . k k
No.4
Y.B. Jiang & Z.S. Gao: NORMAL FAMILIES OF MEROMORPHIC FUNCTIONS
1511
By maximum principle and Montel’s theorem, F is normal at z = 0. (ii) There exists a subsequence of ftk , still denoted as ftk , such that ftk (0) = 0. Since f (z) the multiplicity of every zero of ftk is k + m at least, and Htk = tkzs , then Htk (0) = 0. Thus, there exists 0 < ω < ρ such that Htk is holomorphic in Δω = {z : |z| < ω}. Then Htk converges spherically locally uniformly to a holomorphic function U (z) in Δω = {z : |z| < ω}. Since Htk (0) = 0, then U (0) = 0. Hence, there exists 0 < r < ρ such that U (z) is holomorphic in Δr = {z : |z| < r}, and has a unique zero z = 0 in Δr . Thus, Htk converges spherically locally uniformly to a holomorphic function U (z) in Δr , then ftk converges spherically locally uniformly to a holomorphic function z s U (z) in Δr . Hence F is normal at z = 0. From (i) and (ii) we know F is normal at 0. Case (2) When a(z0 ) = 0. We may assume that z0 = 0. If F is not normal at z0 , then by Lemma 1, there exist points zj → z0 , ρj → 0, fj ∈ F −
nk
such that ρj n+1 fj (zj + ρj ζ) ⇒ g(ζ), g(ζ) being a non-constant meromorphic function in C, and g # (ζ) ≤ 1. Since ∀f ∈ F , the multiplicity of zeros of f is k + m at least, then the multiplicity of zeros of g is k + m at least. Then from Lemma 2, Lemma 3 and Theorem A, g(g (k) )n − a(z0 ) has at least two distinct zeros, and g(g (k) )n − a(z0 ) ≡ 0. Similar to the proof of Case (1.1) we get a contradiction, then F is normal at z0 . Thus F is normal in D as z0 is arbitrary. 2 Proof of Theorem 3 For any point z0 in D, either a(z0 ) = 0 or a(z0 ) = 0. Case (1) When a(z0 ) = 0. We may assume that z0 = 0. Similar to the proof of Case (1) of Theorem 1, and use the same notations as in the proof of Case (1) of Theorem 1, first, by Lemmas 2–4 and Theorem A, we know that H(ζ)(H (k) (ζ))n − ad ζ d in Case (1.1) has at least one zero α0 . Then by Hurwitz’s theorem, there exist a subse(k) (k) quence of fj (ρj ζ)(fj (ρj ζ))n − a(ρj ζ), still denoted as fj (ρj ζ)(fj (ρj ζ))n − a(ρj ζ), and points αj → α0 , such that, when j is sufficiently large, (k)
fj (ρj αj )(fj (ρj αj ))n − a(ρj αj ) = 0. Thus |Fj (αj )| = |
fj (ρj αj )
(k)
s+ nk ρj n+1
n
| ≥ |
1
s+ nk ρj n+1
|M , and H(α0 ) = lim Fj (αj ) = ∞, which contradicts j→∞
αd0 .
with H(α0 )(H (α0 )) = Hence F is normal at z0 . Secondly, by Lemmas 2–4 and Theorem A, we have g(ζ)(g (k) (ζ))n − ad in Case (1.2) has at (k) least one zero β0 . Then by Hurwitz’s theorem, there exist a subsequence of fj (zj +ρj ζ)(fj (zj + (k)
ρj ζ))n − a(zj + ρj ζ), still denoted as fj (zj + ρj ζ)(fj (zj + ρj ζ))n − a(zj + ρj ζ), and points βj → β0 , such that, when j is sufficiently large, (k)
fj (zj + ρj βj )(fj (zj + ρj βj ))n − a(zj + ρj βj ) = 0. nk − n+1 fj (zj +ρj βj ) (zj +ρj βj )s |
Thus |gj (βj )| = |ρj
nk − n+1
≥ |ρj
1 (zj +ρj βj )s |M ,
and g(β0 ) = lim gj (βj ) = ∞, which j→∞
contradicts with g(β0 )(g (k) (β0 ))n = ad . Hence F is normal at z0 . Case (2) When a(z0 ) = 0. We may assume that z0 = 0. If F is not normal at z0 . Then from Lemma 1, there exist points zt → 0, ρt → 0, ft ∈ F nk − n+1
such that gt (ζ) = ρt
ft (zt + ρt ζ) ⇒ g(ζ), g(ζ) being a non-constant meromorphic function
1512
ACTA MATHEMATICA SCIENTIA
Vol.32 Ser.B
in C, and g # (ζ) ≤ 1. Then g(ζ) has only zeros of multiplicity max{k + m, 2m + 2} at least and poles of multiplicity 2 at least. From Lemmas 2–4 and Theorem A we know g(g (k) )n −a(0) has at least one zero, denoted as (k) ζ0 . By Hurwitz’s theorem, there exist a subsequence of ft (zt +ρt ζ)(ft (zt +ρt ζ))n −a(zt +ρt ζ), (k) still denoted as ft (zt +ρt ζ)(ft (zt +ρt ζ))n −a(zt +ρt ζ), and points ζt → ζ0 , such that, when t is nk − n+1
(k)
sufficiently large, ft (zt + ρt ζt )(ft (zt + ρt ζt ))n − a(zt + ρt ζt ) = 0. Thus |gt (ζt )| = |ρt − nk |ρt n+1 |M,
(k)
ft (zt +
ρt ζt )| ≥ so g(ζ0 ) = lim gt (ζt ) = ∞, which contradicts with g(ζ0 )(g (ζ0 )) = a(0). t→∞ Hence F is normal at z0 . Thus, F is normal in D as z0 is arbitrary. 2 Proofs of Theorems 2 and 4 Similar to the proof of Theorem 1 and notice Lemma 3, we can prove Theorem 2. Similar to the proof of Theorem 3, we can prove Theorem 4. n
References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25]
Yang L, Yang C C. Angular distribution of values of f f . Sci China Ser A, 1994, 37(3): 284–294 Yang C C, Yang L,Wang Y F. On the zeros of f (f (k) )n − a. Chin Sci Bull, 1994, 39(8): 617–621 Yang C C, Hu P C. On the value distribution of f f (k) . Kodai Math J, 1996, 19(2): 157–167 Zhang Z F, Song G D. On the zeros of f (f (k) )n − a(z). Chin Ann Math Ser A, 1998, 19(2): 275–282 Alotaibi A. On the zeros of af (f (k) )n − 1 for n ≥ 2, Comput Methods Funct Theory, 2004, 4(1): 227–235 Wang J P. On the value distribution of f f (k) . Kyungpook Math J, 2006, 46(2): 169–180 Hu P C, Meng D W. Normality criteria of meromorphic functions with multiple zeros. J Math Anal Appl, 2009, 357(2): 323–329 Lahiri I. A simple normality criterion leading to a counterexample to the converse of the Bloch principle. New Zealand J Math, 2005, 34(1): 61–65 Li B Q. On the Bloch heuristic principle, normality, and totally ramified values. Arch Math, 2006, 87(5): 417–426 Charak K S, Rieppo J. Two normality criteria and the converse of the Bloch principle. J Math Anal Appl, 2009, 353(1): 43–48 Bergweiler W, Eremenko A. On the singularitoes of the inverse to a meromorphic function of finite order. Rev Mat Iberoamericana, 1995, 11(2): 355–373 Zalcman L. Normal families: new perspectives. Bull Amer Math Soc, 1998, 35(3): 215–230 Fang M L, Zalcman L. A note on normality and shared values. J Aust Math Soc, 2004, 76(1): 141–150 Pang X C, Zalcman L. Normality and shared values. Ark Mat, 2000, 38(1): 171–182 Zhang Q C. Normal families of meromorphic functions concerning shared values. J Math Anal Appl, 2008, 338(1): 545–551 Han M H, Gu Y X. The normal famility of meromorphic functions. Acta Math Sci, 2008, 28B(4): 759–762 Xia J Y, Xu Y. Normal families of meromorphic functions with multiple values. J Math Anal Appl, 2009, 354(1): 387–393 Li Y T, Gu Y X. On normal families of meromorphic functions. J Math Anal Appl, 2009, 354(2): 421–425 Sun D C. Normality criteria for families of meromorphic algebroid functions. Acta Math Sci, 2010, 30B(1): 166–172 Jiang Y B, Gao Z S. Normal families of meromorphic functions sharing values or functions. J Inequal Appl, 2011, 2011: 72 Hayman W K. Meromorphic Functions. Oxford: Clarendon Press, 1964 Schiff J. Normal Families. Berlin: Springer-Verlag, 1993 Yang L. Value Distribution Theory. Berlin: Springer, 1993 Yang C C, Yi H X. Uniqueness Theory of Meromorphic Functions. Netherlands: Kluwer, 2003 Gu Y X, Pang X C, Fang M L. Theory of Normal Families and Its Application. Beijing: Science Press, 2007
NORMAL FAMILIES OF MEROMORPHIC FUNCTIONS SHARING A HOLOMORPHIC FUNCTION AND THE CONVERSE OF THE BLOCH PRINCIPLE∗
Jiang Yunbo (
)
)
Gao Zongsheng (
LMIB and School of Mathematics and Systems Science, Beihang University, Beijing 100191, China E-mail: [email protected]; [email protected]
Abstract In this paper, we investigate normal families of meromorphic functions, prove some theorems of normal families sharing a holomorphic function, and give a counterexample to the converse of the Bloch principle based on the theorems. Key words meromorphic function; holomorphic function; shared function; normal family; Bloch principle 2010 MR Subject Classification
1
30D35; 30D45
Introduction and Main Results
In this paper, we use the standard notations and concepts of the Nevanlinna theory (see [21–25]). Let D be a domain in the complex plane C, and F be a family of meromorphic functions defined in D. F is said to be normal in D, in the sense of Montel, if every sequence fn (z) ∈ F (n = 1, 2, · · ·) has a subsequence fnk (z) (k = 1, 2, · · ·) which converges spherically locally uniformly in D, to a meromorphic functions or ∞ (see [22, 23, 25]). Recently, some results about normal families was gotten (see [12–20, 25]). We use → to stand for convergence, ⇒ to stand for spherical local uniform convergence in C for the convenience of expression. In 1994, Yang and Yang [1] proposed a conjecture: If f is an entire function and k ≥ 2, (k) f f − a (a = 0) has infinitely many zeros. Some work was done to the above conjecture and more general form f (f (k) )n − a (a is a small function of f ), see [2–6]. Zhang and Song [4] proved the following theorem: Theorem A Suppose that f is a transcendental meromorphic function, n, k are two positive integers, then, when n ≥ 2, f (f (k) )n − a(z) has infinitely many zeros, where a(z) ≡ 0 is a small function of f . Alotaibi [5] showed a generalization of Theorem A using a simple proof. ∗ Received
September 16, 2010; revised July 1, 2011. Supported by the NSFC (11171013).
1504
ACTA MATHEMATICA SCIENTIA
Vol.32 Ser.B
There are famous Bloch principle and the converse of Bloch principle in the research of normal families of meromorphic (holomorphic) functions. Bloch principle reads: if a meromorphic (holomorphic) function in the complex plane certainly reduced to a constant when satisfying a property P , then a family of meromorphic (holomorphic) functions in a domain D certainly become a normal family when satisfying the property P . The converse of Bloch principle reads: if a family of meromorphic (holomorphic) functions in a domain D certainly become a normal family when satisfying a property P , then a meromorphic (holomorphic) function in the complex plane certainly reduced to a constant when satisfying the property P . Although Bloch principle is not valid in general, but many normality theorems were got starting form Bloch principle. Influenced by Bloch principle, Hu and Meng [7] proved the following normality theorem corresponding to Theorem A when a is a complex number: Theorem B Take positive integers n and k with n, k ≥ 2 and take a non-zero complex number a. Let F be a family of meromorphic functions in the domain D such that each f ∈ F has only zeros of multiplicity at least k. For each pair (f , g) ∈ F , if f (f (k) )n and g(g (k) )n share a IM, then F is normal in D. Naturally we ask: whether there exist normality theorems corresponding to Theorem A when a is a function. In this paper, we prove the following two theorems. Theorem 1 Let n, k ≥ 2, m ≥ 0 be three integers, and m be divisible by n + 1. Suppose that a(z)(≡ 0) is a holomorphic function with zeros of multiplicity m in a domain D. Let F be a family of holomorphic functions in D, for each f ∈ F , f has only zeros of multiplicity k + m at least. For each pair (f , g) ∈ F , f (f (k) )n and g(g (k) )n share a(z) IM, then F is normal in D. Theorem 2 m, a(z) are supposed as in Theorem 1. Let n ≥ 2m + 2, k ≥ 2 be two positive integers. Let F be a family of meromorphic functions in a domain D, for each f ∈ F , f has only zeros of multiplicity max{k + m,2m + 2} at least. For each pair (f , g) ∈ F , f (f (k) )n and g(g (k) )n share a(z) IM, then F is normal in D. Remark Obviously we can see that Theorem 2 extends Theorem B. Corollary 1 m, a(z) are supposed as in Theorem 1. Let n, k ≥ 2 be two positive integers. Let F be a family of holomorphic functions in a domain D, for each f ∈ F , f has only zeros of multiplicity k + m at least, and f (f (k) )n = a(z)(z ∈ D), then F is normal in D. Corollary 2 m, a(z) are supposed as in Theorem 1. Let n ≥ 2m + 2, k ≥ 2 be two positive integers. Let F be a family of meromorphic functions in a domain D, for each f ∈ F , f has only zeros of multiplicity max{k + m, 2m + 2} at least, and f (z)(f (k) (z))n = a(z)(z ∈ D). Then F is normal in D. In 2005, Lahiri [8] proved the following normality theorem. Theorem C Let F be a family of meromorphic functions in a complex domain D. Let a, b ∈ C and a = 0. Define a =b . Ef = z ∈ D : f (z) + f (z) If there exists a positive constant M such that |f (z)| ≥ M for all f ∈ F whenever z ∈ Ef , then F is a normal family. Recently, Charak and Rieppo [10] proved the following Lahiri type normality theorem.
No.4
Y.B. Jiang & Z.S. Gao: NORMAL FAMILIES OF MEROMORPHIC FUNCTIONS
1505
Theorem D Let F be a family of meromorphic functions in a complex domain D. Let a, b ∈ C and a = 0. Let m1 , m2 , n1 , n2 be nonnegative integers such that m1 n2 − m2 n1 > 0, m1 + m2 ≥ 1, n1 + n2 ≥ 2. Put a = b . Ef = z ∈ D : (f (z))n1 (f (z))m1 + (f (z))n2 (f (z))m2 If there exists a positive constant M such that |f (z)| ≥ M for all f ∈ F whenever z ∈ Ef , then F is a normal family. In this paper, we prove the following Lahiri type normality theorem. Theorem 3 m, a(z) are supposed as in Theorem 1. Let n ≥ 2m + 2, k be two positive integers. Let F be a family of meromorphic functions in a complex domain D, for each f ∈ F , f has only zeros of multiplicity max{k + m, 2m + 2} at least and poles of multiplicity 2 at least. Define Ef = {z ∈ D : f (z)(f (k) (z))n = a(z)}. If there exists a positive constant M such that |f (z)| ≥ M for all f ∈ F whenever z ∈ Ef , then F is a normal family. Theorem 4 m, a(z) are supposed as in Theorem 1. Let n ≥ 2m + 2, k be two positive integers. Let F be a family of meromorphic functions in a complex domain D, for each f ∈ F , f has only zeros of multiplicity max{k + m, 2m + 2} at least and poles of multiplicity 2 at least. Define Ef = {z ∈ D : f (z)(f (k) (z))n = a(z)}. If there exists a complex number b = 0 such that f (z) = b for all f ∈ F whenever z ∈ Ef , then F is a normal family. Lahiri [8] gave a counterexample to the converse of the Bloch principle based on Theorem C, Charak and Rieppo [10] also gave a counterexample to the converse of the Bloch principle based on Theorem D. Li [9] also gave a counterexample to the converse of the Bloch principle based on a theorem he proved in [9]. In this paper, we also give a counterexample to the converse of the Bloch principle based on Theorems 3 and 4, as the following Example 1. Example 1 Let f (z) = e2z , a(z) = e6z , n = 2. Then f (z)(f (z))2 − a(z) = 4e6z − e6z = 3e6z = 0, i.e., there exists a nonconstant meromorphic function f (z) possessing the property described in Theorems 3 and 4.
2
Some Lemmas
Lemma 1 [12] Let F be a family of meromorphic functions on the unit disc Δ, such that all zeros of functions in F have multiplicity greater than or equal to l, and all poles of functions in F have multiplicity greater than or equal to j. Let α be a real number satisfying −l < α < j. Then F is not normal in any neighborhood of z0 ∈ Δ if and only if there exist (i) Points zk ∈ Δ, zk → z0 ; (ii) Positive numbers ρk , ρk → 0 and (iii) Functions fk ∈ F such that ρα k fk (zk + ρk ζ) ⇒ g(ζ), where g(ζ) is a nonconstant meromorphic function, which may be taken to satisfy the normalization g # (z) ≤ g # (0) = 1 (z ∈ C), where g # (z) denotes the spherical derivative of g(z).
1506
ACTA MATHEMATICA SCIENTIA
Vol.32 Ser.B
Lemma 2 Let a(z) be a non-zero polynomial of degree m, and n, k (≥ 2) be two positive integers. If f is a non-constant polynomial with zeros of multiplicity k + m at least, then f (z)(f (k) (z))n − a(z) has at least two distinct zeros, and f (z)(f (k) (z))n − a(z) ≡ 0. Proof Since f is a non-constant polynomial with zeros of multiplicity k +m at least, then the degree of f is k+m at least, thus deg(f (z)(f (k) (z))n ) > deg(a(z)), then f (z)(f (k) (z))n −a(z) has at least one zero, and f (z)(f (k) (z))n − a(z) ≡ 0. If f (z)(f (k) (z))n −a(z) has only one zero. We may assume f (z)(f (k) (z))n −a(z) = λ(z−z0 )l , where λ is a non-zero constant, l is a positive integer. Compare the degrees of a(z) and f (z), we have l = deg(f (z)(f (k) (z))n ) > m + 1, then (f (z)(f (k) (z))n )(m) − λ · l · (l − 1) · · · (l − m + 1)(z − z0 )l−m = a(m) (z) = 0, (f (f (k) )n )(m+1) = λ · l · (l − 1) · · · (l − m)(z − z0 )l−m−1 , thus z0 is the unique zero of (f (f (k) )n )(m+1) . Since f is a non-constant polynomial with zeros of multiplicity k + m at least, then z0 is a zero of f , thus (f (f (k) )n )(m) (z0 ) = 0, it contradicts with (f (f (k) )n )(m) (z0 ) = a(m) (z0 ) = 0. Thus, f (z)(f (k) (z))n − a(z) has at least two distinct zeros. 2 Lemma 3 Let a(z) be a non-zero polynomial of degree m, and n (≥ 2m + 2), k be two positive integers. If f is a non-polynomial rational function, and all the zeros of f have multiplicity 2m + 2 at least, then f (z)(f (k) (z))n − a(z) has two distinct zeros at least, and f (z)(f (k) (z))n − a(z) ≡ 0. Proof Since f is a non-polynomial rational function, then obviously f (z)(f (k) (z))n − a(z) ≡ 0. Let (z − γ1 )p1 (z − γ2 )p2 · · · (z − γs )ps f (z)(f (k) (z))n = C , (2.1) (z − δ1 )q1 (z − δ2 )q2 · · · (z − δt )qt p = p1 + p2 + · · · + ps , q = q1 + q2 + · · · + qt , where C is a non-zero constant, s, t ≥ 1, pi ≥ 2m + 2, qj ≥ n(k + 1) + 1 (i = 1, 2, · · · , s; j = 1, 2, · · · , t) are integers. Differentiate both sides of (2.1), we have (f (f (k) )n ) = C
(z − γ1 )p1 −1 (z − γ2 )p2 −1 · · · (z − γs )ps −1 Q1 (z), (z − δ1 )q1 +1 (z − δ2 )q2 +1 · · · (z − δt )qt +1 (1)
(0)
where Q1 (z) = (p − q)z s+t−1 + · · · + b1 z + b1 (f (f (k) )n )(m+1) = C
(i)
is a polynomial, b1 (i = 0, 1) are constants.
(z − γ1 )p1 −m−1 (z − γ2 )p2 −m−1 · · · (z − γs )ps −m−1 Qm+1 (z), (z − δ1 )q1 +m+1 (z − δ2 )q2 +m+1 · · · (z − δt )qt +m+1 (1)
(2.2) (0)
where Qm+1 (z) = (p − q)(p − q − 1) · · · (p − q − m)z (m+1)(s+t−1) + · · · + bm+1 z + bm+1 is a (i) polynomial, bm+1 (i = 0, 1) are constants. Now we discuss two cases: Case (1) If f (f (k) )n − a(z) has a unique zero, let f (f (k) )n − a(z) =
A(z − z0 )r , (z − δ1 )q1 (z − δ2 )q2 · · · (z − δt )qt
where A is a non-zero constant, r is a positive integer. Subcase (1.1) m ≥ r. Differentiate both sides of (2.3), then (f (f (k) )n ) − a (z) =
(z − δ1
A(z − z0 )r−1 R1 (z) , − δ2 )q2 +1 · · · (z − δt )qt +1
)q1 +1 (z
(2.3)
No.4
1507
Y.B. Jiang & Z.S. Gao: NORMAL FAMILIES OF MEROMORPHIC FUNCTIONS (1)
(0)
where R1 (z) = (r − q)z t + · · · + l1 z + l1 (f (f (k) )n )(r) − a(r) (z) =
(1)
(0)
is a polynomial, l1 , l1
are constants.
A · Rr (z) , (z − δ1 )q1 +r (z − δ2 )q2 +r · · · (z − δt )qt +r (1)
(0)
where Rr (z) = (r − q)(r − q − 1) · · · (r − q − r + 1)z rt + · · · + lr z + lr (0) lr are constants. (f (f (k) )n )(r+1) − a(r+1) (z) =
(1)
is a polynomial, lr ,
A · Rr+1 (z) , (z − δ1 )q1 +r+1 (z − δ2 )q2 +r+1 · · · (z − δt )qt +r+1 (1)
(0)
where Rr+1 (z) = (r − q)(r − q − 1) · · · (r − q − r)z (r+1)t−1 + · · · + lr+1 z + lr+1 is a polynomial, (1) (0) lr+1 , lr+1 are constants. (f (f (k) )n )(m+1) − a(m+1) (z) =
(z − δ1
)q1 +m+1 (z
A · Rm+1 (z) , − δ2 )q2 +m+1 · · · (z − δt )qt +m+1 (1)
(2.4) (0)
where Rm+1 (z) = (r − q)(r − q − 1) · · · (r − q − m)z (m+1)t−(m−r+1) + · · · + lm+1 z + lm+1 is a (1) (0) polynomial, lm+1 , lm+1 are constants. Comparing formula (2.1) with (2.3), we have p > q. From formula (2.2) and (2.4), p − ms− s ≤ (m+ 1)t− (m− r + 1), then r − m ≥ p− ms− s− (m+ 1)t+ 1. Since ms+ s+ (m+ 1)t ≤ p(m+1) q(m+1) p(m+1) p(m+1) 2m+2 + n(k+1) < 2m+2 + n(k+1) < p, thus r − m ≥ 1, which contradicts with m ≥ r. Subcase (1.2) m < r. Differentiate both sides of (2.3), then (f (f (k) )n )(m+1) − a(m+1) (z) =
(z − z0 )r−m−1 ϕ(z) , (z − δ1 )q1 +m+1 (z − δ2 )q2 +m+1 · · · (z − δt )qt +m+1
(2.5)
where ϕ(z) = D(r − q)(r − q − 1) · · · (r − q − m)z (m+1)t + · · · + b1 z + b0 is a polynomial, b1 , b0 are constants. Next, we discuss three subcases: Subcase (1.2.1) when r < q + m. Comparing formula (2.1) with (2.3), we have p > q. From (2.2) and (2.5), we know p − ms − s ≤ (m + 1)t, then p ≤ ms + s + (m + 1)t ≤ p(m+1) q(m+1) 2m+2 + n(k+1) < p, this is a contradiction. Subcase (1.2.2) when r = q + m. If p > q, then, by (2.2) and (2.5), p ≤ ms + q(m+1) s + (m + 1)t ≤ p(m+1) 2m+2 + n(k+1) < p, a contradiction, thus p ≤ q. From (2.2) and (2.5), q(m+1) r − m − 1 ≤ (s + t + 1)(m + 1), then q < r ≤ s(m + 1) + t(m + 1) < p(m+1) 2m+2 + n(k+1) ≤ q, a contradiction. Subcase (1.2.3) when r > q + m. Comparing (2.1) with (2.3) we have p > q. Similar discussion as Subcase (1.2.1) yields a contradiction. Case (2) If f (f (k) )n − a(z) has no zero. Then r = 0 in (2.3), similar discussion to (1) we can get a contradiction. By Cases (1) and (2), we have f (z)(f (k) (z))n − a(z) has two distinct zeros at least. 2 Lemma 4 Let a(z) be a non-zero polynomial of degree m, and n(≥ 2m+2) be an integer. If f is a non-constant rational function with only zeros of multiplicity m+1 at least and poles of multiplicity 2 at least, then f (z)(f (z))n −a(z) has one zero at least, and f (z)(f (z))n −a(z) ≡ 0. Proof If f is a non-constant polynomial, as f has only zeros of multiplicity m + 1 at least, we know deg(f ) ≥ m + 1, deg(f (z)(f (z))n ) ≥ nm + m + 1, then f (z)(f (z))n − a(z) has one zero at least, and f (z)(f (z))n − a(z) ≡ 0.
1508
ACTA MATHEMATICA SCIENTIA
Vol.32 Ser.B
If f is a rational function but not a polynomial, then obviously f (z)(f (z))n − a(z) ≡ 0. Let (z − γ1 )u1 (z − γ2 )u2 · · · (z − γs )us f (z) = B , (z − δ1 )v1 (z − δ2 )v2 · · · (z − δt )vt U = u1 + u2 + · · · + us , V = v1 + v2 + · · · + vt . Then f (z)(f (z))n = B
(z − γ1 )U1 (z − γ2 )U2 · · · (z − γs )Us rn (z) P = , V V V 1 2 t (z − δ1 ) (z − δ2 ) · · · (z − δt ) Q
(2.6)
where r(z) = B(U − V )z s+t−1 + · · · + a1 z + a0 is a polynomial, a1 , a0 are two complex numbers, Ui = (n + 1)ui − n (i = 1, 2, · · · , s), Vj = (n + 1)vj + n (j = 1, 2, · · · , t). Thus, deg(P ) ≤ (n + 1)U + nt − n, deg(Q) = (n + 1)V + nt.
(2.7)
If f (f )n − a(z) has no zero, then f (f )n = a(z) +
P D = , Q Q
(2.8)
where D is a non-zero constant. Then deg(P ) = deg(Q) + m.
(2.9)
Comparing (2.7) with (2.9), we know V < U , so deg(P ) = (n + 1)U + nt − n, then, by (2.9), we get U − V = n+m n+1 . Differentiating (2.6), we get (f (f )n )(m+1) = B
(z − γ1 )U1 −m−1 (z − γ2 )U2 −m−1 · · · (z − γs )Us −m−1 ψ(z) , (z − δ1 )V1 +m+1 (z − δ2 )V2 +m+1 · · · (z − δt )Vt +m+1
(2.10)
where ψ(z) = [(n + m)(s + t − 1) + M − ms − N − mt]z (n+m+1)(s+t−1) + · · · + b1 z + b0 is a polynomial, b1 , b0 are constants, M = U1 + U2 + · · · + Us , N = V1 + V2 + · · · + Vt . Differentiating (2.8), we have (f (f )n )(m+1) =
D(−1)m [N (N + 1) · · · (N + m)]z (m+1)(t−1) + · · · + c1 z + c0 , (z − δ1 )V1 +m+1 (z − δ2 )V2 +m+1 · · · (z − δt )Vt +m+1
(2.11)
where c1 , c0 are constants. Since M − ms − s ≤ (m + 1)(t − 1) and M > (2m + 2)s, so t > s. Thus (3n + 2)t ≤ N = M − n − m + ns + nt ≤ (m + 1)(t − 1) + ms + s − n − m + ns + nt < 2(n + m + 1)t. Hence n < 2m, a contradiction. Then f (z)(f (z))n − a(z) has one zero at least. 2
3
Proofs of Theorems
Proof of Theorem 1 For any point z0 in D, either a(z0 ) = 0 or a(z0 ) = 0. Case (1) When a(z0 ) = 0. We may assume z0 = 0. Thus a(z) = ad z d + ad+1 z d+1 + · · · = d z h(z), where ad (= 0) is a constant, d (≥ 1) is divisible by n + 1. We may assume that d = (n + 1)s, s (≥ 1) being a positive integer.
No.4
1509
Y.B. Jiang & Z.S. Gao: NORMAL FAMILIES OF MEROMORPHIC FUNCTIONS
Let G = {Ht |Ht (z) =
ft (z) z s }.
If G is not normal at 0, then by Lemma 1, there exist points −
nk
still denoted as
zj ρj
zj → 0, positive numbers ρj → 0 and Hj ∈ G such that gj (ζ) = ρj n+1 Hj (zj + ρj ζ) ⇒ g(ζ), where g(ζ) is a non-constant meromorphic function in C, and g # (ζ) ≤ 1. Now, we discuss two cases. Case (1.1) There exists a subsequence of being a finite complex number. Then Fj (ζ) =
fj (ρj ζ) s+ nk ρj n+1
zj ρj ,
zj ρj ))
(ρj ζ)s Hj (zj + ρj (ζ −
=
nk
ρsj ρjn+1
, such that
zj ρj
→ c, c
⇒ ζ s g(ζ − c) = H(ζ),
so (k)
(k)
Fj (ζ)(Fj (ζ))n −
fj (ρj ζ)(fj (ρj ζ))n − a(ρj ζ) a(ρj ζ) = ⇒ H(ζ)(H (k) (ζ))n − ad ζ d . ρdj ρdj
Since for any f ∈ F , the multiplicity of every zero of f is k + m at least, the multiplicity of every zero of H is k + m at least, then from Lemma 2, Lemma 3 and Theorem A, H(ζ)(H (k) (ζ))n − ad ζ d ≡ 0, and H(ζ)(H (k) (ζ))n − ad ζ d has two distinct zeros at least. Assume that ζ1 , ζ2 are two distinct zeros of H(ζ)(H (k) (ζ))n − ad ζ d . We choose σ > 0 properly, such that D(ζ1 , σ) D(ζ2 , σ) = ∅, where D(ζ1 , σ) = {ζ| |ζ − ζ1 | < σ}, D(ζ2 , σ) = {ζ| |ζ − ζ2 | < σ}. (k)
By Hurwitz’s theorem, there exists a subsequence of fj (ρj ζ)(fj (ρj ζ))n − a(ρj ζ), still (k) denoted as fj (ρj ζ)(fj (ρj ζ))n − a(ρj ζ), points ζj → ζ1 , and points ζj → ζ2 , such that, when j is sufficiently large, (k) (k) fj (ρj ζj )(fj (ρj ζj ))n − a(ρj ζj ) = 0, fj (ρj ζj )(fj (ρj ζj ))n − a(ρj ζj ) = 0.
(3.1)
Because, for each pair of functions f and g in F , f (f (k) )n and g(g (k) )n share a(z) in D, thus, for any positive integer r, when j is sufficiently large, fr (ρj ζj )(fr(k) (ρj ζj ))n − a(ρj ζj ) = 0, fr (ρj ζj )(fr(k) (ρj ζj ))n − a(ρj ζj ) = 0.
(3.2)
(k) For fixed r, let j converge to ∞, then ρj ζj → 0, ρj ζj → 0, hence fr (0)(fr (0))n − a(0) = 0. (k) Then, by the isolation property of every zero of fr (ζ)(fr (ζ))n − a(ζ), when j is sufficiently large, ρj ζj = ρj ζj = 0, so when j is sufficiently big, ζ1 = ζ2 = 0, which contradicts with D(ζ1 , σ) D(ζ2 , σ) = ∅. Thus G is normal at 0. zj ρj ,
Case (1.2) There exists a subsequence of Then (k) fj (zj
s
+ ρj ζ) = (zj + ρj ζ)
(k) Hj (zj
+ ρj ζ) +
still denoted as
k
zj ρj
, such that
(k−i)
ci (zj + ρj ζ)s−i Hj
zj ρj
→ ∞.
(zj + ρj ζ)
i=1 nk
= (zj + ρj ζ)s ρjn+1
−k (k) gj (ζ)
+
k
nk
ci (zj + ρj ζ)s−i ρjn+1
i=1
where ci = d(d − 1) · · · (d − i + 1)Ckd when d ≥ i, ci = 0 when d < i.
−(k−i) (k−i) gj (ζ),
1510
ACTA MATHEMATICA SCIENTIA
Since any term in the expansion of (
nk
ρj n+1
−k)(n−l)
(k)
(gj (ζ))n−l (
k
i=1
1 [(zj (zj +ρj ζ)d nk
ci (zj + ρj ζ)s−i ρjn+1
Vol.32 Ser.B n
nk
+ ρj ζ)s ρjn+1 gj (ζ)
−k+i (k−i) gj (ζ))l ]
l=1
Cnl (zj + ρj ζ)s(n−l)
can be stated as
lk l
nk ρjn+1 (k) l n−l s−iν n+1 −k+iν (k−iν ) gj (ζ)Cn (gj (ζ)) (ciν (zj + ρj ζ) ρj gj (ζ)) (zj + ρj ζ)sl ν=1 l
1 (k) (k−iν ) l n−l g = (ζ)C (g (ζ)) (c g (ζ)) ⇒ 0, j iν j n j l
z
( ρjj + ζ)ν=1
iν
ν=1
here l ∈ {1, 2, · · · , n}, all iν ∈ {1, 2, · · · , k}(ν ∈ {1, 2, · · · , l}), then (k)
ad fj (zj + ρj ζ)(fj (zj + ρj ζ))n − ad a(zj + ρj ζ) nk ad 1 (k) gj (ζ)(gj (ζ))n + (zj + ρj ζ)s ρjn+1 gj (ζ) = d h(zj + ρj ζ) (zj + ρj ζ)
l n k nk ( nk −k)(n−l) (k) −k+i (k−i) Cnl (zj + ρj ζ)s(n−l) ρj n+1 (gj (ζ))n−l ci (zj + ρj ζ)s−i ρjn+1 gj (ζ) i=1
l=1
−ad ⇒ g(ζ)(g
(k)
(ζ)) − ad , n
where convergence is in C − {ζ|g(ζ) = ∞}. Then, by Lemma 2, Lemma 3 and Theorem A, we have g(ζ)(g (k) (ζ))n − ad has two distinct zeros at least, and g(ζ)(g (k) (ζ))n − ad ≡ 0. Suppose that ζ3 , ζ4 are two distinct zeros of g(ζ)(g (k) (ζ))n − ad . We choose γ > 0 properly, such that D(ζ3 , γ) D(ζ4 , γ) = ∅, where D(ζ3 , γ) = {ζ| |ζ − ζ3 | < γ}, D(ζ4 , γ) = {ζ| |ζ − ζ4 | < γ}. (k) By Hurwitz’s theorem, there exist a subsequence of fj (zj + ρj ζ)(fj (zj + ρj ζ))n − a(zj + (k)
ρj ζ), still denoted as fj (zj + ρj ζ)(fj (zj + ρj ζ))n − a(zj + ρj ζ), points ζj → ζ3 and points ζj∗ → ζ4 , such that, when j is sufficiently large, (k)
fj (zj + ρj ζj )(fj (zj + ρj ζj ))n − a(zj + ρj ζj ) = 0, (k)
fj (zj + ρj ζj∗ )(fj (zj + ρj ζj∗ ))n − a(zj + ρj ζj∗ ) = 0.
(3.3) (3.4)
Similar to the proof of Case (1.1) we get a contradiction. Then G is normal at 0. From Cases (1.1) and (1.2) we have G is normal at 0, so there exists Δρ = {z : |z| < ρ} and a subsequence Htk of Ht , such that Htk converges spherically locally uniformly to a meromorphic function U (z) or ∞(k → ∞) in Δρ . Now, we discuss two cases (i) and (ii): (i) When k is sufficiently big, ftk (0) = 0. So U (0) = ∞. Then, for arbitrary constant R > 0, ∃σ ∈ (0, ρ), when z ∈ Δσ , we have |U (z)| > R. Hence, for sufficiently large k, |Htk (z)| > R2 . So 1 σ ft is holomorphic in Δσ , and when |z| = 2 , k
2s+1 1 1 = ft (z) Ht (z)z s ≤ Rσ s . k k
No.4
Y.B. Jiang & Z.S. Gao: NORMAL FAMILIES OF MEROMORPHIC FUNCTIONS
1511
By maximum principle and Montel’s theorem, F is normal at z = 0. (ii) There exists a subsequence of ftk , still denoted as ftk , such that ftk (0) = 0. Since f (z) the multiplicity of every zero of ftk is k + m at least, and Htk = tkzs , then Htk (0) = 0. Thus, there exists 0 < ω < ρ such that Htk is holomorphic in Δω = {z : |z| < ω}. Then Htk converges spherically locally uniformly to a holomorphic function U (z) in Δω = {z : |z| < ω}. Since Htk (0) = 0, then U (0) = 0. Hence, there exists 0 < r < ρ such that U (z) is holomorphic in Δr = {z : |z| < r}, and has a unique zero z = 0 in Δr . Thus, Htk converges spherically locally uniformly to a holomorphic function U (z) in Δr , then ftk converges spherically locally uniformly to a holomorphic function z s U (z) in Δr . Hence F is normal at z = 0. From (i) and (ii) we know F is normal at 0. Case (2) When a(z0 ) = 0. We may assume that z0 = 0. If F is not normal at z0 , then by Lemma 1, there exist points zj → z0 , ρj → 0, fj ∈ F −
nk
such that ρj n+1 fj (zj + ρj ζ) ⇒ g(ζ), g(ζ) being a non-constant meromorphic function in C, and g # (ζ) ≤ 1. Since ∀f ∈ F , the multiplicity of zeros of f is k + m at least, then the multiplicity of zeros of g is k + m at least. Then from Lemma 2, Lemma 3 and Theorem A, g(g (k) )n − a(z0 ) has at least two distinct zeros, and g(g (k) )n − a(z0 ) ≡ 0. Similar to the proof of Case (1.1) we get a contradiction, then F is normal at z0 . Thus F is normal in D as z0 is arbitrary. 2 Proof of Theorem 3 For any point z0 in D, either a(z0 ) = 0 or a(z0 ) = 0. Case (1) When a(z0 ) = 0. We may assume that z0 = 0. Similar to the proof of Case (1) of Theorem 1, and use the same notations as in the proof of Case (1) of Theorem 1, first, by Lemmas 2–4 and Theorem A, we know that H(ζ)(H (k) (ζ))n − ad ζ d in Case (1.1) has at least one zero α0 . Then by Hurwitz’s theorem, there exist a subse(k) (k) quence of fj (ρj ζ)(fj (ρj ζ))n − a(ρj ζ), still denoted as fj (ρj ζ)(fj (ρj ζ))n − a(ρj ζ), and points αj → α0 , such that, when j is sufficiently large, (k)
fj (ρj αj )(fj (ρj αj ))n − a(ρj αj ) = 0. Thus |Fj (αj )| = |
fj (ρj αj )
(k)
s+ nk ρj n+1
n
| ≥ |
1
s+ nk ρj n+1
|M , and H(α0 ) = lim Fj (αj ) = ∞, which contradicts j→∞
αd0 .
with H(α0 )(H (α0 )) = Hence F is normal at z0 . Secondly, by Lemmas 2–4 and Theorem A, we have g(ζ)(g (k) (ζ))n − ad in Case (1.2) has at (k) least one zero β0 . Then by Hurwitz’s theorem, there exist a subsequence of fj (zj +ρj ζ)(fj (zj + (k)
ρj ζ))n − a(zj + ρj ζ), still denoted as fj (zj + ρj ζ)(fj (zj + ρj ζ))n − a(zj + ρj ζ), and points βj → β0 , such that, when j is sufficiently large, (k)
fj (zj + ρj βj )(fj (zj + ρj βj ))n − a(zj + ρj βj ) = 0. nk − n+1 fj (zj +ρj βj ) (zj +ρj βj )s |
Thus |gj (βj )| = |ρj
nk − n+1
≥ |ρj
1 (zj +ρj βj )s |M ,
and g(β0 ) = lim gj (βj ) = ∞, which j→∞
contradicts with g(β0 )(g (k) (β0 ))n = ad . Hence F is normal at z0 . Case (2) When a(z0 ) = 0. We may assume that z0 = 0. If F is not normal at z0 . Then from Lemma 1, there exist points zt → 0, ρt → 0, ft ∈ F nk − n+1
such that gt (ζ) = ρt
ft (zt + ρt ζ) ⇒ g(ζ), g(ζ) being a non-constant meromorphic function
1512
ACTA MATHEMATICA SCIENTIA
Vol.32 Ser.B
in C, and g # (ζ) ≤ 1. Then g(ζ) has only zeros of multiplicity max{k + m, 2m + 2} at least and poles of multiplicity 2 at least. From Lemmas 2–4 and Theorem A we know g(g (k) )n −a(0) has at least one zero, denoted as (k) ζ0 . By Hurwitz’s theorem, there exist a subsequence of ft (zt +ρt ζ)(ft (zt +ρt ζ))n −a(zt +ρt ζ), (k) still denoted as ft (zt +ρt ζ)(ft (zt +ρt ζ))n −a(zt +ρt ζ), and points ζt → ζ0 , such that, when t is nk − n+1
(k)
sufficiently large, ft (zt + ρt ζt )(ft (zt + ρt ζt ))n − a(zt + ρt ζt ) = 0. Thus |gt (ζt )| = |ρt − nk |ρt n+1 |M,
(k)
ft (zt +
ρt ζt )| ≥ so g(ζ0 ) = lim gt (ζt ) = ∞, which contradicts with g(ζ0 )(g (ζ0 )) = a(0). t→∞ Hence F is normal at z0 . Thus, F is normal in D as z0 is arbitrary. 2 Proofs of Theorems 2 and 4 Similar to the proof of Theorem 1 and notice Lemma 3, we can prove Theorem 2. Similar to the proof of Theorem 3, we can prove Theorem 4. n
References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25]
Yang L, Yang C C. Angular distribution of values of f f . Sci China Ser A, 1994, 37(3): 284–294 Yang C C, Yang L,Wang Y F. On the zeros of f (f (k) )n − a. Chin Sci Bull, 1994, 39(8): 617–621 Yang C C, Hu P C. On the value distribution of f f (k) . Kodai Math J, 1996, 19(2): 157–167 Zhang Z F, Song G D. On the zeros of f (f (k) )n − a(z). Chin Ann Math Ser A, 1998, 19(2): 275–282 Alotaibi A. On the zeros of af (f (k) )n − 1 for n ≥ 2, Comput Methods Funct Theory, 2004, 4(1): 227–235 Wang J P. On the value distribution of f f (k) . Kyungpook Math J, 2006, 46(2): 169–180 Hu P C, Meng D W. Normality criteria of meromorphic functions with multiple zeros. J Math Anal Appl, 2009, 357(2): 323–329 Lahiri I. A simple normality criterion leading to a counterexample to the converse of the Bloch principle. New Zealand J Math, 2005, 34(1): 61–65 Li B Q. On the Bloch heuristic principle, normality, and totally ramified values. Arch Math, 2006, 87(5): 417–426 Charak K S, Rieppo J. Two normality criteria and the converse of the Bloch principle. J Math Anal Appl, 2009, 353(1): 43–48 Bergweiler W, Eremenko A. On the singularitoes of the inverse to a meromorphic function of finite order. Rev Mat Iberoamericana, 1995, 11(2): 355–373 Zalcman L. Normal families: new perspectives. Bull Amer Math Soc, 1998, 35(3): 215–230 Fang M L, Zalcman L. A note on normality and shared values. J Aust Math Soc, 2004, 76(1): 141–150 Pang X C, Zalcman L. Normality and shared values. Ark Mat, 2000, 38(1): 171–182 Zhang Q C. Normal families of meromorphic functions concerning shared values. J Math Anal Appl, 2008, 338(1): 545–551 Han M H, Gu Y X. The normal famility of meromorphic functions. Acta Math Sci, 2008, 28B(4): 759–762 Xia J Y, Xu Y. Normal families of meromorphic functions with multiple values. J Math Anal Appl, 2009, 354(1): 387–393 Li Y T, Gu Y X. On normal families of meromorphic functions. J Math Anal Appl, 2009, 354(2): 421–425 Sun D C. Normality criteria for families of meromorphic algebroid functions. Acta Math Sci, 2010, 30B(1): 166–172 Jiang Y B, Gao Z S. Normal families of meromorphic functions sharing values or functions. J Inequal Appl, 2011, 2011: 72 Hayman W K. Meromorphic Functions. Oxford: Clarendon Press, 1964 Schiff J. Normal Families. Berlin: Springer-Verlag, 1993 Yang L. Value Distribution Theory. Berlin: Springer, 1993 Yang C C, Yi H X. Uniqueness Theory of Meromorphic Functions. Netherlands: Kluwer, 2003 Gu Y X, Pang X C, Fang M L. Theory of Normal Families and Its Application. Beijing: Science Press, 2007