Further results on normal families of meromorphic functions concerning shared values
aa r X i v : . [ m a t h . C V ] A ug Further results on normal families of meromorphic functionsconcerning shared values
Xiao-Bin Zhang a ∗ and Jun-Feng Xu b a College of Science, Civil Aviation University of China, Tianjin 300300, China b Department of Mathematics, Wuyi University, Jiangmen, Guangdong 529020, China
Abstract
In this paper, we prove two normality criteria for families of some functions concern-ing shared values, the results generalize those given by Hu and Meng. Some examplesare given to show the sharpness of our results.
Keywords and phrases:
Normal family; meromorphic function; shared value.
Let C denote the complex plane and f ( z ) be a non-constant meromorphic function in C .It is assumed that the reader is familiar with the standard notation used in the Nevanlinnavalue distribution theory such as the characteristic function T ( r, f ), the proximity function m ( r, f ), the counting function N ( r, f ) (see, e.g. [6, 16, 17]), and S ( r, f ) denotes any quantitythat satisfies the condition S ( r, f ) = o ( T ( r, f )) as r → ∞ outside of a possible exceptionalset of finite linear measure.Let f ( z ) and g ( z ) be two non-constant meromorphic functions, a be a finite complexnumber, if f − a and g − a have the same zeros (Ignoring multiplicities), then we say that f and g share a .Let F be a family of meromorphic functions defined in a domain D ⊂ C . F is said to benormal in D , in the sense of Montel, if for any sequence f n ∈ F , there exists a subsequence f n j such that f n j converges spherically locally uniformly in D , to a meromorphic functionor ∞ (see [6, 17]).According to Bloch’s principle, every condition which reduces a meromorphic function in C to a constant, makes a family of meromorphic functions in a domain D normal. Althoughthe principle is false in general, many authors proved normality criteria for families ofmeromorphic functions by starting from Picard type theorems. For instance, Theorem A. [5]
Let n ≥ be an integer, a , b ∈ C and a = 0 . If, for a meromorphicfunction f , f ′ + af n = b for all z ∈ C , then f must be a constant. ∗ Correspoding author: E-mail: [email protected](X.B. Zhang); [email protected](J.F. Xu) heorem B. [10, 11] Let n ≥ be an integer, a , b ∈ C , a = 0 and F be a family ofmeromorphic functions in a domain D . If f ′ + af n = b for all f ∈ F , then F is a normalfamily. In 2008, Zhang [18] improved theorem B by the idea of shared values, he got
Theorem C.
Let F be a family of meromorphic functions in D, n be a positive integer and a , b be two constants such that a = 0 , ∞ and b = ∞ . If n ≥ and for each pair of functions f and g ∈ F , f ′ − af n and g ′ − ag n share the value b , then F is normal in D . In 1998, Wang and Fang [13] proved
Theorem D.
Let k , n ≥ k + 1 be positive integers and f be a transcendental meromorphicfunction, then ( f n ) ( k ) assumes every finite nonzero value infinitely often. Using the idea of shared values, Li and Gu [9] obtained a corresponding normalitycriteria.
Theorem E.
Let F be a family of meromorphic functions defined in a domain D . Let k , n ≥ k + 2 be positive integers and a = 0 be a finite complex number. If ( f n ) ( k ) and ( g n ) ( k ) share a in D for every pair of functions f , g ∈ F , then F is normal in D . In 2004, Alotaibi [1] got
Theorem F.
Suppose that f is a transcendental meromorphic function in the plane. Let a be a small function of f , then af ( f ( k ) ) n − has infinitely many zeros. Using the idea of shared values, Hu and Meng [8] obtained a corresponding normalitycriteria.
Theorem G.
Take positive integers n and k with n , k ≥ and take a nonzero complexnumber a . Let F be a family of meromorphic functions in the plane domain D such thateach 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 , then F is normal in D . In 1996, Yang and Hu [15] got
Theorem H.
Take nonnegative integers n, n , · · · , n k with n ≥ , n + n + · · · + n k ≥ and define d = n + n + n + · · · + n k . Let f be a transcendental meromorphic func-tion with the deficiency δ (0 , f ) > / (3 d + 1) . Then for any nonzero value c , the function f n ( f ′ ) n · · · ( f ( k ) ) n k − c has infinitely many zeros. Moreover, if n ≥ , the deficient conditioncan be omitted. It’s natural to ask whether there exists normality criteria corresponding to Theorem H.We consider this problem and obtain
Theorem 1.1.
Let a = 0 be a constant, n ≥ , k ≥ , n k ≥ , n j ( j = 1 , , · · · , k − benonnegative integers. Let F be a family of meromorphic functions in the plane domain D such that each f ∈ F has only zeros of multiplicity at least k . For each pair ( f, g ) ∈ F , if f n ( f ′ ) n · · · ( f ( k ) ) n k and g n ( g ′ ) n · · · ( g ( k ) ) n k share a , then F is normal in D . xample 1.1. Let D = { z : | z | < } and F = { f m } where f m := e mz , and for every pairof functions f, g ∈ F , ( f n ) ( k ) and ( g n ) ( k ) share in D , it is easy to verify that F is notnormal at the point z = 0 . Example 1.2.
Let D = { z : | z | < } and F = { f m } where f m := mz + mk ( k +1)! . Then ( f k +1 m ) ( k ) = m k +1 ( k + 1)! z + 1 , and for every pair of functions f, g ∈ F , ( f k +1 ) ( k ) and ( g k +1 ) ( k ) share in D , it is easy to verify that F is not normal at the point z = 0 . Remark 1.2.
In Theorem 1.1, let k = n k = 1, then f n ( f ′ ) n · · · ( f ( k ) ) n k = ( f n +1 ) ′ n +1 and g n ( g ′ ) n · · · ( g ( k ) ) n k = ( g n +1 ) ′ n +1 , this case is a corollary of Theorem E. Examples 1.1 and 1.2given by Li and Gu show that the condition a = 0 in Theorem E is inevitable and n ≥ k + 2in Theorem E is sharp. The examples also show that the conditions in Theorem 1.1 aresharp, at least for the case k = n k = 1.If n = 1, we have Theorem 1.3.
Let a = 0 be a constant, k > , n k ≥ , n j ( j = 1 , , · · · , k − benonnegative integers such that n + · · · + n k − ≥ . Let F be a family of meromorphicfunctions in the plane domain D such that each f ∈ F has only zeros of multiplicity at least k . For each pair ( f, g ) ∈ F , if f ( f ′ ) n · · · ( f ( k ) ) n k and g ( g ′ ) n · · · ( g ( k ) ) n k share a , then F isnormal in D . Remark 1.4.
In Theorem 1.3, if n ≥
2, the theorem still holds for k = 2. Corollary 1.5.
Let a = 0 be a constant, n , k , n k be positive integers such that nk ≥ and n j ( j = 1 , , · · · , k − be nonnegative integers. Let F be a family of holomorphicfunctions in the plane domain D such that each f ∈ F has only zeros of multiplicity at least k . For each pair ( f, g ) ∈ F , if f n ( f ′ ) n · · · ( f ( k ) ) n k and g n ( g ′ ) n · · · ( g ( k ) ) n k share a , then F is normal in D . Remark 1.6.
Examples 1.2 shows that Corollary 1.5 fails if n = k = 1 and thus thecondition nk ≥ Lemma 2.1 ([12]) . Let F be a family of meromorphic functions on the unit disc ∆ , allof whose zeros have the multiplicity at least k , and suppose that there exists A ≥ suchthat | f ( k ) ( z ) | ≤ A wherever f ( z ) = 0 , f ∈ F . Then if F is not normal, there exist, for each ≤ α ≤ k :( a ) a number r, < r < , ( b ) points z n , | z n | < r, ( c ) functions f n ∈ F , and ( d ) positive numbers ρ n → such that ρ − αn f n ( z n + ρ n ξ ) = g n ( ξ ) → g ( ξ ) locally uniformly with respect to the sphericalmetric, where g ( ξ ) is a non-constant meromorphic function on C , all of whose zeros havemultiplicity at least k , such that g ♯ ( ξ ) ≤ g ♯ (0) = kA + 1 . In particular, if g is an entirefunction, it is of exponential type. Here, as usual, g ♯ ( z ) = | g ′ ( z ) | / (1+ | g ( z ) | ) is the sphericalderivative. emma 2.2 ([4]) . Let f be an entire function and M a positive integer. If f ♯ ( z ) ≤ M forall z ∈ C , then f has the order at most one. Lemma 2.3 ([19]) . Take nonnegative integers n, n , · · · , n k with n ≥ , n k ≥ and define d = n + n + n + · · · + n k . Let f be a transcendental meromorphic function whose zeros havemultiplicity at least k . Then for any nonzero value c , the function f n ( f ′ ) n · · · ( f ( k ) ) n k − c has infinitely many zeros, provided that n + n + · · · + n k − ≥ and k > if n = 1 .Specially, if f is transcendental entire, the function f n ( f ′ ) n · · · ( f ( k ) ) n k − c has infinitelymany zeros. Lemma 2.4.
Take nonnegative integers n , · · · , n k , k with n k ≥ , k ≥ , n + n + · · · + n k − ≥ and define d = 1 + n + n + · · · + n k . Let f be a non-constant rational func-tion whose zeros have multiplicity at least k . Then for any nonzero value c , the function f ( f ′ ) n · · · ( f ( k ) ) n k − c has at least two distinct zeros. Proof.
We shall divide our argument into two cases.
Case 1.
Suppose that f ( f ′ ) n · · · ( f ( k ) ) n k − c has exactly one zero. Case 1.1. If f is a non-constant polynomial, since the zeros of f have multiplicity at least k ,we know that f n ( f ′ ) n · · · ( f ( k ) ) n k is also a non-constant polynomial, so f ( f ′ ) n · · · ( f ( k ) ) n k − c has at least one zero, suppose that f ( f ′ ) n · · · ( f ( k ) ) n k = c + B ( z − z ) l , (2.1)where B is a nonzero constant and l > f ( f ′ ) n · · · ( f ( k ) ) n k has only simple zeros, which contradicts the assumption that the zeros of f have multiplicityat least k ≥ Case 1.2. If f is a non-constant rational function but not a polynomial. Set f ( z ) = A ( z − a ) m ( z − a ) m · · · ( z − a s ) m s ( z − b ) l ( z − b ) l · · · ( z − b t ) l t , (2.2)where A is a nonzero constant and m i ≥ k ( i = 1 , , · · · , s ) , l j ≥ j = 1 , , · · · , t ).Then f ( k ) ( z ) = A ( z − a ) m − k ( z − a ) m − k · · · ( z − a s ) m s − k g k ( z )( z − b ) l + k ( z − b ) l + k · · · ( z − b t ) l t + k . (2.3)For simplicity, we denote m + m + · · · + m s = M ≥ ks, (2.4) l + l + · · · + l t = N ≥ t. (2.5)It is easily obtained that deg( g k ) ≤ k ( s + t − , (2.6)4oreover, g k ( z ) = ( M − N )( M − N − · · · ( M − N − k +1) z k ( s + t − + c m z k ( s + t − − + · · · + c .Combining (2.6) and (2.7) yields f ( f ′ ) n · · · ( f ( k ) ) n k = A d ( z − a ) dm − P kj =1 jn j · · · ( z − a s ) dm s − P kj =1 jn j g ( z )( z − b ) dl + P kj =1 jn j · · · ( z − b t ) dl t + P kj =1 jn j = P ( z ) Q ( z ) , (2.7)where P ( z ), Q ( z ), g ( z ) are polynomials and g ( z ) = Q kj =1 g n j j ( z ) with deg( g ) ≤ P kj =1 jn j ( s + t − g ( z ) = ( M − N ) d − ( M − N − d − − n · · · ( M − N − k +1) n k z P kj =1 jn j ( s + t − + d m z P kj =1 jn j ( s + t − − + · · · + d .Then from (2.10) we obtain( f ( f ′ ) n · · · ( f ( k ) ) n k ) ′ = A d ( z − a ) dm − P kj =1 jn j − · · · ( z − a s ) dm s − P kj =1 jn j − h ( z )( z − b ) dl + P kj =1 jn j +1 · · · ( z − b t ) dl t + P kj =1 jn j +1 , (2.8)where h ( z ) is a polynomial with deg( h ) ≤ ( P kj =1 jn j + 1)( s + t − f ( f ′ ) n · · · ( f ( k ) ) n k − c has exactly one zero, we obtain from (2.11) that f ( f ′ ) n · · · ( f ( k ) ) n k = c + B ( z − z ) l ( z − b ) dl + P kj =1 jn j · · · ( z − b t ) dl t + P kj =1 jn j , (2.9)where B is a nonzero constant. Then( f ( f ′ ) n · · · ( f ( k ) ) n k ) ′ = B ( z − z ) l − H ( z )( z − b ) dl + P kj =1 jn j +1 · · · ( z − b t ) dl t + P kj =1 jn j +1 , (2.10)where H ( z ) is a polynomial of the form H ( z ) = B ( l − dN − k X j =1 jn j t ) z t + B t − z t − + · · · + b . (2.11) Case 1.2.1. If l = dN + P kj =1 jn j t .From (2.11) we get deg( p ) ≥ deg( Q ), namely dM − k X j =1 jn j s + deg( g ) ≥ dN + k X j =1 jn j t. (2.12)In view of deg( g ) ≤ P kj =1 jn j ( s + t − d ( M − N ) ≥ k X j =1 jn j , (2.13)which implies M > N .Combining (2.12), (2.14) and (2.15) yields dM − ( k X j =1 jn j + 1) s ≤ deg( H ) = t ≤ N < M, (2.14)5hich implies that k X j =1 n j M − k X j =1 jn j s < s, (2.15)this together with (2.8) implies that k − X j =1 ( k − j ) n j < . (2.16)which is a contradiction since n + n + · · · + n k − ≥ Case 1.2.2. If l = dN + P kj =1 jn j t .If M > N , with similar discussion as above, we get the same contradiction.If M ≤ N , combining (2.12) and (2.14) yields l − ≤ deg( h ) ≤ ( k X j =1 jn j + 1)( s + t − , (2.17)we deduce from (2.21) that dN ≤ k X j =1 jn j s + s + t − k X j =1 jn j < k X j =1 jn j s + s + N. (2.18)Note that M ≤ N , from (2.22) we get( d − M − k X j =1 jn j s < s. (2.19)Similar to the end of the proof in Case 1.2.1, we get a contradiction. Case 1 has been ruledout. Case 2.
Suppose that f ( f ′ ) n · · · ( f ( k ) ) n k − c has no zero.Similar to the proof of Case 1.1, we deduce that f is not a polynomial. If f is a non-constantrational function but not a polynomial. Similar to the proceeding of proof in Case 1.2.1,we get a contradiction.Hence f n ( f ′ ) n · · · ( f ( k ) ) n k − c has at least two distinct zeros.This proves Lemma 2.4.Using the similar proof of Lemma 2.4, we get Lemma 2.5.
Take nonnegative integers n , n , · · · , n k , k with n ≥ , n k ≥ , k ≥ . Let f be a non-constant rational function whose zeros have multiplicity at least k . Then for anynonzero value c , the function f n ( f ′ ) n · · · ( f ( k ) ) n k − c has at least two distinct zeros. Lemma 2.6.
Take nonnegative integers n , n , · · · , n k , k with n ≥ , n k ≥ , k ≥ . Let f be a non-constant polynomial whose zeros have multiplicity at least k . Then for any nonzerovalue c , the function f n ( f ′ ) n · · · ( f ( k ) ) n k − c has at least two distinct zeros. Proof of Theorem 1.3
Without loss of generality, we may assume D = ∆ = { z : | z | < } . For simplicity, we denote f j ( z j + ρ j ξ ) by f j and set γ M = 1 + n + · · · + n k , Γ M = k X j =1 jn j Suppose that F is not normal in D . By Lemma 2.1, for 0 ≤ α < k , there exist: r < z j → j → ∞ ), f j ∈ F and ρ j → + such that g j ( ξ ) = ρ − Γ M /γ M j f j ( z j + ρ j ξ ) → g ( ξ ) locallyuniformly with respect to the spherical metric, where g ( ξ ) is a non-constant meromorphicfunction on C , all of whose zeros have multiplicity at least k , such that g ♯ ( ξ ) ≤ g ♯ (0) = kA + 1.On every compact subset of C which contains no poles of g , we have uniformly f j ( f ′ j ) n · · · ( f ( k ) j ) n k − a = g j ( ξ )( g ′ j ( ξ )) n · · · ( g ( k ) j ( ξ )) n k − a → g ( g ′ ) n · · · ( g ( k ) ) n k − a. (3.1)If g n ( g ′ ) n · · · ( g ( k ) ) n k ≡ a , then g has no zeros and no poles, thus g is an entire function.By Lemma 2.2, g is of order at most 1. Moreover, g ( ξ ) = e c ξ + c , where c ( = 0) and c areconstants. Thereby, we get g ( ξ )( g ′ ( ξ )) n · · · ( g ( k ) ( ξ )) n k = c Γ M e γ M ( c ξ + c ) , (3.2)which contradicts the case g n ( g ′ ) n · · · ( g ( k ) ) n k ≡ a .Since g is a non-constant meromorphic function, by Lemmas 2.3 and 2.4, we deduce that g ( ξ )( g ′ ( ξ )) n · · · ( g ( k ) ( ξ )) n k − a has at least two distinct zeros.We claim that g ( ξ )( g ′ ( ξ )) n · · · ( g ( k ) ( ξ )) n k − a has just a unique zero.Let ξ and ξ ⋆ be two distinct zeros of g ( ξ )( g ′ ( ξ )) n · · · ( g ( k ) ( ξ )) n k − a . We choose a positivenumber δ small enough such that D T D = ∅ , and such that g ( g ′ ) n · · · ( g ( k ) ) n k − a has noother zeros in D S D except for ξ and ξ ⋆ , where D = { ξ ∈ C | | ξ − ξ | < δ } , D = { ξ ∈ C | | ξ − ξ ⋆ | < δ } . By (3.1) and Hurwitz’s theorem, for sufficiently large j there exist points ξ j ∈ D , ξ ⋆j ∈ D such that f j ( z j + ρ j ξ j )( f ′ j ( z j + ρ j ξ j )) n · · · ( f ( k ) j ( z j + ρ j ξ j )) n k − a = 0 ,f j ( z j + ρ j ξ ⋆j )( f ′ j ( z j + ρ j ξ ⋆j )) n · · · ( f ( k ) j ( z j + ρ j ξ ⋆j )) n k − a = 0 . By the assumption in Theorem 1.1, f ( f ′ ) n · · · ( f ( k )1 ) n k and f j ( f ′ j ) n · · · ( f ( k ) j ) n k share a foreach j , it follows that f ( z j + ρ j ξ j )( f ′ ( z j + ρ j ξ j )) n · · · ( f ( k )1 ( z j + ρ j ξ j )) n k − a = 0 , ( z j + ρ j ξ ⋆j )( f ′ ( z j + ρ j ξ ⋆j )) n · · · ( f ( k )1 ( z j + ρ j ξ ⋆j )) n k − a = 0 . Let j → ∞ , and note that z j + ρ j ξ j → z j + ρ j ξ ⋆j →
0, we get f (0)( f ′ (0)) n · · · ( f ( k )1 (0)) n k − a = 0 . Since the zeros of f ( f ′ ) n · · · ( f ( k )1 ) n k − a have no accumulation points, in fact, for sufficientlylarge j , we have z j + ρ j ξ j = 0 , z j + ρ j ξ ⋆j = 0 . Thus ξ j = − z j ρ j , ξ ⋆j = − z j ρ j . This contradicts the fact that ξ j ∈ D ( ξ , δ ), ξ ⋆j ∈ D ( ξ ⋆ , δ ) and D ( ξ , δ ) T D ( ξ ⋆ , δ ) = ∅ .So g ( ξ )( g ′ ( ξ )) n · · · ( g ( k ) ( ξ )) n k − a has just a unique zero, which contradicts the fact that g ( ξ )( g ′ ( ξ )) n · · · ( g ( k ) ( ξ )) n k − a has at least two distinct zeros.This completes the proof of Theorem 1.3.By Theorem H, Lemmas 2.3, 2.5 and 2.6, the proofs of Theorem 1.1 and Corollary 1.5can be carried out in the line of Theorem 1.3, we omit the process here. In Theorem 1.3, if n = n = · · · = n k − = 0, then f ( f ′ ) n · · · ( f ( k ) ) n k = f ( f ( k ) ) n k , g ( g ′ ) n · · · ( g ( k ) ) n k = g ( g ( k ) ) n k . This case is the same as the case in Theorem G. Hu andMeng proved that Theorem G holds for the case k ≥ , n k ≥
2. They also gave an example[8, Example 1.3] to show that Theorem G is not valid for the case k = 1. It’s natural toask whether Theorem 1.3 holds for the case n k = 1, n + · · · + n k − = 0. Actually, this isan open problem as follows. Problem 4.1.
Let F be a family of meromorphic functions in a domain D , let k be apositive integer and b be a finite nonzero value. If, for every f ∈ F , all zeros of f havemultiplicity at least k , and f ( z ) f ( k ) ( z ) = b , is F normal in D ? Xu and Cao [14] gave a partial answer to Problem 4.1.
Theorem 4.1.
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