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Normal families of meromorphic functions sharing one function

Abstract

Suppose p(z) is a holomorphic function, the multiplicity of its zeros is at most d, P(z) is a nonconstant polynomial. Let be a family of meromorphic functions in a domain D, all of whose zeros and poles have multiplicity at least max{ k 2 +d+1,k+d}. If for each pair of functions f and g in , P(f) f ( k ) and P(g) g ( k ) share a holomorphic function p(z), then is normal in D. It generalizes and extends the results of Jiang, Gao and Wu, Xu.

MSC:30D35, 30D45.

1 Introduction and results

Let D be a domain in , let be a family of meromorphic functions in D. is said to be normal in D, in the sense of Montel, if for any sequence { f n }F contains a subsequence { f n j } such that f n j converges spherically locally uniformly in D to a meromorphic function or ∞ [13].

Let aC{}, let f and g be two nonconstant meromorphic functions in D. If f(z)a and g(z)a have the same zeros (ignoring multiplicity), we say f and g share the value a in D.

In 1959, Hayman [1] proved that if f is a transcendental meromorphic function, then f n f assumes every finite nonzero complex number infinitely often for any positive integer n3. He [4] conjectured that this remains valid for n=1 and n=2. Further, the case of n=2 was confirmed by Mues [5] in 1979. The case n=1 was considered and settled by Clunie [6].

In 1994, Yang and Yang [7] proposed a conjecture: If f is an entire function and k2, then ( f f ( k ) ) n a(z) (a(z)0) has infinitely many zeros.

Zhang and Song [8] proved the following theorem.

Theorem A Suppose that f is a transcendental meromorphic function, n, k are two positive integers, then when n2, ( f f ( k ) ) n a(z) has infinitely many zeros, where a(z)0 is a small function of f.

In 2005, Wang [9] proved the following theorem.

Theorem B Let f be a transcendental meromorphic function, let c(z)0 be a small function of f, and let n, k be two positive integers. If n3, then f n f ( k ) c(z) has infinitely many zeros.

In the case of f f ( k ) , Yang and Yang [7] proposed a conjecture: If f is transcendental, then f f ( k ) assumes every finite nonzero complex number infinitely often. In 2006, Wang [10] proved that this conjecture holds when f has only zeros of multiplicity at least k+1 (k2).

In 2011, Meng and Hu [11] obtained the following theorem.

Theorem C Take a positive integer k and a nonzero complex number a. Let be a family of meromorphic functions in a domain DC such that each fF has only zeros of multiplicity at least k+1. For each pair (f,g)F, if f f ( k ) and g g ( k ) share a, then is normal in D.

In 2011, Jiang and Gao [12] obtained the following theorem.

Theorem D Suppose that d (≥0) is an integer, p(z) is an analytic function in D, and the multiplicity of its all zeros is at most d. Let be a family of meromorphic functions in D, let n be a positive integer. If n2d+2 and for each pair of functions f and g in , f n f and g n g share p(z) in D, then is normal in D.

In 2012, Wu and Xu [13] got the following theorem.

Theorem E Let k be a positive integer, let b0 be a finite complex number, let P be a polynomial with either degP3 or degP=2 and P having only one distinct zero, and let be a family of meromorphic functions in D, all of whose zeros have multiplicity at least k. If for each pair of functions f and g in , P(f) f ( k ) and P(g) g ( k ) share b in D, then is normal in D.

It is natural to ask whether Theorem E can be improved by the idea of sharing a holomorphic function. In this paper, we study the problem and obtain the following theorems.

Theorem 1.1 Suppose that d0 is an integer, p(z)0 is a holomorphic function in D, and the multiplicity of its all zeros is at most d. Let be a family of meromorphic functions in D, the multiplicity of all zeros and poles of fF is at least max{ k 2 +d+1,k+d}. If for each pair of functions f and g in , f f ( k ) and g g ( k ) share p(z) in D, then is normal in D.

Remark 1.1 Theorem 1.1 still holds when p(z) is a nonzero finite constant.

Theorem 1.2 Suppose that d0 is an integer, p(z)0 is a holomorphic function in D, and the multiplicity of its all zeros is at most d. Let P be a nonconstant polynomial, be a family of meromorphic functions in D, the multiplicity of all zeros and poles of fF is at least max{ k 2 +d+1,k+d}. If for each pair of functions f and g in , P(f) f ( k ) and P(g) g ( k ) share p(z) in D, then is normal in D.

2 Some lemmas

Lemma 2.1 (see [14])

Let k be a positive integer, let be a family of meromorphic functions in D such that each function fF has only zeros with multiplicities at least k, and suppose that there exists A1 such that | f ( k ) (z)|A whenever f(z)=0, fF. If is not normal at z 0 D, then for each α, 0αk, there exists a sequence of complex numbers z n D, z n z 0 , a sequence of positive numbers ρ n 0, and a sequence of functions f n F such that

g n (ξ)= f n ( z n + ρ n ξ ) ρ n α g(ξ)

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

Lemma 2.2 (see [15])

Let f(z) be a meromorphic function and k be a positive integer. If f ( k ) 0, then

N ( r , 1 f ( k ) ) N ( r , 1 f ) +k N ¯ (r,f)+S(r,f).

Lemma 2.3 (see [1])

Let f 1 (z), f 2 (z) be two meromorphic functions defined in D={z:|z|<R}, then

N(r, f 1 f 2 )N ( r , 1 f 1 f 2 ) =N(r, f 1 )+N(r, f 2 )N ( r , 1 f 1 ) N ( r , 1 f 2 ) .

Lemma 2.4 (see [16])

Let f be a transcendental meromorphic function, let P f (z), Q f (z) be two differential polynomials of f. If f n P f = Q f holds and the degree of Q f is at most n, then m(r, P f )=S(r,f).

Lemma 2.5 Let d (≥0) be an integer, let k be a positive integer, and let p(z)= a d z d + a d 1 z d 1 ++ a 1 z+ a 0 be a polynomial, where a d 0, a d 1 ,, a 0 are constants. Suppose that f is a transcendental meromorphic function, all of whose zeros and poles have multiplicity at least 2, p(z) is a small function of f(z), then f f ( k ) (z)p(z) has infinitely many zeros.

Proof Let

ψ(z)=f f ( k ) p(z).
(2.1)

Suppose f f ( k ) p(z) has only finitely many zeros, then N(r, 1 ψ ( z ) )=S(r,f). By (2.1), then

( ψ p ) = f f ( k ) p + f f ( k + 1 ) p + ( 1 p ) f f ( k ) .
(2.2)

Let

ψ 1 = ψ p .

Since the multiplicity of zeros of f(z) is at least 2, we can get from (2.2) that

N ( r , 1 f ) N ( r , 1 ( ψ 1 ) ) +S(r,f).
(2.3)

By Lemma 2.2, we know that

N ( r , 1 ( ψ 1 ) ) N ( r , 1 ψ 1 ) + N ¯ (r,f)+S(r,f).
(2.4)

We can get from (2.2) that

f f ( k ) p ( ψ 1 ) ψ 1 f f ( k ) p f f ( k + 1 ) p ( 1 p ) f f ( k ) = ( ψ 1 ) ψ 1 ,

i.e.,

f ( f ( k ) p ( ψ 1 ) ψ 1 f f ( k ) f p f ( k + 1 ) p ( 1 p ) f ( k ) ) = ( ψ 1 ) ψ 1 .
(2.5)

Let

fH= ( ψ 1 ) ψ 1 ,
(2.6)

where H= f ( k ) p ( ψ 1 ) ψ 1 f f ( k ) f p f ( k + 1 ) p ( 1 p ) f ( k ) . By Lemma 2.4, we get m(r,H)=S(r,f).

From (2.5) and Lemma 2.3, we obtain that

m ( r , 1 f ) m ( r , H ) + m ( r , ψ 1 ( ψ 1 ) ) N ( r , ( ψ 1 ) ψ 1 ) N ( r , ψ 1 ( ψ 1 ) ) + m ( r , ( ψ 1 ) ψ 1 ) + S ( r , f ) N ( r , ( ψ 1 ) ) + N ( r , 1 ψ 1 ) N ( r , 1 ( ψ 1 ) ) N ( r , ψ 1 ) + S ( r , f ) N ¯ ( r , f ) + N ( r , 1 ψ ) N ( r , 1 ( ψ 1 ) ) + S ( r , f ) .
(2.7)

We can get from (2.6) that

f f ( k ) ( ( ψ 1 ) p ψ 1 f f p f ( k + 1 ) p f ( k ) ( 1 p ) ) = ( ψ 1 ) ψ 1 .
(2.8)

Let

f f ( k ) G= ( ψ 1 ) ψ 1 ,
(2.9)

where G= ( ψ 1 ) p ψ 1 f f p f ( k + 1 ) p f ( k ) ( 1 p ) . By Lemma 2.4, then m(r,G)=S(r,f).

By (2.9), we have that

m ( r , f ( k ) ) m ( r , ( ψ 1 ) ψ 1 ) + m ( r , 1 f ) + m ( r , 1 G ) m ( r , 1 f ) + N ( r , G ) N ( r , 1 G ) + S ( r , f ) .
(2.10)

Since ( ψ 1 ) ψ 1 has only simple poles, and by (2.9) we know that the poles of f are impossible G’s. Hence the poles of G are only possible from the zeros and poles of p(z) or the zeros of ψ 1 , f and f ( k ) .

Hence by (2.8) and (2.9), we obtain that

N ( r , G ) N ¯ ( r , 1 ψ 1 ) + N ¯ ( r , 1 f ) + N ¯ ( r , 1 f ( k ) ) + N ( r , 1 p ) N ¯ ( r , 1 ψ ) + N ¯ ( r , 1 f ) + N ¯ ( r , 1 f ( k ) ) + S ( r , f ) .
(2.11)

Since ( ψ 1 ) ψ 1 has only simple poles, so by (2.9) we know that

N ( r , 1 G ) N(r,f)+N ( r , f ( k ) ) N ¯ (r,f).
(2.12)

Combining (2.7) and (2.10)-(2.12), we have

m ( r , f ( k ) ) { N ¯ ( r , 1 ψ ) + N ¯ ( r , 1 f ) + N ¯ ( r , 1 f ( k ) ) } { N ( r , f ) + N ( r , f ( k ) ) N ¯ ( r , f ) } + { N ¯ ( r , f ) + N ( r , 1 ψ ) N ( r , 1 ( ψ 1 ) ) } + S ( r , f ) .

Hence

T ( r , f ( k ) ) N ¯ ( r , 1 ψ ) + N ¯ ( r , 1 f ) + N ¯ ( r , 1 f ( k ) ) N ( r , f ) + N ¯ ( r , f ) + N ¯ ( r , f ) + N ( r , 1 ψ ) N ( r , 1 ( ψ 1 ) ) + S ( r , f ) .

Since the multiplicity of the zeros and poles of f(z) is at least 2, by an elementary calculation and combing with Lemma 2.2, (2.3) and (2.4), the above inequality yields

T ( r , f ( k ) ) N ( r , 1 f ( k ) ) + 2 N ( r , 1 ψ ) + S ( r , f ) N ( r , 1 f ) + k N ¯ ( r , f ) + 2 N ( r , 1 ψ ) + S ( r , f ) ( k + 1 ) N ¯ ( r , f ) + 3 N ( r , 1 ψ ) + S ( r , f ) .
(2.13)

Since the multiplicity of the poles of f(z) is at least 2, we can get from (2.13) that

T ( r , f ( k ) ) ( 1 1 k + 2 ) N ( r , f ( k ) ) + 3 N ( r , 1 ψ ) + S ( r , f ) ( 1 1 k + 2 ) N ( r , f ( k ) ) + S ( r , f ) .

This implies T(r, f ( k ) )=S(r,f), then f ( k ) is a rational function, thus f is a rational function which contradicts with f is transcendental. Hence f f ( k ) (z)p(z) has infinitely many zeros. □

Remark 2.1 When p(z) is a nonzero finite constant or a small function of f(z), similarly we can get the same conclusion.

Lemma 2.6 Let d (≥0) be an integer, let k be a positive integer, and let p(z)= a d z d + a d 1 z d 1 ++ a 1 z+ a 0 be a polynomial, where a d 0, a d 1 ,, a 0 are constants. If f(z) is a nonconstant polynomial, all of whose zeros and poles have multiplicity at least k+d, then f f ( k ) (z)p(z) has at least two distinct zeros, and f f ( k ) (z)p(z)0.

Proof We discuss the following two cases.

Case 1. If f f ( k ) p(z)0, then f f ( k ) p(z)C, where C is a nonzero constant. So f f ( k ) p(z)+C. Since the multiplicity of all the zeros of f is at least k+d, thus deg(f f ( k ) )k+2d, which contradicts with deg(p(z))=d.

Case 2. If f f ( k ) p(z) has only one zero z 0 , we assume f f ( k ) p(z)A ( z z 0 ) l , where A is a nonzero constant, l is a positive integer.

We discuss the following two cases.

  1. (i)

    If ld+1, then f f ( k ) p(z)+A ( z z 0 ) l . Since deg(f f ( k ) )k+2d, the degree of the right of the equation is at most d+1, which is smaller than the degree of the left of the equation. We get a contradiction.

  2. (ii)

    If l>d+1, then f f ( k ) p(z)+A ( z z 0 ) l . So ( f f ( k ) ) ( d ) a d +Al(ld+1) ( z z 0 ) l d . Since a d 0, so ( f f ( k ) ) ( d ) has only simple zeros, which contradicts with the multiplicity of all the zeros of f is at least k+d.

By Case 1 and Case 2, f f ( k ) p(z) has at least two distinct zeros.

If f f ( k ) p(z)0, then similar to the proof of Case 1, we get a contradiction. Hence f f ( k ) p(z)0. □

Lemma 2.7 Let d (≥0) be an integer, let k be a positive integer, and let p(z)= a d z d + a d 1 z d 1 ++ a 1 z+ a 0 be a polynomial, where a d 0, a d 1 ,, a 0 are constants. If f(z) is a nonconstant rational function and not a polynomial, and the multiplicity of whose zeros and poles is at least k 2 +d+1, then f f ( k ) (z)p(z) has at least two distinct zeros, and f f ( k ) (z)p(z)0.

Proof Since f(z) is a nonconstant rational function and not a polynomial, then obviously f f ( k ) (z)p(z)0. Let

f(z)=B ( z α 1 ) m 1 ( z α 2 ) m 2 ( z α s ) m s ( z β 1 ) n 1 ( z β 2 ) n 2 ( z β t ) n t ,
(2.14)

where B is a nonzero constant. Since the multiplicity of the zeros and poles of f is at least k 2 +d+1, we have m i k 2 +d+1 (i=1,2,,s), n j k 2 +d+1 (j=1,2,,t). For simplicity, we denote

m 1 + m 2 ++ m s =m ( k 2 + d + 1 ) s, n 1 + n 2 ++ n t =n ( k 2 + d + 1 ) t.

By (2.19), we get

f ( k ) (z)=B ( z α 1 ) m 1 k ( z α 2 ) m 2 k ( z α s ) m s k g ( z ) ( z β 1 ) n 1 + k ( z β 2 ) n 2 + k ( z β t ) n t + k ,
(2.15)

where g(z)=(mn)(mn1)(mnk+1) z k ( s + t 1 ) ++ c 1 z+ c 0 is a polynomial, c i (i=0,1) are constants and deg(g(z))k(s+t1). Thus (2.14) together with (2.15) implies

f f ( k ) (z)=B ( z α 1 ) 2 m 1 k ( z α 2 ) 2 m 2 k ( z α s ) 2 m s k g ( z ) ( z β 1 ) 2 n 1 + k ( z β 2 ) 2 n 2 + k ( z β t ) 2 n t + k .
(2.16)

By (2.16), we obtain

( f f ( k ) ( z ) ) ( d + 1 ) =B ( z α 1 ) 2 m 1 k d 1 ( z α 2 ) 2 m 2 k d 1 ( z α s ) 2 m s k d 1 g 1 ( z ) ( z β 1 ) 2 n 1 + k + d + 1 ( z β 2 ) 2 n 2 + k + d + 1 ( z β t ) 2 n t + k + d + 1 ,
(2.17)

where deg( g 1 (z))(k+d+1)(s+t1).

Next, we discuss the following two cases.

Case 1. If f f ( k ) p(z) has only one zero z 0 , then let

f f ( k ) (z)p(z)=C ( z z 0 ) l ( z β 1 ) 2 n 1 + k ( z β 2 ) 2 n 2 + k ( z β t ) 2 n t + k .
(2.18)

Subcase 1.1. When dl.

Combining (2.16) and (2.18), we get d+2n+kt=deg(g(z))+2mksk(s+t1)+2mks. That is, 2(mn)k+d, and then m>n.

Differentiating both sides of (2.18), we have

( f f ( k ) ( z ) ) ( d + 1 ) p ( d + 1 ) ( z ) = C g 2 ( z ) ( z β 1 ) 2 n 1 + k + d + 1 ( z β 2 ) 2 n 2 + k + d + 1 ( z β t ) 2 n t + k + d + 1 ,
(2.19)

where deg( g 2 (z))t(d+1)+ld1.

By (2.17) and (2.19), we know 2m(k+d+1)sdeg( g 2 (z))t(d+1)+ld1. Thus 2m(k+d+1)st(d+1)ld1. Since 2m(k+d+1)st(d+1) 2m(k+d+1) m k / 2 + d + 1 (d+1) n k / 2 + d + 1 >0, then 0<ld1, which contradicts with dl.

Subcase 1.2. When d<l.

Differentiating both sides of (2.18), we have

( f f ( k ) ( z ) ) ( d + 1 ) p ( d + 1 ) ( z ) = D ( z z 0 ) l d 1 g 3 ( z ) ( z β 1 ) 2 n 1 + k + d + 1 ( z β 2 ) 2 n 2 + k + d + 1 ( z β t ) 2 n t + k + d + 1 ,
(2.20)

where g 3 (z)=(l2nkt)(l2nktd) z t ( d + 1 ) ++ d 1 z+ d 0 , where d i (i=0,1) are constants and deg( g 3 (z))t(d+1).

Differentiating both sides of (2.18) step by step for d times, we have z 0 is a zero of ( f f ( k ) ( z ) ) ( d ) p ( d ) (z), as p ( d ) (z)= a d 0 and the multiplicity of all the zeros of f(z) is at least k 2 +d+1, thus α i z 0 (i=1,2,,s). When p(z) is a constant, from (2.18) we can also get α i z 0 (i=1,2,,s).

Here, we discuss three subcases as follows.

Subcase 1.2.1. When l<2n+kt+d.

Combining (2.16) and (2.18), we get d+2n+kt=deg(g(z))+2mksk(s+t1)+2mks. That is, 2(mn)k+d, and then m>n.

Since α i z 0 (i=1,2,,s), by (2.17) and (2.20), we have t(d+1)deg( g 3 (z))2m(k+d+1)s. Thus 2m(k+d+1)s+t(d+1)(k+d+1) m k / 2 + d + 1 +(d+1) n k / 2 + d + 1 <2m, which is impossible.

Subcase 1.2.2. When l=2n+kt+d.

If m>n, by a similar discussion to Subcase 1.2.1, we can get a contradiction. Thus mn. Since α i z 0 (i=1,2,,s), by (2.17) and (2.20), we have ld1deg( g 1 (z))(k+d+1)(s+t1), since l=2n+kt+d, thus 2n+kt+dd1(k+d+1)(s+t1). Then 2n(k+d+1)s+(d+1)t(k+d)<(k+d+1) m k / 2 + d + 1 +(d+1) n k / 2 + d + 1 2n, which is impossible.

Subcase 1.2.3. When l>2n+kt+d.

By (2.16) and (2.18), we get l=deg(g(z))+2mksk(s+t1)+2mks=2m+ktk. If m>n, by a similar discussion to Subcase 1.2.1, we get a contradiction. Thus mn.

Case 2. If f f ( k ) p(z) has no zero. Then l=0 in (2.18), by a similar discussion to Subcase 1.1, we get a contradiction.

By Case 1 and Case 2, we get f f ( k ) p(z) has at least two distinct zeros. □

3 Proof of Theorem 1.1

For any point z 0 in D, either p( z 0 )=0 or p( z 0 )0.

Case 1. When p( z 0 )=0. We may assume z 0 =0. Then p(z)= a d z d + a d + 1 z d + 1 += z d h(z), where a d (0), a d + 1 , are constants, d1, h( z 0 )0, without loss of generality, let h( z 0 )= a d , where h(z) is a holomorphic function.

Let F 1 ={ F j | F j = f j z d / 2 , f j F}. If F 1 is not normal at 0, then by Lemma 2.1, there exists a sequence of complex numbers z j 0, a sequence of positive numbers ρ j 0 and a sequence of functions F j F 1 such that G j (ξ)= ρ j k 2 F j ( z j + ρ j ξ)G(ξ) spherically locally uniformly in , where G(ξ) is a nonconstant meromorphic function in , and the multiplicity of the zeros and poles of G(ξ) is at least max{ k 2 +d+1,k+d}. Here, we discuss two cases as follows.

Case 1.1. There exists a subsequence of z j ρ j , we may denote it as z j ρ j such that z j ρ j c, c is a finite complex number. Then

ϕ j (ξ)= f j ( ρ j ξ ) ρ j d + k 2 = ( ρ j ξ ) d 2 F j ( z j + ρ j ( ξ z j ρ j ) ) ρ j d 2 ρ j k 2 ξ d 2 G(ξc)=H(ξ)

spherically locally uniformly in , so

ϕ j (ξ) ϕ j ( k ) (ξ) p ( ρ j ξ ) ρ j d = f j ( ρ j ξ ) f j ( k ) ( ρ j ξ ) p ( ρ j ξ ) ρ j d H(ξ) H ( k ) (ξ) a d ξ d

spherically locally uniformly in .

Since fF, the multiplicity of whose zeros and poles is at least max{ k 2 +d+1,k+d}, then the multiplicity of all zeros and poles of H is at least max{ k 2 +d+1,k+d}, by Lemmas 2.5-2.7, we get H(ξ) H ( k ) (ξ) a d ξ d 0, and H(ξ) H ( k ) (ξ) a d ξ d has at least two distinct zeros.

Suppose ξ 0 , ξ 0 are two distinct zeros of H(ξ) H ( k ) (ξ) a d ξ d . We may choose a proper σ>0 such that D( ξ 0 ,σ)D( ξ 0 ,σ)=, where D( ξ 0 ,σ)={ξ||ξ ξ 0 |<σ}, D( ξ 0 ,σ)={ξ||ξ ξ 0 |<σ}.

By Hurwitz’s theorem, there exists a subsequence of f j ( ρ j ξ) f j ( k ) ( ρ j ξ)p( ρ j ξ), we may still denote it as f j ( ρ j ξ) f j ( k ) ( ρ j ξ)p( ρ j ξ), then there exist points ξ j D( ξ 0 ,σ) and points ξ j D( ξ 0 ,σ) such that for sufficiently large j, f j ( ρ j ξ j ) f j ( k ) ( ρ j ξ j )p( ρ j ξ j )=0, f j ( ρ j ξ j ) f j ( k ) ( ρ j ξ j )p( ρ j ξ j )=0.

Since f j f j ( k ) and g j g j ( k ) share p(z) in D, it follows that for any positive integer m, f m ( ρ j ξ j ) f m ( k ) ( ρ j ξ j )p( ρ j ξ j )=0, f m ( ρ j ξ j ) f m ( k ) ( ρ j ξ j )p( ρ j ξ j )=0.

Fix m, let j and note ρ j ξ j 0, ρ j ξ j 0, we obtain f m (0) f m ( k ) (0)p(0)=0.

Since the zeros of f m (0) f m ( k ) (0)p(0) have no accumulation points, in fact when j is large enough, we have ρ j ξ j = ρ j ξ j =0. Thus, when j is large enough, ξ 0 = ξ 0 =0, which contradicts with D( ξ 0 ,σ)D( ξ 0 ,σ)=. Thus, F 1 is normal at 0.

Case 1.2. There exists a subsequence of z j ρ j , we may denote it as z j ρ j such that z j ρ j . Then

f j ( z j + ρ j ξ ) f j ( k ) ( z j + ρ j ξ ) = ( z j + ρ j ξ ) d 2 F j ( z j + ρ j ξ ) [ ( z j + ρ j ξ ) d 2 ( F j ( z j + ρ j ξ ) ) ( k ) + i = 1 k c i ( z j + ρ j ξ ) d 2 i ( F j ( z j + ρ j ξ ) ) ( k i ) ] = ( z j + ρ j ξ ) d G j ( ξ ) G j ( k ) ( ξ ) + i = 1 k c i ( z j + ρ j ξ ) d i ρ j i G j ( ξ ) G j ( k i ) ( ξ ) ,

where c i = d 2 ( d 2 1)( d 2 i+1) C d / 2 i when d 2 i, and c i =0 when d 2 <i.

Thus, we have

a d f j ( z j + ρ j ξ ) f j ( k ) ( z j + ρ j ξ ) p ( z j + ρ j ξ ) a d = ( G j ( ξ ) G j ( k ) ( ξ ) + i = 1 k c i G j ( ξ ) G j ( k i ) ( ξ ) ( z j ρ j + ξ ) i ) a d h ( z j + ρ j ξ ) a d G ( ξ ) G ( k ) ( ξ ) a d ,

spherically locally uniformly in C{ξ|G(ξ)=}.

Since the multiplicity of all zeros and poles of G is at least max{ k 2 +d+1,k+d} and by Lemmas 2.5-2.7, we have G(ξ) G ( k ) (ξ) a d 0, and G(ξ) G ( k ) (ξ) a d has at least two distinct zeros.

Suppose ξ 1 , ξ 1 are two distinct zeros of G(ξ) G ( k ) (ξ) a d . We may choose a proper δ>0 such that D( ξ 1 ,δ)D( ξ 1 ,δ)=, where D( ξ 1 ,δ)={ξ||ξ ξ 1 |<σ}, D( ξ 1 ,δ)={ξ||ξ ξ 1 |<δ}.

By Hurwitz’s theorem, there exists a subsequence of a d f j ( z j + ρ j ξ) f j ( k ) ( z j + ρ j ξ) a d p( z j + ρ j ξ), we may still denote it as a d f j ( z j + ρ j ξ) f j ( k ) ( z j + ρ j ξ) a d p( z j + ρ j ξ), then there exist points ξ j D( ξ 1 ,δ) and points ξ j D( ξ 1 ,δ) such that for sufficiently large j, a d f j ( z j + ρ j ξ) f j ( k ) ( z j + ρ j ξ) a d p( z j + ρ j ξ)=0, a d f j ( z j + ρ j ξ) f j ( k ) ( z j + ρ j ξ) a d p( z j + ρ j ξ)=0.

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

By Case 1.1 and Case 1.2, we know F 1 is normal at 0. Hence there exists ρ ={z:|z|<ρ} and a subsequence of F j k of F j such that F j k converges spherically locally uniformly to a meromorphic function F(z) or ∞ (k) in ρ .

Here, we discuss the following two cases.

Case i. When k is large enough, f j k 0. Then F(0)=. Thus, for constant R>0, σ(0,ρ), we have |F(z)|>R when z ρ . Thus, for sufficiently large k, | F j k (z)|> R 2 , 1 f j k is a holomorphic function in ρ . Hence when |z|= σ 2 ,

| 1 f j k | = | 1 F j k z d / 2 | 2 d / 2 + 1 R σ d / 2 .

By the maximum principle and Montel’s theorem, is normal at z=0.

Case ii. There exists a subsequence of f j k , we may still denote it as f j k , such that f j k (0)=0. Since fF, the multiplicity of whose zeros is at least max{ k 2 +d+1,k+d}, then F(0)=0. Thus, there exists 0<r<ρ such that F(z) is holomorphic in r ={z:|z|<r} and has a unique zero z=0 in r . Then F j k converges spherically locally uniformly to a holomorphic function F(z) in r , f j k converges spherically locally uniformly to a holomorphic function z d 2 F(z) in r . Hence is normal at z=0.

By Case i and Case ii, we obtain is normal at z=0.

Case 2. When p( z 0 )0.

Suppose that is not normal at z 0 . Then by Lemma 2.1, there exists a sequence of complex numbers z t z 0 , a sequence of positive numbers ρ t 0 and a sequence of functions f t F such that g t (ξ)= ρ t k 2 f t ( z t + ρ t ξ)g(ξ) spherically locally uniformly in , where g(ξ) is a nonconstant meromorphic function in , and the multiplicity of the zeros and poles of g(ξ) is at least max{ k 2 +d+1,k+d}.

Hence by Lemmas 2.5-2.7, we have g(ξ) g ( k ) (ξ)p( z 0 )0, and g(ξ) g ( k ) (ξ)p( z 0 ) has at least two distinct zeros. Similar to the proof of Case 1.1, we get a contradiction. Thus, is normal at z 0 .

Hence, is normal in D as z 0 is arbitrary. The proof is complete.

4 Proof of Theorem 1.2

Because P(z) has at least one zero, we may assume, with no loss of generality, that P(z)= z n + a n 1 z n 1 ++ a q z q , where q1 is a positive integer and a q 0. Suppose that is not normal in D. Then similar to the proof of Theorem 1.1, we can get a contradiction. Hence is normal in D. The proof is complete.

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Qiu, L., Hu, F. Normal families of meromorphic functions sharing one function. J Inequal Appl 2013, 288 (2013). https://doi.org/10.1186/1029-242X-2013-288

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