# Normal Criterion Concerning Shared Values

Authors: Wei Chen; Yingying Zhang; Jiwen Zeng; Honggen Tian
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

We study normal criterion of meromorphic functions shared values, we obtain the following. LetF be a family of meromorphic functions in a domain D, such that function f∈F has zeros of multiplicity at least 2, there exists nonzero complex numbers bf,cf depending on f satisfying (i)bf/cf is a constant;  (ii)min{σ(0,bf),σ(0,cf),σ(bf,cf)≥m} for some m>0;  (iii)(1/cfk-1)(f′)k(z)+f(z)≠bfk/cfk-1 or (1/cfk-1)(f′)k(z)+f(z)=bfk/cfk-1⇒f(z)=bf, then F is normal. These results improve some earlier previous results.


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## Body

## 1. Introduction and Main Results

We useC to denote the open complex plane, C^(=C∪{∞}) to denote the extended complex plane and D to denote a domain in C. A family F of meromorphic functions defined in D⊂C is said to be normal, if for any sequence {fn}⊂F contains a subsequence which converges spherically, and locally, uniformly in D to a meromorphic function or ∞. Clearly F is said to be normal in D if and only if it is normal at every point of D see [1].LetD be a domain in C. For f meromorphic on C and a∈C, set
(1.1)Ef¯(a)=f-1({a})∩D={z∈D:f(z)=a}.Two meromorphic functionsf and g on D are said to share the value a if Ef¯(a)=Eg¯(a). Let a and b be complex numbers. If g(z)=b whenever f(z)=a, we write
(1.2)f(z)=a⇒g(z)=b.
If f(z)=a⇒g(z)=b and g(z)=b⇒f(z)=a, we write
(1.3)f(z)=a⟺g(z)=b.According to Bloch’s principle [2], every condition which reduces a meromorphic function in the plane C to a constant forces a family of meromorphic functions in a domain D normal. Although the principle is false in general (see [3]), many authors proved normality criterion for families of meromorphic functions by starting from Liouville-Picard type theorem (see [4]). It is also more interesting to find normality criteria from the point of view of shared values. In this area, Schwick [5] first proved an interesting result that a family of meromorphic functions in a domain is normal if in which every function shares three distinct finite complex numbers with its first derivative. And later, more results about normality criteria concerning shared values have emerged [6–9]. In recent years, this subject has attracted the attention of many researchers worldwide.In this paper, we useσ(x,y) to denote the spherical distance between x and y and the definition of the spherical distance can be found in [10].In2008, Fang and Zalcman [11] proved the following results.Theorem 1.1 (see [11]).
Letf be a transcendental function. Let a(≠0) and b be complex numbers, and let n(≥2),k be positive integers, then f+a(f′)n assumes every value b∈C infinitely often.Theorem 1.2 (see [11]).
LetF be a transcendental function. Let a(≠0) and b be complex numbers, and let n(≥2),k be positive integers. If for every f∈F has multiple zeros, and f+a(f′)n≠b, then F is normal in D.In 2009, Xu et al. [12] proved the following results.Theorem 1.3 (see [12]).
Letf be a transcendental function. Let a(≠0) and let b be complex numbers, and n,k be positive integers, which satisfy n≥k+1, then f+a(f(k))n assumes each value b∈C infinitely often.Theorem 1.4 (see [12]).
Letf be a transcendental function. Let a(≠0) and b be complex numbers, and let n,k be positive integers, which satisfy n≥k+1. If for every f∈F has only zeros of multiplicity at least k+1, and satisfies f+a(f(k))n≠b, then F is normal in D.In Theorems1.2 and 1.4, the constants are the same for each f∈F. Now we will prove the condition for the constants be the same can be relaxed to some extent.Theorem A.
LetF be a family of meromorphic functions in the unit disc Δ, and k be a positive integer and k≥3. For every f∈F, such that all zeros of f have multiplicity at least 2, there exist finite nonzero complex numbers bf,cf depending on f satisfying that(i)
bf/cf is a constant;(ii)
min{σ(0,bf),σ(0,cf),σ(bf,cf)≥m} for some m>0;(iii)
(1/cfk-1)(f′)k(z)+f(z)≠bfk/cfk-1.Then F is normal in Δ.Theorem B.
LetF be a family of meromorphic functions in the unit disc Δ, and k(≥3) be a positive integer. For every f∈F, such that all zeros of f have multiplicity at least 2, there exist finite nonzero complex numbers bf,cf depending on f satisfying that(i)
bf/cfis a constant;(ii)
min{σ(0,bf),σ(0,cf),σ(bf,cf)≥m} for some m>0;(iii)
(1/cfk-1)(f′)k(z)+f(z)=bfk/cfk-1⇒f(z)=bf.Then F is normal in Δ.


## 2. Some Lemmas

In order to prove our theorems, we require the following results.Lemma 2.1 (see [7]).
LetF be a family of meromorphic functions in a domain D, and 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
(2.1)gn(ζ)=fn(zn+ρnς)ρnα→g(ζ)
locally uniformly with respect to the spherical metric in C, where g is a nonconstant meromorphic function, all of whose zeros have multiplicity at least k, such that g#(ζ)≤g#(0)=kA+1. Morever,  g has order at most 2.
Here as usual,g#(ζ)=|g′(ζ)|/(1+|g(ζ)|2) is the spherical derivative.Lemma 2.2 (see [10]).
Letm be any positive number. Then, Möbius transformation g satisfies σ(g(a),g(b))≥m,σ(g(b),g(c))≥m,σ(g(c),g(a))≥m, for some constants a,b, and c also satisfy the uniform Lipschitz condition
(2.2)σ(g(z),g(w))≤kmσ(z,w),
where km is a constant depending on m.


## 3. Proof of Theorems

Proof of Theorem A.
LetM=bf/cf. We can find nonzero constants b and c satisfying M=b/c. For each f∈F, define a Möbius map gf by gf=cfz/c, thus gf-1=cz/cf.
Next we will showG={(gf-1∘f)∣f∈F} is normal in Δ. Suppose to the contrary, G is not normal in Δ. Then by Lemma 2.1. We can find gn∈G,zn∈Δ, and ρn→0+, such that Tn(ζ)=gn(zn+ρnζ)/ρn1/(k+1) converges locally uniformly with respect to the spherical metric to a nonconstant meromorphic function T(ζ) whose zeros of multiplicity at least 2 and spherical derivative is limited and T has order at most 2.
We now consider three cases.
Case1. If (1/ck-1)(T′)k(ζ)≡bk/ck-1, then T(ζ) is a polynomial with degree at most 1, a contradiction.
Case2. If there exists ζ0 such that (1/ck-1)(T′)k(ζ0)=bk/ck-1. Noting that ρnTn(ζ)+(1/ck-1)(Tn′)k(ζ)-(bk/ck-1)→(1/ck-1)(T′)k(ζ)-(bk/ck-1). By Hurwitz’s theorem, there exist a sequence of points ζn→ζ0 such that (for large enough n)
(3.1)0=ρnTn(ζn)+1ck-1(Tn′)k(ζn)-bkck-1=gn(zn+ρnζn)+1ck-1(gn′)k(zn+ζn)-bkck-1=ccffn(zn+ρnζn)+1ck-1ckcfk(fn′)k(zn+ζn)-bkck-1.
Hence fn(zn+ρnζn)+(1/cfk-1)(fn′)k(zn+ζn)=bfk/cfk-1. This contradicts with the suppose of Theorem A.
Case3. If (1/ck-1)(T′)k(ζ)≠bk/ck-1. Let c1,c2,…,ck be the solution of the equation wk=ck, then T′(ζ)≠ci(i=1,2,…,k). When T(ζ) is a rational function, then T′(ζ) is also a rational function. By Picard Theorem we can deduce that T′(ζ) is a constant (k≥3). Hence T(ζ) is a polynomial with degree at most 1. This contradicts with T(ζ) has zeros of multiplicity at least 2. When T(ζ) is a transcendental function, combining with the second main theorem, we have
(3.2)T(r,T′)≤N-(r,T′)+∑i=1k‍N-(r,1T′-ci)+s(r,T′)≤N-(r,T′)+s(r,T′)≤12N(r,T′)+s(r,T′)≤12T(r,T′)+s(r,T′).
Hence, T(r,T′)≤s(r,T′), a contradiction.
HenceG={(gf-1∘f)∣f∈F} is normal and equicontinuous in Δ. There given (ε/km>0), where km is the constant of Lemma 2.2, there exists δ>0 such that for the spherical distance σ(x,y)<δ,
(3.3)σ((gf-1∘f)(x),(gf-1)(y))<εkm
for each f∈F. Hence by Lemma 2.2. 							(3.4)σ(f(x),f(y))=σ((gf∘gf-1∘f)(x),(gf∘gf-1∘f)(y))=kmσ((gf-1∘f)(x),(gf-1∘f)(y))<ε.
Therefore, the family is equicontinuous in Δ. This completes the proof of Theorem A.Proof of Theorem B.
LetM=bf/cf. We can find nonzero constants b and c satisfying M=b/c. For each f∈F, define a Möbius map gf by gf=cfz/c, thus gf-1=cz/cf.
Next we will showG={(gf-1∘f)∣f∈F} is normal in Δ. Suppose to the contrary, G is not normal in Δ. Then by Lemma 2.1. We can find gn∈G,zn∈Δ, and ρn→0+, such that Tn(ζ)=gn(zn+ρnζ)/ρn1/(k+1) converges locally uniformly with respect to the spherical metric to a nonconstant meromorphic function T(ζ) whose spherical derivate is limited and T has order at most 2.
We will also consider three cases.
Case1. If (1/ck-1)(T′)k(ζ)≡bk/ck-1, then T(ζ) is a polynomial with degree at most 1, a contradiction.
Case2. If there exists ζ0 such that (1/ck-1)(T′)k(ζ0)=bk/ck-1. Noting that ρnTn(ζ)+(1/ck-1)(Tn′)k(ζ)-(bk/ck-1)→(1/ck-1)(T′)k(ζ)-(bk/ck-1). By Hurwitz’s theorem, there exist a sequence of points ζn→ζ0 such that (for large enough n)
(3.5)0=ρnTn(ζn)+1ck-1(Tn′)k(ζn)-bkck-1=gn(zn+ρnζn)+1ck-1(gn′)k(zn+ζn)-bkck-1=ccffn(zn+ρnζn)+1ck-1ckcfk(fn′)k(zn+ζn)-bkck-1.
Hence fn(zn+ρnζn)+(1/cfk-1)(fn′)k(zn+ζn)=bfk/cfk-1, then we have fn(zn+ρnζn)=bf by the condition (iii)(1/cfk-1)(f′)k(z)+f(z)=bfk/cfk-1⇒f(z)=bf.
Thus(3.6)T(ζ0)=limn→∞gn(zn+ρnζn)ρn=limn→∞cf(zn+ρnζn)cfρn=limn→∞bρn=∞.
This is a contradiction.
Case3. If (1/ck-1)(T′)k(ζ)≠bk/ck-1. Let c1,c2,…,ck be the solution of the equation wk=ck, then T′(ζ)≠ci(i=1,2,…,k). When T(ζ) is a rational function, then T′(ζ) is also a rational function. By Picard theorem we can deduce that T′(ζ) is a constant (k≥3). Hence T(ζ) is a polynomial with degree at most 1. This contradicts with T(ζ) has zeros of multiplicity at least 2. When T(ζ) is a transcendental function, combining with the second main theorem, we have
(3.7)T(r,T′)≤N-(r,T′)+∑i=1k‍N-(r,1T′-ci)+s(r,T′)≤N-(r,T′)+s(r,T′)≤12N(r,T′)+s(r,T′)≤12T(r,T′)+s(r,T′).
Hence, T(r,T′)≤s(r,T′), a contradiction.
HenceG={(gf-1∘f)∣f∈F} is normal and equicontinuous in Δ. There given (ε/km>0), where km is the constant of Lemma 2.2, there exists δ>0 such that for the spherical distance σ(x,y)<δ,
(3.8)σ((gf-1∘f)(x),(gf-1)(y))<εkm
for each f∈F. Hence by Lemma 2.2. 							(3.9)σ(f(x),f(y))=σ((gf∘gf-1∘f)(x),(gf∘gf-1∘f)(y))=kmσ((gf-1∘f)(x),(gf-1∘f)(y))<ε.
Therefore, the family is equicontinuous in Δ. This completes the proof of Theorem B.Remark 3.1.
Using the similar argument, if the condition (iii)f(z)=bf when (1/cfk-1)(f′)k(z)+f(z)=bfk/cfk-1 is replaced by (iii)|f(z)|≥|bf| when (1/cfk-1)(f′)k(z)+f(z)=bfk/cfk-1, then F is normal too.



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