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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 ∞ .

Let $a∈C∪{∞}$, 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  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 $n≥3$. He  conjectured that this remains valid for $n=1$ and $n=2$. Further, the case of $n=2$ was confirmed by Mues  in 1979. The case $n=1$ was considered and settled by Clunie .

In 1994, Yang and Yang  proposed a conjecture: If f is an entire function and $k≥2$, then $( f f ( k ) ) n −a(z)$ ($a(z)≢0$) has infinitely many zeros.

Zhang and Song  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.

In 2005, Wang  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 $n≥3$, then $f n f ( k ) −c(z)$ has infinitely many zeros.

In the case of $f f ( k )$, Yang and Yang  proposed a conjecture: If f is transcendental, then $f f ( k )$ assumes every finite nonzero complex number infinitely often. In 2006, Wang  proved that this conjecture holds when f has only zeros of multiplicity at least $k+1$ ($k≥2$).

In 2011, Meng and Hu  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 $D∈C$ such that each $f∈F$ 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  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 $n≥2d+2$ 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  got the following theorem.

Theorem E Let k be a positive integer, let $b≠0$ be a finite complex number, let P be a polynomial with either $degP≥3$ 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 $d≥0$ 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 $f∈F$ 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 $d≥0$ 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 $f∈F$ 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 )

Let k be a positive integer, let be a family of meromorphic functions in D such that each function $f∈F$ has only zeros with multiplicities at least k, and suppose that there exists $A≥1$ such that $| f ( k ) (z)|≤A$ whenever $f(z)=0$, $f∈F$. If is not normal at $z 0 ∈D$, then for each α, $0≤α≤k$, 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 )

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 )

Let $f 1 (z)$, $f 2 (z)$ be two meromorphic functions defined in $D={z:|z|, 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 )

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 $l≤d+1$, 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 $( f f ( k ) ) ( d ) ≡ a d +Al⋯(l−d+1) ( z − z 0 ) l − d$. 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 $g(z)=(m−n)(m−n−1)⋯(m−n−k+1) z k ( s + t − 1 ) +⋯+ c 1 z+ c 0$ is a polynomial, $c i$ ($i=0,1$) are constants and $deg(g(z))≤k(s+t−1)$. 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 $deg( g 1 (z))≤(k+d+1)(s+t−1)$.

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 $d≥l$.

Combining (2.16) and (2.18), we get $d+2n+kt=deg(g(z))+2m−ks≤k(s+t−1)+2m−ks$. That is, $2(m−n)≥k+d$, 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 $deg( g 2 (z))≤t(d+1)+l−d−1$.

By (2.17) and (2.19), we know $2m−(k+d+1)s≤deg( g 2 (z))≤t(d+1)+l−d−1$. Thus $2m−(k+d+1)s−t(d+1)≤l−d−1$. Since $2m−(k+d+1)s−t(d+1)≥$ $2m−(k+d+1) m k / 2 + d + 1 −(d+1) n k / 2 + d + 1 >0$, then $0, which contradicts with $d≥l$.

Subcase 1.2. When $d.

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 $g 3 (z)=(l−2n−kt)⋯(l−2n−kt−d) z t ( d + 1 ) +⋯+ d 1 z+ d 0$, 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 $d+2n+kt=deg(g(z))+2m−ks≤k(s+t−1)+2m−ks$. That is, $2(m−n)≥k+d$, 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 $m≤n$. Since $α i ≠ z 0$ ($i=1,2,…,s$), by (2.17) and (2.20), we have $l−d−1≤deg( g 1 (z))≤(k+d+1)(s+t−1)$, since $l=2n+kt+d$, thus $2n+kt+d−d−1≤(k+d+1)(s+t−1)$. 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 $l=deg(g(z))+2m−ks≤k(s+t−1)+2m−ks=2m+kt−k$. If $m>n$, by a similar discussion to Subcase 1.2.1, we get a contradiction. Thus $m≤n$.

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, $d≥1$, $h( z 0 )≠0$, 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

$ϕ j (ξ)= f j ( ρ j ξ ) ρ j d + k 2 = ( ρ j ξ ) d 2 F j ( z j + ρ j ( ξ − z j ρ j ) ) ρ j d 2 ρ j k 2 → ξ d 2 G(ξ−c)=H(ξ)$

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 $∀f∈F$, 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 .

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 $∀f∈F$, the multiplicity of whose zeros is at least $max{ k 2 +d+1,k+d}$, then $F(0)=0$. Thus, there exists $0 such that $F(z)$ is holomorphic in $△ r ={z:|z| 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 $q≥1$ 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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## Acknowledgements

The authors wish to thank the referees and editors for their very helpful comments and useful suggestions.

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Correspondence to FeiFei Hu.

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LQ and FH performed and drafted manuscript. All authors read and approved the final manuscript.

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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 