# Normal Families concerning Polynomials and Shared Values

Authors: Xin-Li Wang; Ning Cui
Journal: International Journal of Analysis
 (2013)

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## Abstract

We study the problem of normal families of meromorphic functions concerning polynomials and shared values. We prove that a familyℱ of meromorphic functions in a domain D is normal if, for each function f∈ℱ, Pfzfkz=a⇔fkz=b, where P is a polynomial with the origin as zero, k is a positive integer, and a≠0, b are two finite constants.


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## Body

## 1. Introduction and Main Results

Let D be a domain inℂ and let ℱ be a family of meromorphic functions defined in D. ℱ is said to be normal in D, in the sense of Montel, if, for any sequence {fn}∈ℱ, there exists a subsequence {fnj} such that {fnj} converges spherically locally uniformly in D to a meromorphic function or ∞ (see [1]).Letf and g be meromorphic functions in D, and let a and b be complex numbers. If g(z)=b whenever f(z)=a, we write
(1)f(z)=a⟹g(z)=b.
If f(z)=a⇒g(z)=b and g(z)=b⇒f(z)=a, we write
(2)f(z)=a⟺g(z)=b.
If f(z)=a⇔g(z)=a, we say that f and g share a on D.In 1992, Schwick [2] firstly found a connection between normality criteria and shared values as follows.Theorem A.
Letℱ be a family of meromorphic functions in D, and let a1,a2,a3 be distinct complex numbers. If, for each f∈ℱ, f and f′ share ai(i=1,2,3), then ℱ is normal in D.In recent years, more results about normality criteria concerning shared values have been found (see [3–5]). The next theorem was proved by Fang and Zalcman in [4], which was an important normality criteria concerning shared values.Theorem B.
Letℱ be a family of meromorphic functions in D; let k be a positive integer; let a≠0 and b be two finite values. If, for every f∈ℱ, all zeros of f have multiplicity at least k and f(z)f(k)(z)=a⇔f(k)(z)=b, then ℱ is normal in D.Based on the idea in [3], we will improve and extend Theorem B in the following results.Theorem 1.
Letℱ be a family of meromorphic functions in D; let k be a positive integer; let a≠0 and b be two finite values. Let P be a polynomial with the origin as zero. If, for every f∈ℱ, all zeros of f have multiplicity at least k and P(f(z))f(k)(z)=a⇔f(k)(z)=b, then ℱ is normal in D.Example 2.
LetD={z:|z|<1}, P(z)=z2, ℱ={fn(z):fn(z)=(1/n2)(enz+e-nz-2)}. Then, all zeros of fn(z) have multiplicity at least 2. For each fn(z)∈ℱ, we have fn′(z)=(1/n)(enz-e-nz), fn′′(z)=enz+e-nz. Thus, P(fn(z))fn′′(z)=(1/n4)(enz+e-nz-2)2(enz+e-nz). P(fn(z))fn′′(z)=0⇒fn′′(z)=2, but ℱ is not normal in D. This shows that the condition a≠0 is necessary.Theorem 3.
Letℱ be a family of meromorphic functions in D; let k be a positive integer; let a≠0 and b be two finite values. Let P be a polynomial with the origin as zero and degP≥2, which has at least one multiple zero. If, for every f∈ℱ, all zeros of f have multiplicity at least k and P(f(z))f(k)(z)=a⇒f(k)(z)=b, then ℱ is normal in D.


## 2. Some Lemmas

In order to prove our results, we need the following lemmas.Lemma 1 (see [6]).
Letℱ be a family of meromorphic functions on the unit disc Δ satisfying all zeros of functions in ℱ which have multiplicity ≥p and all poles of functions in ℱ≥q. Let α be a real number satisfying -p<α<q. Then, ℱ is not normal at a point z0∈Δ if and only if there exist(i)
pointszn∈Δ, zn→z0;(ii)
positive numbersρn, ρn→0;(iii)
functionsfn∈ℱ such that gn(ξ)=ρnαfn(zn+ρnξ)→g(ξ) spherically uniformly on each compact subset of ℂ, where g is a nonconstant meromorphic function satisfying the zeros of g which are of multiplicities ≥p and the poles of g which are of multiplicities ≥q. Moreover, the order of g is not greater than 2.Here,g#=|g′(ξ)|/(1+|g(ξ)|2) is the spherical derivative.Lemma 2 (see [7]).
Letℱ be a family of meromorphic functions on the complex plane ℂ. If the spherical derivative f# of f is bounded on ℂ, then the order of f is at most two. Especially, when f is an entire function on the complex plane ℂ, then the order of f is at most one.Lemma 3 (see [8]).
Letf be a transcendental meromorphic function on the complex plane ℂ, and let n be a positive integer. Then fnf′ assumes every nonzero finite complex value infinitely often.Lemma 4 (see [9]).
Letn≥2 and k be two positive integers, and let f be a transcendental meromorphic function. Then fnf(k) assumes every nonzero finite complex value infinitely often.Lemma 5 (see [5]).
Letb be a nonzero finite complex number, let n,k be two positive integers, and let f(z) be a nonconstant rational function; all of whose zeros have multiplicities at least k. For the polynomial case, if n≥2 or k≥2, then fnf(k)-b has at least two distinct zeros; for the nonpolynomial rational case, the conclusion holds if n≥2.


## 3. Proof of Theorems

We only prove Theorem1, and the proof of Theorem 3 is similar to the second case in the proof of Theorem 1.Next, we consider two cases.Case 1.
Suppose the origin is a simple zero ofP(z). We assume that P(z)=zn+an-1zn-1+⋯+a1z, where a1≠0. Without loss of generality, we may assume that D=Δ={z:|z|<1}. Since normality is a local property, it is enough to show that ℱ is normal at each point of D. Suppose, on the contrary, that ℱ is not normal at z0∈D. By Lemma 1 (with α=-k/2), there exist functions fj∈ℱ, points zj→z0, and positive numbers ρj→0 such that
(3)gj(ξ)=ρj-k/2fj(zj+ρjξ)⟶g(ξ)
converges spherically uniformly on compact subsets of ℂ, where g(ξ) is a nonconstant meromorphic function on ℂ and all of whose zeros have multiplicity at least k. Moreover, the order of g(ξ) is at most 2. Therefore, we have
(4)P(fj(zj+ρjξ))fj(k)(zj+ρjξ)-a=[fjn(zj+ρjξ)+an-1fjn-1(zj+ρjξ)+⋯+a1fj(zj+ρjξ)]fj(k)(zj+ρjξ)-a=[ρj(n-1)k/2gjn(ξ)+an-1ρj(n-2)k/2gjn-1(ξ)+⋯+a1gj(ξ)]gj(k)(ξ)-a⟶a1gg(k)-a
uniformly on compact subsets of ℂ disjoint from the poles of g.
We claim (i)a1gg(k)≠a; (ii) g(k)≠0.(i)
Suppose thata1g(ξ0)g(k)(ξ0)=a, then g(ξ0)≠∞. If a1gg(k)≡a, g is an entire function and g≠0. By Lemma 2, the order of g is at most 1. Assume g(ξ)=ecξ+d, where c≠0 and d are two constants. But a1gg(k)=a1cke2cξ+2d≢a, which is a contradiction. Therefore, we have a1gg(k)≢a. According to Hurwitz’s theorem, there exists a sequence ξj, ξj→ξ0 such that P(fj(zj+ρjξj))fj(k)(zj+ρjξj)=a for j sufficiently large. It follows that fj(k)(zj+ρjξj)=b, so that
(5)gj(k)(ξj)=ρjk/2fj(k)(zj+ρjξj)=ρjk/2b.Thus(6)g(k)(ξ0)=limj→∞gj(k)(ξj)=0,which contradictsa1g(ξ0)g(k)(ξ0)=a≠0. This proves (i).(ii)
Suppose thatg(k)(ξ0)=0, then g(ξ0)≠∞. If g(k)≡0, g is a polynomial whose degree is less than k-1, which contradicts that all zeros of g have multiplicity at least k. So that g(k)≢0. Since gj(k)(ξ)-ρjk/2b=ρjk/2(fj(k)(zj+ρjξ)-b)→g(k)(ξ), by Hurwitz’s theorem, there exists a sequence ξj,ξj→ξ0 such that fj(k)(zj+ρjξj)=b for j sufficiently large. It follows that P(fj(zj+ρjξj))fj(k)(zj+ρjξj)=a, so that
(7)limj→∞P(fj(zj+ρjξj))fj(k)(zj+ρjξj)=a1g(ξ0)g(k)(ξ0)=a≠0,which contradictsg(k)(ξ0)=0. This proves (ii).
Next, we assume thatg is a transcendental meromorphic function. When k=1, by Lemma 3, gg(k)-(a/a1) assumes every nonzero finite complex value infinitely often, a contradiction. When k≥2, by Nevanlinna's second fundamental theorem,
(8)T(r,gg(k))≤N¯(r,g)+N¯(r,1gg(k))+N¯(r,1gg(k)-(a/a1))≤N¯(r,g)+N¯(r,1g)+S(r,g),
where S(r,g)=o(T(r,g)).
By Nevanlinna’s first fundamental theorem, we have(9)T(r,gg(k))≥12[N(r,gg(k))+N(r,1gg(k))]+O(1)≥k+22N¯(r,g)+N¯(r,1g)+S(r,g).
Combining (8) with (9), we have
(10)N¯(r,g)=S(r,g).
According to logarithmic derivative theorem,
(11)m(r,1g2)=m(r,g(k)g1gg(k))≤m(r,g(k)g)+m(r,1gg(k))≤m(r,1gg(k))+S(r,g),
we have
(12)m(r,1g2)+m(r,1gg(k)-(a/a1))≤m(r,1gg(k))+m(r,1gg(k)-(a/a1))+S(r,g)≤m(r,1gg(k)+(1/(gg(k)-(a/a1))))+S(r,g)≤m(r,[((gg(k))′gg(k)+(gg(k))′gg(k)-(a/a1))1(gg(k))′])+S(r,g)≤m(r,1(gg(k))′)+S(r,g)≤T(r,(gg(k))′)-N(r,1(gg(k))′)+S(r,g)≤N¯(r,g)+T(r,(gg(k))′)-N(r,1(gg(k))′)+S(r,g).
It follows that
(13)2T(r,g)≤N¯(r,g)+N(r,1g2)+N(r,1gg(k)-(a/a1))-N(r,1(gg(k))′)+S(r,g)≤N¯(r,g)+(k+1)N¯(r,1g)+N¯(r,1gg(k)-(a/a1))+S(r,g).
Since gg(k)≠a/a1 and the zeros of g have multiplicity exactly k(≥2), we have by (10) and (13)
(14)2T(r,g)≤k+1kN(r,1g)+S(r,g)≤32N(r,1g)+S(r,g)≤32T(r,g)+S(r,g).
Thus T(r,g)=S(r,g), a contradiction. Hence g is a rational function.
Now, suppose thatg is a nonpolynomial rational function. Let
(15)g(ξ)=anξn+an-1ξn-1+⋯+a1ξ+a0+q(ξ)p(ξ),
where a0,a1,…,an are constants such that an≠0 and p and q(≢0) are coprime polynomials with degq<degp and degp≥1. If n≥k,
(16)g(k)(ξ)=ann(n-1)⋯(n-k+1)ξn-k+⋯+akk!+qk(ξ)pk(ξ),
where pk(ξ) and qk(ξ) are coprime polynomials and, by induction, degqk<degpk. Then, by (16), g(k)(ξ)=0 has solutions, a contradiction. Hence n<k. Thus, we have
(17)g(ξ)=alξl+al-1ξl-1+⋯+a1ξ+a0+q(ξ)p(ξ)(l≤k-1,al≠0),g(k)(ξ)=qk(ξ)pk(ξ).
Since g(k)≠0, qk(ξ) is a nonzero constant A. Since degP≥1, g must have a pole in the finite plane. It follows that g(k) has a finite pole of order at least k+1; thus, degpk≥k+1,
(18)a1g(ξ)g(k)(ξ)=a1(alξl+al-1ξl-1+⋯+a1ξ+a0+q(ξ)p(ξ))Apk(ξ)=Aa1(alξl+al-1ξl-1+⋯+a1ξ+a0)p(ξ)+q(ξ)p(ξ)pk(ξ).
Then, the equation a1g(ξ)g(k)(ξ)=a has solutions, which contradict a1g(ξ)g(k)(ξ)≠a.
Finally, suppose thatg is a polynomial. Since g(k)≠0 and the zeros of g are of multiplicity k, we have g(k)(ξ)=c1(ξ-c2)k, where c1≠0. Thus, a1g(ξ)g(k)(ξ)-a=0 has solutions, a contradiction.Case 2.
Suppose the origin is a multiple zero ofP(z). Without loss of generality, we assume that D=Δ={z:|z|<1}. Let P(z)=zn+an-1zn-1+⋯+aqzq, where q≥2 is a positive integer and an-1,…,aq(≠0) are constants. Suppose ℱ is not normal at z0∈D. By Lemma 1 (with α=-k/(q+1)), there exist functions fj∈ℱ, points zj→z0, and positive numbers ρj→0 such that
(19)gj(ξ)=ρj-k/(q+1)fj(zj+ρjξ)⟶g(ξ)
converges spherically uniformly on compact subsets of ℂ, where g(ξ) is a nonconstant meromorphic function on ℂ, and all of whose zeros have multiplicity at least k. Moreover, the order of g(ξ) is at most 2. We have
(20)P(fj(zj+ρjξ))fj(k)(zj+ρjξ)-a=[fjn(zj+ρjξ)+an-1fjn-1(zj+ρjξ)+⋯+aqfjq(zj+ρjξ)]fj(k)(zj+ρjξ)-a=[ρj(n-q)k/(q+1)gjn(ξ)+an-1ρj(n-q-1)k/(q+1)gjn-1(ξ)+⋯+aqgjq(ξ)]gj(k)(ξ)-a⟶aqgqg(k)-a
uniformly on compact subsets of ℂ disjoint from the poles of g. If aqgqg(k)≡a, g has no zeros and poles. That is to say, g is an entire function and g≠0. Since the spherical derivative of g is bounded, by Lemma 2, we have ord(g)≤1. Hence suppose g(ξ)=ecξ+d, where c(≠0) and d are constants. Thus, aqgqg(k)=aqcke(q+1)cξ+(q+1)d≢a, a contradiction. By Lemmas 4 and 5, aqgqg(k)-a has zeros. Suppose aqgq(ξ0)g(k)(ξ0)=a. Since aqgqg(k)≢a, by Hurwitz’s theorem, there exists a sequence ξj, ξj→ξ0 such that (for j sufficiently large)
(21)P(fj(zj+ρjξj))fj(k)(zj+ρjξj)-a=[fjn(zj+ρjξj)+an-1fjn-1(zj+ρjξj)+⋯+aqfjq(zj+ρjξj)]fj(k)(zj+ρjξj)-a=0.
It follows that fj(k)(zj+ρjξj)=b, so that
(22)gj(k)(ξj)=ρjqk/(q+1)fj(k)(zj+ρjξj)=ρjqk/(q+1)b.
Thus,
(23)g(k)(ξ0)=limj→∞gj(k)(ξj)=0,
which contradicts aqgq(ξ0)g(k)(ξ0)=a≠0. This completes the proof.



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