# Normal Families and Growth of Meromorphic Functions with TheirKth Derivatives

Authors: Jianming Qi; Fanning Meng; Wenjun Yuan
Journal: Journal of Function Spaces
 (2018)

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## Abstract

Relying on the normal family theory, we mainly study uniqueness problems of meromorphic functions and theirkth derivatives and estimate sharply the growth order of their meromorphic functions. Our theorems improve some previous results.


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## Body

## 1. Introduction and Main Results

In this paper, a meromorphic function means a function that is meromorphic in the whole complex planeC. We will use the fundamental results and the standard notation of the Nevanlinna theory of meromorphic functions such as T(r,f), m(r,f), and N(r,f) as explained in [1].LetR be a rational function (or a polynomial) that belongs asymptotically to krα as r→∞, where k≠0,α are constants. The degree of R at infinity is defined as deg⁡R≔deg∞R≔max{0,α}.Suppose thatf(z) and g(z) are two nonconstant meromorphic functions. We define a(z) as a meromorphic function or a finite complex number. In this paper, we define g(z)-a(z)=0 whenever f(z)-a(z)=0; then we write f(z)=a(z)⇒g(z)=a(z). If f(z)=a(z)⇒g(z)=a(z) and g(z)=a(z)⇒f(z)=a(z), then we write f(z)=a(z)⇔g(z)=a(z) and say that f(z) and g(z) share a(z) IM (ignoring multiplicity). If f-a and g-a have the same zeros with the same multiplicities, we write f(z)=a(z)⇌g(z)=a(z) and say that f and g share a CM (counting multiplicity).The subject on sharing values between entire functions and their derivatives was first studied by Rubel and Yang (cf. [2]). In 1977, they proved a result that if a nonconstant entire function f and f′ share two distinct finite numbers a,b CM, then f≡f′. Since then, sharing value problems have been studied by many authors and a number of profound results have been obtained (cf. [1, 3, 4]). To state our main results, we also need the following concepts.Definition 1.
The orderρ(f) and the hyperorder ρ2(f) of a meromorphic function f are both defined as follows: (1)ρf=lim⁡supr→∞⁡log⁡Tr,flog⁡r,ρ2f=lim⁡supr→∞⁡log⁡log⁡Tr,flog⁡r.
On estimating the order or hyperorder of meromorphic functions, there has been a lot of work on the growth order of meromorphic function to certain types of complex differential equations and complex difference equations (or complex functional equations) (cf. [3, 5–8]).
In 2011, Lü and Yi ([9]) investigated the estimating order of meromorphic functions with their derivatives about sharing values and obtained the following results.Theorem A.
Letf be a nonconstant meromorphic function with finitely many poles, and let α=Q1eQ and β=Q2eQ, where Q1,Q2(≢Q1) and Q are three polynomials. Let k be a positive integer. If all the zeros of f-α have multiplicity at least k and (2)fz=αz⟹fkz=αz,fz=βz⟺fkz=βz,then ρ(f)<+∞.Theorem B.
LetQ1(≢0) and Q2 be two distinct polynomials, let k be a positive integer, and let f be a transcendental entire function with all the zeros of f-Q2 having multiplicity at least k. If f(z)=Q1(z)⇌f(k)(z)=Q1(z) and f(z)=Q2(z)⇒f(k)(z)=Q2(z), then one of the following cases must occur: (1)
f=f(k).(2)
f(z)=Q2(z)+Aeλz and (μ-1)Q1=μQ2-Q2(k), where A and λk=μ≠1 are two nonzero constants.In this paper, we continue to investigate the growth order and hyperorder of meromorphic functions with their derivatives about sharing functions and improve Theorems A and B. Now, we state our results as follows.Theorem 2.
Letf be a nonconstant meromorphic function with finitely many poles, and let α=R1eγ and β=R2eγ, where R1,R2(≢R1) are two rational functions and γ is an entire function. Let k be a positive integer. If all the zeros of f-α have multiplicity at least k and (3)fz=αz⟹fkz=αz,fz=βz⟺fkz=βz,then ρ2(f)≤ρ2(α)=ρ(γ).Remark 3.
The following examples show that the conclusionρ2(f)≤ρ(γ) is sharp.Example 4.
Letf(z)=Aez, where A is a nonzero constant. Let α(z)=ee-z+z and β(z)=zee-z+z. Noting that f=f(k), one has (4)fz=αz⟹fkz=αz,fz=βz⟺fkz=βz.Thus, it satisfies the assumptions of Theorem 2 and ρ2(f)=0<ρ2(α)=1.Example 5.
Letf(z)=zez+Ae3z, α(z)=zez, and β(z)=(z-1/2)ez, where A is a nonzero constant. Differentiating f yields f′z=(z+1)ez+3Ae3z, and then f(z)=α(z)⇒f′z=α(z) and f(z)=β(z)⇔f′z=β(z). Thus, ρ2(f)≤ρ2(α).Theorem 6.
Letf be a nonconstant meromorphic function with finitely many poles, and let α=Q1eQ and β=Q2eQ, where Q1,Q2(≢Q1), and Q are three polynomials with deg⁡Q1=q1, deg⁡Q2=q2, and deg⁡Q=q, where q1,q2,q are three nonnegative integers. Let k be a positive integer. If all the zeros of f-α have multiplicity at least k and (5)fz=αz⟹fkz=αz,fz=βz⟺fkz=βz,then ρ(f)≤maxq,q-q2-q1/k.Remark 7.
The next two examples can illustrate that the conclusion of Theorem6 is meaningful.Example 8.
Letf(z)=z2ez+1+Ae2z, α(z)=z2ez+1, and β(z)=(z2-4/3z-2/3)ez+1, where A is a nonzero constant. Differentiating f twice yields f′(z)=(z2+2z)ez+1+2Ae2z and f′′(z)=(z2+4z+2)ez+1+4Ae2z, and then f(z)=α(z)⇒f′′(z)=α(z) and f(z)=β(z)⇔f′′z=β(z). Thus, it satisfies the assumption of Theorem 6. Noting that k=2, λ=2, and μ=4, we obtain that ρ(f)=1≤max⁡{q=1,q-q2-q1/k=1-2-2/2=1} holds.Example 9.
Letf(z)=z2ez2+1+Ae2z, α(z)=z2ez2+1, and β(z)=(-4z4-8z2-2)ez2+1, where A is a nonzero constant. Differentiating f twice yields f′z=(2z3+2z)ez2+1+2Ae2z and f′′z=(4z4+10z2+2)ez2+1+2Ae2z, and then f(z)=α(z)⇒f′′(z)=α(z) and f(z)=β(z)⇔f′′z=β(z). Thus, it satisfies the assumption of Theorem 6. We have k=2,λ=2,μ=2. We obtain that ρ(f)=2≤max{q=2,q-q2-q1/k=2-4-2/2=1} holds.Theorem 10.
Letf be a transcendental meromorphic function with finitely many poles, and let α=Q1eQ and β=Q2eQ, where Q1,Q2(≢Q1), and Q are three polynomials. Let k be a positive integer. If all the zeros of f-α have multiplicity at least k and ρ(β)<ρ(f) and (6)fz=αz⟹fkz=αz,fz=βz⇌fkz=βz,then the conclusions of Theorem B still hold, and Q(z) must be a constant.The next example will show iff is rational; then Theorem 10 fails.Example 11.
Letf(z)=z3, Q2(z)=(3/2)z3-(3/2)z2, and Q1=Q=0. Note that (7)f′-Q2f-Q2=3,f=0⟹f′=0.However, it does not satisfy any case of Theorem B.Remark 12.
The conditionρ(β)<ρ(f) plays an important pole in the proof of Theorem 10. Unfortunately, we do not know whether the condition could be weakened.


## 2. Some Lemmas

In order to prove our theorems, we need the following lemmas.A familyF of meromorphic functions defined on a domain Ω⊂C^ (here C^=C∪{∞}) is said to be normal in Ω if every sequence of elements of F contains a subsequence that converges locally uniformly in Ω with respect to the spherical metric to a meromorphic function or ∞ (cf. [10]).Lemma 13 (see [3, 11]).
Let{fn} be a family of meromorphic (analytic) functions on the unit disc Δ. If an→a, |a|<1, and f♯(an)→∞, then there exist(a)
a subsequence offn (which we still write as {fn});(b)
pointszn→z0, |z0|<1;(c)
positive numbersρn→0,
such thatfn(zn+ρnξ)=gn(ξ)→g(ξ) locally uniformly, where g is a nonconstant meromorphic (entire) function on C, such that (8)ρn≤Mfn♯an,where M is a constant that is independent of n.
Here, as usual,f♯(ξ)=|f′(ξ)|/1+|f(ξ)|2 is the spherical derivative.The next lemma is an extending result obtained by Lü and Qi in [12].Lemma 14.
Letf be a meromorphic function of hyperorder ρ2(f)>0. Then, for any ϵ>0, there exists a sequence zn→∞ such that (9)f♯zn>eznd-ϵfor large enough n, where d=ρ2(f) if ρ2(f)<∞ or d is an arbitrary positive number.Lemma 15 (cf. [5]).
Letf(z) be an entire function; ρ(f)>1. Then, for each 0<μ<ρ(f)-1, there exist points an→∞(n→∞), such that (10)limn→∞f♯ananμ=+∞.Lemma 16 (cf. [1]).
Suppose thatf(z) and g(z) are two nonconstant meromorphic functions in the complex plane with ρ(f) and ρ(g) as their orders, respectively. Then (11)ρfg≤max⁡ρf,ρg,ρf+g≤max⁡ρf,ρg.Lemma 17 (cf. [13]).
Letf(z) and a(z) be meromorphic functions of finite order such that both of f(z) and a(z) have only finite many poles, f(z) and a(z) have no common poles, and the order of a(z) is less than the order of f(z). If f(z) and f(k)(z) share a(z) CM, then f(k)(z)-a(z)=c(f(z)-a(z)) for some nonzero constant c.


## 3. The Proof of Theorem2

By the method in [3, 12, 14–16], we will present our proof as follows.Note thatα=R1eγ; then ρ2(α)=ρ(γ). It suffices to show ρ2(f)≤ρ(γ).By reduction to absurdity, we assume thatρ2(f)>ρ(γ). Set ρ2(f)=d>c≔ρ(γ), and set H≔f-α; then(i)
H=0⇒H(k)=α-α(k),(ii)
H=β-α⇔H(k)=β-α(k),whereα(k)=ϕeγ and ϕ is a differential polynomial of R1 and γ. Let (12)Υ=β-α=R2-R1eγ≕Reγ,θ=β-αk=R2-ϕeγ≕Ψeγ.We may set F=H/Υ and get ρ2(F)=ρ2(f)=d. By Lemma 14, for 0<ϵ<d-c/2, there exists a sequence wn→∞ as n→∞ such that (13)F♯wn>ewnρ2F-ϵ=ewnd-ϵ.Note that Υ=β-α=(R2-R1)eγ has at most finitely many zeros and poles and f-α has at most finitely many poles. Then there exists a positive number r such that F has no poles in D={z:|z|>r}.In view ofwn→∞ as n→∞, without loss of generality, we assume that |wn|≥r+1 for all n. Define D1={z:|z|<1} and (14)Fnz≔Fwn+z=Hwn+zΥwn+z,and then every Fn is analytic in D1. Also Fn♯(0)=F♯(wn)→∞ as n→∞. It follows from Marty’s criterion that Fn is not normal at z=0.By Lemma13 and choosing an appropriate subsequence of Fn if necessary, we may assume that there exist sequences zn and ρn with |zn|<r<1 and ρn→0 such that the sequence gn is defined by(15)gnζ=Fnzn+ρnζ=Hwn+zn+ρnζΥwn+zn+ρnζ⟶gζlocally uniformly in C, where g is a nonconstant entire function of order at most 1, whose zeros have multiplicity at least k and(16)ρn≤MFn♯0=MF♯wn≤Me-wnd-ϵfor a positive number M.We assert that(17)ρnkHkwn+zn+ρnζΥwn+zn+ρnζ⟶gkζ.We prove the assertion by induction. By (15), the assertion holds for k=0. We assume that the assertion is right for k=l; that is, (18)ρnlHlwn+zn+ρnζΥwn+zn+ρnζ⟶glζ.Now we will prove that the assertion is also right for k=l+1. Let (19)Gnζ=ρnlHlwn+zn+ρnζΥwn+zn+ρnζ,and then(20)Gn′ζ=ρnl+1Hl+1wn+zn+ρnζΥwn+zn+ρnζ-ρnl+1Hlwn+zn+ρnζΥ′wn+zn+ρnζΥ2wn+zn+ρnζ=ρnl+1Hl+1wn+zn+ρnζΥwn+zn+ρnζ-ρnGnζΥ′wn+zn+ρnζΥwn+zn+ρnζ⟶gl+1ζ.Recall the above definition of degree of rational function; we have(21)Υ′Υz=wn+zn+ρnζ=R′+Rγ′Rz=wn+zn+ρnζ≤wnqMwn+zn+ρnζ,γ′≤wnqM2wn,γ′≤wnqeAwnc+ϵ,and here A is a positive constant and q is an integer. Noting that 0<ϵ<d-c/2, we have d-ϵ>c+ϵ. Then, combining (16) and (21), we conclude that(22)ρnGnζΥ′wn+zn+ρnζΥwn+zn+ρnζ≤MGnζwnqeAwnc+ϵ-wnd-ϵ⟶0,asn⟶∞.Then (23)ρnl+1Hl+1wn+zn+ρnζΥwn+zn+ρnζ⟶gl+1ζ.Hence, we complete the proof of the assertion.Obviously,g(k)≢0. Otherwise g would be a polynomial of degree less than k and could not have zeros of multiplicity at least k. In the following, we will prove that(a)
g=0⇒g(k)=0,(b)
g=1⇔g(k)=0.Firstly, we prove (a). Suppose thatg(ζ0)=0. By Hurwitz’s theorem, there exists a sequence (ζn)n, ζn→ζ0, such that (for n being sufficiently large)(24)gnζn=Hwn+zn+ρnζnΥwn+zn+ρnζn=0.Thus, H(wn+zn+ρnζn)=0 and H(k)(wn+zn+ρnζn)=(α-α(k))(wn+zn+ρnζn)=(R1-ϕ)eγ(wn+zn+ρnζn).By (16), (17), (21), and (22), we derive that(25)gkζ0=limn→∞ρnkHkwn+zn+ρnζnΥwn+zn+ρnζn=limn→∞ρnkR1-ϕwn+zn+ρnζnR2-R1wn+zn+ρnζn=0.Thus, g(ζ)=0⇒g(k)(ζ)=0. So (a) holds.Similarly, we can proveg=1⇒g(k)=0. Now we prove (26)gk=0⟹g=1.From (16) and (17), we have(27)ρnkHkwn+zn+ρnζ-θwn+zn+ρnζΥwn+zn+ρnζ⟶gkζ.Suppose that g(k)(η0)=0. Then g(k)≢0. Hence, by (27) and Hurwitz’s theorem, there exists a sequence (ηn)n, ηn→η0, such that (for n being sufficiently large) (28)Hkwn+zn+ρnηn=θwn+zn+ρnηn.Then, it follows from the assumption that we get H(wn+zn+ρnηn)=Υ(wn+zn+ρnηn). It implies that (29)gη0=limr→∞Hwn+zn+ρnηnΥwn+zn+ρnηn=1.Thus, we prove that g(k)=0⇒g=1 and obtain (b).From (a) and (b), it is easy to deduce thatg≠0. Thus, g=Aeλz; it contradicts with g=1⇒g(k)=0. So ρ2(f)≤ρ(γ). Therefore, we complete the proof of Theorem 2.


## 4. The Proof of Theorem6

Using some ideas of [3, 14], we give the following proof of Theorem 6.LetH=f-α. Then all the zeros of H have multiplicities at least k and (30)H=0⟹Hk=α-αk,H=β-α⟺Hk=β-αk,where α(k)=P1eQ and P1 is a polynomial. A careful calculation leads us to obtain deg⁡P1=kdeg⁡Q′+deg⁡Q1=k(q-1)+q1. Let (31)γ=β-α=Q2-Q1eQ≔PeQ,θ=β-αk=Q2-P1eQ≔P2eQ,where P2 is a polynomial. Set F=H/γ.Ifρ(F)>max{q,q-q2-q1/k}, by Lemma 15, for each 0<μ<ρ(F)-1, there exist wn→∞ such that, for n→∞,(32)limn→∞F♯wnwnμ=+∞.In view of wn→∞ as n→∞, without loss of generality, we may assume that |wn|≥r+1 for all n. Define D1={z:|z|<1} and (33)Fnz≔Fwn+z=Hwn+zγwn+z.Note that γ=β-α=(Q2-Q1)eQ has at most finitely many zeros and f-α has at most finitely poles; then there exists a positive number r such that F has no poles in D={z:|z|>r}, so Fn(z)=F(wn+z) is analytic in D1. Therefore, all Fn(z) are analytic in D1. Thus, we have structured a family (Fn)n of holomorphic functions.AlsoFn♯(0)=F♯(wn)→∞ as n→∞. It follows from Marty’s criterion that Fn is not normal at z=0. Therefore, by applying Lemma 13 and choosing an appropriate subsequence of Fn if necessary, we may assume that there exist subsequences zn and ρn with |zn|<r<1 and ρn→0 such that the sequence gn is defined by(34)gnζ≔Fnzn+ρnζ=Hwn+zn+ρnζγwn+zn+ρnζ⟶gζlocally uniformly in C, where g is a nonconstant entire function of order at most 1, whose zeros have multiplicity at least k and(35)ρn≤MFn♯0=MF♯wn≤Mwn-μfor a positive number M.We assert that(36)ρnkHkwn+zn+ρnζγwn+zn+ρnζ⟶gkζ.We will use the mathematical induction to prove the assertion. By (34), the assertion is right for k=0. We assume that the assertion is right for k=l; that is, (37)ρnlHlwn+zn+ρnζγwn+zn+ρnζ⟶glζ.Now we prove that the assertion is also right when k=l+1. Let (38)Gnζ=ρnlHlwn+zn+ρnζγwn+zn+ρnζ,and then(39)Gn′ζ=ρnl+1Hl+1wn+zn+ρnζγwn+zn+ρnζ-ρnl+1Hlwn+zn+ρnζγ′wn+zn+ρnζγ2wn+zn+ρnζ=ρnl+1Hl+1wn+zn+ρnζγwn+zn+ρnζ-ρnGnζγ′wn+zn+ρnζγwn+zn+ρnζ⟶gl+1ζ.Since γ=PeQ, we have (40)ρnGnζγ′wn+zn+ρnζγwn+zn+ρnζ=ρnGnζP′+PQ′wn+zn+ρnζPwn+zn+ρnζ.In view of the degree of a polynomial function, one has(41)γ′γz=wn+zn+ρnζ=P′+PQ′Pz=wn+zn+ρnζ=P′P+Q′z=wn+zn+ρnζ=Ownl1asn⟶∞,where l1≔deg⁡Q′ is a nonnegative integer. Then, combining (35) and (41) yields(42)ρnGnζγ′wn+zn+ρnζγwn+zn+ρnζ≤MGnζwnl1-μ⟶0,asn⟶∞.Furthermore, (43)ρnl+1Hl+1wn+zn+ρnζγwn+zn+ρnζ⟶gl+1ζ.So we complete the proof of the assertion.Obviously,g(k)≢0. Otherwise, g would be a polynomial of degree less than k and so could not have zeros of multiplicity at least k. In the following, we will prove(c)g=0⇒g(k)=0,(d)g=1⇔g(k)=0.Firstly, we prove (c). Suppose thatg(ζ0)=0. Then, by Hurwitz’s theorem, we get that there exists a sequence (ζn)n, ζn→ζ0, such that (for n being sufficiently large)(44)gnζn=Hwn+zn+ρnζnγwn+zn+ρnζn=0.Thus, H(wn+zn+ρnζn)=0 and H(k)(wn+zn+ρnζn)=(Q1-P1)eQ(wn+zn+ρnζn). By (36) and (39), we derive that(45)gkζ0=limn→∞ρnkHkwn+zn+ρnζnγwn+zn+ρnζn=limn→∞ρnkQ1-P1wn+zn+ρnζnPwn+zn+ρnζn.Also,(46)Q1-P1Pz=wn+zn+ρnζ=Q1-P1Q2-Q1z=wn+zn+ρnζ=1-P1/Q1Q2/Q1-1z=wn+zn+ρnζ=Ownl2,asn⟶∞where l2≔max{k(q-1)-(q2-q1),0} is a nonnegative integer. Then, combining (35) and (46) yields(47)ρnkQ1-P1wn+zn+ρnζnPwn+zn+ρnζn≤Mkwnl2-kμ⟶0,asn⟶∞.So,(48)gkζ0=limn→∞ρnkHkwn+zn+ρnζnγwn+zn+ρnζn=limn→∞ρnkQ1-P1wn+zn+ρnζnPwn+zn+ρnζn=0.Thus, g(ζ)=0⇒g(k)(ζ)=0 and (c) holds.Next, we will prove thatg=1⇒g(k)=0.Similar to what was stated above, suppose thatg(ζ0)=1. Then, by Hurwitz’s theorem, there exists a sequence (ζn)n, ζn→ζ0, such that (for n being sufficiently large)(49)gnζn=Hwn+zn+ρnζnγwn+zn+ρnζn=1.Thus, H(wn+zn+ρnζn)=γ(wn+zn+ρnζn) and H(k)(wn+zn+ρnζn)=(Q2-P1)eQ(wn+zn+ρnζn). By (35) and (36), we deduce that(50)gkζ0=limn→∞ρnkHkwn+zn+ρnζnγwn+zn+ρnζn=limn→∞ρnkQ2-P1wn+zn+ρnζnQ2-Q1wn+zn+ρnζn.By calculation,(51)Q2-P1Pz=wn+zn+ρnζ=Q2-P1Q2-Q1z=wn+zn+ρnζ=Q2/Q1-P1/Q1Q2/Q1-1z=wn+zn+ρnζ=Ownl3,asn⟶∞where l3≔max{k(q-1)-(q2-q1),0}=l2 is a nonnegative integer. Then, combining (35) and (51) shows that(52)ρnkQ2-P1wn+zn+ρnζnPwn+zn+ρnζn≤Mkwnl3-kμ⟶0,asn⟶∞.So,(53)gkζ0=limn→∞ρnkHkwn+zn+ρnζnγwn+zn+ρnζn=limn→∞ρnkQ2-P1wn+zn+ρnζnPwn+zn+ρnζn=0.It implies that g(ζ)=1⇒g(k)(ζ)=0. Now, we prove (54)gk=0⟹g=1.From (35), (36), (51), and (52), we have(55)ρnkHkwn+zn+ρnζ-θwn+zn+ρnζγwn+zn+ρnζ⟶gkζ.Suppose that g(k)(η0)=0. Since g(k)≢0, by (55) and Hurwitz’s theorem, there exists a sequence ηn, ηn→η0, such that (for n being sufficiently large) (56)Hkwn+zn+ρnηn=θwn+zn+ρnηn.By the assumption, we get H(wn+zn+ρnηn)=γ(wn+zn+ρnηn) and (57)gη0=limr→∞Hwn+zn+ρnηnγwn+zn+ρnηn=1.Thus, we prove g(k)=0⇒g=1 and obtain (d).From (c) and (d), it is easy to deduce thatg≠0. Thus, g=Aeλz. But it contradicts with g=1⇒g(k)=0. So ρ(F)≤max{q,q-q2-q1/k}.Next, we will proveρ(f)≤ρ(F)≤max{q,q-q2-q1/k}.Ifρ(α)<ρ(f), noting the fact that ρ(β-α)≤ρ(α) and Lemma 16, then ρ(F(β-α))≤max{ρ(F),ρ(α)}. Rewrite the function F=f-α/β-α as f=α+F(β-α). Thus, by Lemma 16, we conclude that ρ(f)≤max{ρ(α),ρ(F(β-α))}≤max{ρ(α),ρ(F)}. Furthermore, it follows from ρ(α)<ρ(f) that ρ(f)≤ρ(F)≤max⁡q,q-q2-q1/k.Ifρ(α)=ρ(f), noting that ρ(α)=deg⁡Q, then ρ(f)=ρ(α)=deg⁡Q=q≤max⁡{q,q-q2-q1/k}.Ifρ(α)>ρ(f), noting that F=f-α/β-α=f-Q1eQ/(Q2-Q1)eQ=f/PeQ-Q1/P and Q1/P is a rational function, then ρ(Q1/P)=0 and ρ(α)=deg⁡Q=ρ(PeQ)=q>ρ(f). So, ρ(f)<deg⁡Q=ρ(F)≤max{q,q-q2-q1/k}.Therefore,ρ(f)≤ρ(F)≤max{q,q-q2-q1/k} holds. This completes the proof of Theorem 6.


## 5. The Proof of Theorem10

Based on some ideas in [9], we will prove the theorem.By Theorem6, Lemma 17, and the assumption of Theorem 10, we have(58)fk-βf-β=μ,where μ is a nonzero constant. It follows that(59)fk=μf+1-μQ2eQ.If μ=1, then f=f(k); this is result (1) in Theorem B. Assuming that μ≠1 and z0 is a zero of f-Q1eQ, we have f(k)(z0)-Q1(z0)eQ(z0)=0. Substituting z0 into (59), we have(60)Q2z0-Q1z0=0.Noting that Q2-Q1≠0, we know that all the zeros of f-Q1eQ are the zeros of Q2-Q1, so f-Q1eQ has only finitely many zeros. Set f=Q1eQ+P1eQ⋆, where P1 is a nonzero polynomial and Q⋆ is a polynomial. Then f(k)=P2eQ+(P1Q⋆′k+P3)eQ⋆, where P2=(Q1eQ)(k)/eQ is a polynomial with deg⁡P2=deg⁡Q1Q′k and P3 is a polynomial with deg⁡P3<deg⁡P1Q⋆′k. Substituting them into (59), we have (61)P2eQ+P1Q⋆′k+P3eQ⋆=μQ1eQ+μP1eQ⋆+1-μQ2eQ.Then,(62)P2=μQ1+1-μQ2,(63)P1Q⋆′k+P3=μP1.From (63), we get Q⋆′k=μ and P3=0. Thus Q⋆′ is a constant, say Q⋆′=λ. So Q⋆(z)=λz+b and μ=λk, where b is a constant. Furthermore,(64)fz=Q1zeQz+P1zeλz+b,(65)fkz=P2eQ+P1λk+P3eλz+b.We deduce that deg⁡P3=deg⁡P1′. Since P3=0, we get that P1 is a constant. Set A=P1eb as a constant. Due to the fact that ρ(β)<ρ(f) and (64), Q(z) must be a constant. Hence, we complete the proof of Theorem 10.



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*Source: 1298343-2018-06-03.xml*