# Uniqueness of Entire Functions concerning Difference Operator

Authors: Chun Wu
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We deal with a uniqueness question of entire functions sharing a nonzero value with their difference operators and obtain some results, which improve the results of Qi et al. (2010) and Zhang (2011).


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## Body

## 1. Introduction and Main Results

In this paper, a meromorphic function will mean meromorphic in the whole complex plane. We will use the standard notations of Nevanlinna’s value distribution theory such asT(r,f), N(r,f), N¯(r,f), and m(r,f), as explained in Hayman [1], Yang [2], and Yang and Yi [3]. We denote by S(r,f) any quantity satisfying S(r,f)=o(T(r,f)), as r→∞ possibly outside a set of finite linear measures. For f meromorphic in ℂ, denote by S(f) the family of all meromorphic functions a(z) that satisfy T(r,a)=o(T(r,f)) for r→∞ outside a possible exceptional set of finite linear measure. In addition, we denote by ρ(f) and ρ2(f) the order of f and the hyperorder of f [3, 4]. Moreover, we define difference operators by Δcf=f(z+c)-f(z) where c is a nonzero constant. If c=1, we use the usual difference notation Δcf=Δf.Letf and g be two nonconstant meromorphic functions and a be a finite complex number. We say that f, g share the value a CM (counting multiplicities) if f, g have the same a-points with the same multiplicities, and we say that f, g share the value a IM (ignoring multiplicities) if we do not consider the multiplicities. We denote by N¯L(r,1/(f-a)) the counting function for a-points of both f and g about which f has larger multiplicity than g, with multiplicity not being counted. Similarly, we have the notation N¯L(r,1/(g-a)). Next, we denote by N0(r,1/F′) the counting function of those zeros of F′ that are not the zeros of F(F-1) and denote by N11(r,1/(f-a)) the counting function for common simple 1-point of both f and g. In addition, we need the following three definitions.Definition 1.
Letk be a positive integer. Let f and g be two nonconstant meromorphic functions such that f and g share the value 1 IM. Let z0 be a 1-point of f with multiplicity p and a 1-point of g with multiplicity q. We denote by N¯f>k(r,1/(g-1)) the reduced counting function of those 1-points of f and g such that p>q=k. N¯g>k(r,1/(f-1)) is defined analogously.Definition 2 (see [5]).
Letk be a nonnegative integer or infinity. For a∈ℂ∪∞, we denote by Ek(a,f) the set of all a-points of f, where an a-point of multiplicity m is counted m times if ≤k and k+1 times if m>k. If Ek(a,f)=Ek(a,g), we say that f,g share the value a with weight k.
The definition implies that iff, g share a value a with weight k, then z0 is an a-point of f with multiplicity m(≤k) if and only if it is an a-point of g with multiplicity m(≤k) and z0 is an a-point of f with multiplicity m(>k) if and only if it is an a-point of g with multiplicity n(>k), where m is not necessarily equal to n.
We write thatf, g share (a,k) to mean that f, g share the value a with weight k. Clearly if f, g share (a,k), then f, g share (a,p) for any integer p, 0≤p<k. Also we note that f, g share a value a IM or CM if and only if f, g share (a,0) or (a,∞), respectively.Definition 3.
Letf be a nonconstant meromorphic function, and let p be a positive integer and a∈C∪{∞}. Then, by Np)(r,1/(f-a)), we denote the counting function of those a-points of f (counted with proper multiplicities) whose multiplicities are not greater than p, and by N¯p)(r,1/(f-a)), we denote the corresponding reduced counting function (ignoring multiplicities). By N(p(r,1/(f-a)), we denote the counting function of those a-points of f (counted with proper multiplicities) whose multiplicities are not less than p, and by N¯(p(r,1/(f-a)), we denote the corresponding reduced counting function (ignoring multiplicities), where Np)(r,1/(f-a)), N¯p)(r,1/(f-a)), N(p(r,1/(f-a)), and N¯(p(r,1/(f-a)) mean Np)(r,f), N¯p)(r,f), N(p(r,f), and N¯(p(r,f), respectively, if a=∞.
In 2010, Qi et al. [6] proved the following uniqueness theorem.Theorem A.
Letf and g be transcendental entire functions of finite order, let c be a nonzero complex constant, and let n≥6 be an integer. If f(z)nf(z+c) and g(z)ng(z+c) share z CM, then f=tg for a constant t that satisfies tn+1=1.In 2011, Zhang et al. [7] complemented the above theorem and obtained the following result.Theorem B.
Letf and g be nonconstant entire functions of finite order, and let n≥5 be an integer. Suppose that c is a nonzero complex constant such that Δcf≢0 and Δcg≢0. If fnΔcf and gnΔcg share z CM, and g(z+c) and g(z) share 0 CM then f=tg, where t is a constant satisfying tn+1=1.In this paper, we complement Theorems A and B and obtain the following results which generalize the above theorems.Theorem 4.
Letf be a transcendental entire function of finite order and Δcf≢0, let a≠0 be a small function with respect to f, and let c be a nonzero complex constant. Then for n≥2, f(z)n(f(z+c)-1)Δcf-a has infinitely many zeros.Theorem 5.
Letf(z) and g(z) be transcendental entire functions of ρ2<1, n≥2k+7. Suppose that c is a nonzero complex constant such that Δcf≢0 and Δcg≢0. If [fnΔcf](k) and [gnΔcg](k) share 1 CM, then f=tg for a constant t with tn+1=1.Theorem 6.
Letf and g be transcendental entire functions of ρ2<1, n≥5k+13. c is a nonzero complex constant such that Δcf≢0 and Δcg≢0. If [fnΔcf](k) and [gnΔcg](k) share 1 IM, then f=tg for a constant t with tn+1=1.


## 2. Some Lemmas

Lemma 7 (see [8]).
Letf be a nonconstant meromorphic function of finite order σ, and let c be a nonzero constant. Then, for each ε>0,
(1)T(r,f(z+c))=T(r,f(z))+O(rσ-1+ε)+O(logr).Lemma 8 (see [9]).
Letf be a meromorphic function of finite order, and let c∈ℂ and δ∈(0,1). Then
(2)m(r,f(z+c)f(z))+m(r,f(z)f(z+c))=o(T(r,f)rδ)=S(r,f).Lemma 9 (see [10]).
Letf1, f2, and f3 be nonconstant meromorphic functions such that f1+f2+f3=1. If f1, f2, and f3 are linearly independent, then
(3)T(r,f1)≤∑j=13‍N2(r,1fj)+∑j=13‍N¯(r,fj)+o(T(r)),
where T(r)=max1≤j≤3T(r,fj), r∉E, and E denote a set of positive real numbers of finite linear measure.Lemma 10.
Letf be transcendental entire functions of finite order, let c be a nonzero complex constant, and set F(z)=f(z)nΔcf; then
(4)nT(r,f)+S(r,f)≤T(r,F)≤(n+1)T(r,f)+S(r,f).Proof.
Since(5)T(r,F)=T(r,f(z)nΔcf)≤nT(r,f)+T(r,Δcf)T(r,F)≤nT(r,f)+m(r,Δcf)≤nT(r,f)T(r,F)=+m(r,f)+S(r,f)T(r,F)=(n+1)T(r,f)+S(r,f),
then
(6)(n+1)T(r,f)=T(r,f(z)n+1)=m(r,f(z)n+1)(n+1)T(r,f)≤m(r,f(z)n+1F)+m(r,F)+S(r,f)(n+1)T(r,f)≤m(r,f(z)Δcf)+m(r,F)+S(r,f)(n+1)T(r,f)≤T(r,f(z)Δcf)+T(r,F)+S(r,f)(n+1)T(r,f)≤T(r,Δcff(z))+T(r,F)+S(r,f)(n+1)T(r,f)=m(r,Δcff(z))+N(r,Δcff(z))(n+1)T(r,f)=+T(r,F)+S(r,f)(n+1)T(r,f)≤T(r,F)+N(r,1f(z))+S(r,f)(n+1)T(r,f)≤T(r,F)+T(r,f)+S(r,f).
That is,
(7)nT(r,f)+S(r,f)≤T(r,F)≤(n+1)T(r,f)+S(r,f).Lemma 11 (see [11]).
Letf1 and f2 be two nonconstant meromorphic functions. If c1f1+c2f2=c3, where c1, c2, and c3 are nonzero constants, then
(8)T(r,f1)≤N¯(r,f1)+N¯(r,1f1)+N¯(r,1f2)+S(r,f1).Lemma 12 (see [12]).
Letf(z) be a nonconstant meromorphic function, and let k be a positive integer. Suppose that f(k)≢0; then
(9)N(r,1f(k))≤N(r,1f)+kN¯(r,f)+S(r,f).Lemma 13 (see [13]).
Letf, g share (1,0). Then(i)
N¯f>1(r,1/(g-1))≤N¯(r,1/f)+N¯(r,f)-N0(r,1/f′)+S(r,f),(ii)
N¯g>1(r,1/(f-1))≤N¯(r,1/g)+N¯(r,g)-N0(r,1/g′)+S(r,g).Lemma 14.
Letf(z) and g(z) be two nonconstant entire functions. If f and g share 1 IM, then one of the following cases holds: (i)
T(r,g)≤N2(r,1/g)+N2(r,1/f)+N¯(r,1/f)+2N¯(r,1/g)+S(r,f)+S(r,g), the same inequality holding for T(r,f);(ii)
f≡(Ag+B)/(Cg+D), where A, B, C, and D are finite complex numbers satisfying AD≠BC.Proof.
Let(10)Φ(z)=f′′f′-2f′f-1-g′′g′+2g′g-1.
Clearly m(r,Φ)=S(r,f)+S(r,g). We consider the cases Φ(z)≢0 and Φ(z)≡0.
IfΦ(z)≢0, then if z0 is a common simple 1-point of f′ and g′, substituting their Taylor series at z0 into (10), we see that z0 is a zero of Φ(z). Thus, we have
(11)N11(r,1f-1)=N11(r,1g-1)≤N¯(r,1Φ)N11(r,1f-1)≤T(r,Φ)+O(1)N11(r,1f-1)≤N(r,Φ)+S(r,f)+S(r,g).
Our assumptions are thatΦ(z) has poles; all are simple only at zeros of f′and g′ and poles of f and g, and 1-points of f whose multiplicities are not equal to the multiplicities of the corresponding 1-points of g. Thus, we deduce from (10) that
(12)N(r,Φ)≤N¯(2(r,1f)+N¯(2(r,1g)+N0(r,1f′)N(r,Φ)≤+N0(r,1g′)+N¯L(r,1f-1)+N¯L(r,1g-1),
where N0(r,1/f′) is the counting function which only counts those points such that f′=0, but f(f-1)≠0. By the second fundamental theorem, we have
(13)T(r,g)≤N¯(r,1g)+N¯(r,1g-1)T(r,g)≤-N0(r,1g′)+S(r,g),
since
(14)N¯(r,1g-1)=N11(r,1g-1)+N¯(2(r,1f-1)N¯(r,1g-1)=+N¯g>1(r,1f-1).
Thus, we deduce from (11)–(14) that
(15)T(r,g)≤N¯(r,1g)+N¯(2(r,1f)T(r,g)≤+N¯(2(r,1g)+N0(r,1f′)T(r,g)≤+N¯(2(r,1f-1)+N¯L(r,1f-1)T(r,g)≤+N¯L(r,1g-1)+N¯g>1(r,1f-1)T(r,g)≤+S(r,f)+S(r,g).
From the definition ofN0(r,1/f′), we see that
(16)N0(r,1f′)+N¯(2(r,1f-1)+N(2(r,1f)N0(r,1f′)-N¯(2(r,1f)≤N(r,1f′).
The above inequality and Lemma 12 give
(17)N0(r,1f′)+N¯(2(r,1f-1)≤N(r,1f′)-N(2(r,1f)+N¯(2(r,1f)≤N(r,1f)-N(2(r,1f)+N¯(2(r,1f)+S(r,f)≤N¯(r,1f)+S(r,f).
Substituting (17) in (15), we get
(18)T(r,g)≤N¯(r,1g)+N¯(2(r,1f)+N¯(2(r,1g)T(r,g)≤+N¯(r,1f)+N¯L(r,1f-1)+N¯L(r,1g-1)T(r,g)≤+N¯g>1(r,1f-1)+S(r,f)+S(r,g)T(r,g)≤N2(r,1g)+N¯(2(r,1f)+N¯(r,1f)T(r,g)≤+N¯L(r,1f-1)+N¯L(r,1g-1)T(r,g)≤+N¯g>1(r,1f-1)+S(r,f)+S(r,g),
since
(19)N¯L(r,1f-1)≤N(r,1f-1)-N¯(r,1f-1)N¯L(r,1f-1)≤N(r,ff′)≤N(r,f′f)+S(r,f)N¯L(r,1f-1)≤N¯(r,1f)+S(r,f).
Similarly,
(20)N¯L(r,1g-1)≤N¯(r,1g)+S(r,g).
Combining the above inequalities, Lemma13, and (18), we obtain
(21)T(r,g)≤N2(r,1g)+N2(r,1f)+N¯(r,1f)+2N¯(r,1g)T(r,g)=-N0(r,1g′)+S(r,f)+S(r,g)T(r,g)≤N2(r,1g)+N2(r,1f)+N¯(r,1f)T(r,g)=+2N¯(r,1g)+S(r,f)+S(r,g).
Thus, we obtain (i).
IfΦ(z)≡0, then by (10), we have
(22)f′′f′-2f′f-1≡g′′g′-2g′g-1.
By integrating two sides of the above equality, we obtain
(23)f≡Ag+BCg+D,
where A, B, C, and D are finite complex numbers satisfying AD≠BC. This proves the lemma.Lemma 15 (see [14]).
Letf(z) be a nonconstant meromorphic function, s, k be two positive integers; then (24)Ns(r,1f(k))≤T(r,f)(k)-T(r,f)Ns(r,1f(k))≤+Ns+k(r,1f)+S(r,f),Ns(r,1f(k))≤kN¯(r,f)Ns(r,1f(k))≤+Ns+k(r,1f)+S(r,f).
Clearly, N¯(r,1/f(k))=N1(r,1/f(k)).Lemma 16 (see [15]).
Leta0(z),a1(z),…,an(z),b(z) be polynomials such that a0(z)an(z)≢0; let cj be constants and
(25)deg(∑degaj=d‍aj)=d,
where d=max0≤j≤n{degaj}. If f(z) is a transcendental meromorphic solution of
(26)∑j=0n‍aj(z)f(z+cj)=b(z),
then ρ(f)≥1.


## 3. Proof of Theorems

### 3.1. Proof of Theorem4

LetG(z)=f(z)n(f(z+c)-1)Δcf. Since f is a transcendental entire function of finite order, from Lemma 7, we have
(27)(n+2)T(r,(r,f(z)))≤T(r,f(z)n+1(f(z+c)-1))+S(r,f)≤m(r,f(z)n+1(f(z+c)-1))+S(r,f)≤m(r,f(z)n+1(f(z+c)-1)G)+m(r,G)+S(r,f)≤T(r,G)+S(r,f).
By the second main theorem, we deduce that
(28)T(r,G)≤N¯(r,G)+N¯(r,1G)+N¯(r,1G-a)+S(r,G)T(r,G)≤N¯(r,1G-a)+N¯(r,1f)+N¯(r,1f(z+c)-1)T(r,G)=+N¯(r,1Δcf)+S(r,f)T(r,G)≤N¯(r,1G-a)+N¯(r,1f)+T(r,f(z+c)-1)T(r,G)=+T(r,Δcf)+S(r,f)T(r,G)≤N¯(r,1G-a)+N¯(r,1f)+T(r,f(z+c)-1)T(r,G)=+m(r,Δcff·f)+S(r,f)T(r,G)≤N¯(r,1G-a)+N¯(r,1f)+T(r,f(z+c)-1)T(r,G)=+m(r,Δcff)+m(r,f)+S(r,f)T(r,G)≤N¯(r,1G-a)+3T(r,f)+S(r,f).
According to (27) and (28), we have
(29)(n-1)T(r,f)≤N¯(r,1G-a)+S(r,f).
Noting that n≥2, we get that G-a has infinitely many zeros.This completes the proof of Theorem4.


### 3.2. Proof of Theorem5

Since[f(z)nΔcf](k) and [g(z)nΔcg](k) share 1 CM, we have
(30)[f(z)nΔcf](k)-1[g(z)nΔcg](k)-1=eh(z),
where h(z) is a polynomial. Set F=f(z)nΔcf, G=g(z)nΔcg,
(31)F1=F(k),F2=-eh(z)G(k),F3=eh(z),ssssssssssssssssssssthenF1+F2+F3=1,T(r)=max1≤j≤3T(r,Fj),S(r)=o(T(r)).Next, we will prove thatF1, F2, and F3 are linearly dependent and either F2 or F3 is a constant.Now, we suppose that neitherF2 nor F3 is a constant and F1, F2, and F3 are linearly independent; then by Lemma 9, we have
(32)T(r,F1)≤∑j=13‍N2(r,1Fj)+∑j=13‍N¯(r,Fj)+o(T(r)).
Since Fj(j=1,2,3) are entire functions, by the above inequality, we get
(33)T(r,F1)≤N2(r,1F(k))+N2(r,1G(k))+o(T(r)).
From (33) and the first main theorem, we have
(34)T(r,1F(k))=T(r,F(k))+O(1)=T(r,F1)+O(1)T(r,1F(k))≤N2(r,1F(k))+N2(r,1G(k))+o(T(r))T(r,1F(k))≤N(r,1F(k))T(r,1F(k))=-[N(3(r,1F(k))-2N¯(3(r,1F(k))]T(r,1F(k))=+N(r,1G(k))T(r,1F(k))=-[N(3(r,1G(k))-2N¯(3(r,1G(k))]T(r,1F(k))=+o(T(r)).
Assuming that z0 is zero of f(z) (or g(z)) with multiplicity p, if z0 is zero of f(z+c) (or g(z+c)) with multiplicity q(≥1), let m=min{p,q}, then z0 is a zero of F(k) (or G(k)) with multiplicity np+m-k≥np-k≥3, and if z0 is not zero of f(z+c) (or g(z+c)), then z0 is a zero of F(k) (or G(k)) with multiplicity np-k≥3. Therefore, we get that
(35)N(3(r,1F(k))-2N¯(3(r,1F(k))≥(n-k-2)N(r,1f),(36)N(3(r,1G(k))-2N¯(3(r,1G(k))≥(n-k-2)N(r,1g),
since
(37)nm(r,1f)=m(r,1fn)=m(r,ΔcfF)nm(r,1f)≤m(r,1F)+m(r,Δcff·f)nm(r,1f)≤m(r,F(k)F·1F(k))+m(r,Δcff)nm(r,1f)=+m(r,f)+S(r,f)nm(r,1f)≤m(r,1F(k))+T(r,f)+S(r,f)nm(r,1f)=T(r,1F(k))-N(r,1F(k))+T(r,f)+S(r,f).
Therefore, from (34), (35), (36), (37), and Lemma 12,
(38)(n-1)T(r,f)≤(k+2)N(r,1f)+(k+2)N(r,1g)(n-1)T(r,f)=+T(r,g)+o(T(r)).
On the other hand, from (30), we have G(k)+e-h-e-hF(k)=1. Obviously, according to our assumptions, neither e-h nor e-hF(k) is a constant and F1, F2, and F3 are linearly independent. Similarly, we have
(39)(n-1)T(r,g)≤(k+2)N(r,1g)+(k+2)N(r,1f)(n-1)T(r,g)≤+T(r,f)+o(T(r)).
From (38) and (39), we obtain that
(40)[n-2k-6](T(r,f)+T(r,g))≤o(T(r)),
which is a contradiction to n≥2k+7.Therefore,F1, F2, and F3 are linearly dependent, and there exist constants C1, C2, C3 which are not all equal to zero such that
(41)C1F1+C2F2+C3F3=0.Suppose thatC1=0; we have C2F2+C3F3=0. If C2≠0, we get F2=-(C3/C2)F3; that is, G(k)=C3/C2; thus g(z) is a polynomial; it is impossible. Similarly, if C2=0, we also deduce a contradiction.Suppose thatC1≠0, from (41); we know that (C2,C3)≠(0,0). If C2≠0, from (41), we have
(42)(1-C2C1)F2+(1-C3C1)F3=1
and C1≠C2, C1≠C3. That is,
(43)(1-C2C1)G(k)+1eh=1-C3C1.
From Lemma 11, we have
(44)T(r,G(k))≤N¯(r,1G(k))+N¯(r,G(k))+N¯(r,eh)+S(r,g)T(r,G(k))=N¯(r,1G(k))+S(r,g)≤N(r,1G(k))T(r,G(k))=-[N(2(r,1G(k))-N¯(2(r,1G(k))]+S(r,g).
By the similar argument in (37), we have
(45)nm(r,1g)≤T(r,1G(k))-N(r,1G(k))nm(r,1g)≤+T(r,g)+S(r,g).Fromn≥2k+7>k+2, if z0 is zero of g(z) with multiplicity p, then z0 is a zero of G(k) with multiplicity np-k≥2, and we get
(46)N(2(r,1G(k))-N¯(2(r,1G(k))≥(n-k-1)N(r,1g).
According to (44), (45), and (46), we have
(47)(n-1)T(r,g)≤(k+1)N(r,1g)+S(r,g),
which is a contradiction to n≥2k+7.Therefore,C2=0, C3≠0, which gives (1-C1/C3)F1+F2=1. Similarly, we derive a contradiction by calculation.Hence, we deduce that eitherF2 or F3 is a constant.SupposeF2=c≠1; from F1+F2+F3=1, we have F(k)+eh=1-c; in the same manner as above, we get a contradiction. Therefore, c=1; that is, F2=1. Suppose F3=c≠1; similarly as above, we get c=1; that is, F3=1.Therefore, we conclude thatF2=1 or F3=1.IfF2=1, since F1+F2+F3=1, we have F1=-F3=-eh(z). That is
(48)[fnΔcf](k)·[gnΔcg](k)≡1.
Since n≥2k+7 and f and g are transcendental entire functions with hyperorder less than one, we get that f and g have no zeros. Thus,
(49)f(z)=ea(z),g(z)=eb(z),
where a(z), b(z) are nonzero polynomials.Substitute (49) into (48); we have
(50)[ena(z)(ea(z+c)-ea(z))](k)·[enb(z)(eb(z+c)-eb(z))](k)≡1.
Let na(z)+a(z+c)=A1, na(z)+a(z)=A2, nb(z)+b(z+c)=B1, and nb(z)+b(z)=B2. If k=1, we have
(51)(A1′eA1-A2′eA2)·(B1′eB1-B2′eB2)≡1.
From (51), we know that A1′eA1-A2′eA2=eA2(A1′eA1-A2-A2′)≠0; If A1′≠0, then we have A2′=0; thus, A2 must be a constant. By Lemma 16, we have ρ(a(z))≥1; thus, ρ2(f)≥1, which is a contradiction. If A1′=0, then A1 must be a constant; similarly, we also deduce a contradiction.Ifk=2, by calculation, we have
(52)A1′′eA1+(A1′)2eA1-A2′′eA2-(A2′)2eA2=eA2[eA1-A2(A1′′+(A1′)2)-(A2′′+(A2′)2)]≠0.
If A1′′+(A1′)2≠0, then A2′′+(A2′)2=0. If A2 is transcendental entire, then we have
(53)m(r,A2′)=m(A2′′A2′)=S(r,A2′),
which is a contradiction to A2′ being transcendental entire. If A2 is a polynomial, from Lemma 16, which induces that ρ2(f)≥1, we get a contradiction. If A1′′+(A1′)2=0, similar as above, we get a contradiction. For k≥3, using the similar Method as above, we also deduce a contradiction. Therefore, There are not transcendental entire functions f(z) and g(z) satisfying (48).IfF3=1, that is, eh(z)=1, from (30), we get
(54)[fnΔcf](k)≡[gnΔcg](k).
From (54), we have
(55)fnΔcf≡gnΔcg+p(z),
where p(z) is a polynomial of degree at most k-1. Suppose p(z)≢0; then we get
(56)fnΔcfp(z)=gnΔcgp(z)+1.
Therefore, from the second main theorem, we have
(57)(n+1)T(r,f)≤T(fnΔcfp(z))+S(r,f)(n+1)T(r,f)≤N¯(fnΔcfp(z))+N¯(p(z)fnΔcf)(n+1)T(r,f)=+N¯(p(z)gnΔcg)+S(r,f)(n+1)T(r,f)≤N¯(1f)+N¯(1Δcf)+N¯(1g)(n+1)T(r,f)=+N¯(1Δcg)+S(r,f)(n+1)T(r,f)≤2T(r,f)+2T(r,g)+S(r,f).
Similarly, we have
(58)(n+1)T(r,g)≤2T(r,f)+2T(r,g)+S(r,f).
Therefore,
(59)(n+1)[T(r,f)+T(r,g)]≤4[T(r,f)+T(r,g)]+S(r,f)+S(r,g),
which is a contradiction to n≥2k+7. Thus, p(z)≡0, which implies that
(60)fnΔcf≡gnΔcg.Letf/g=h; if h is not a constant, then by (60), we have
(61)hn+1≡fΔcf·Δcgg.
Thus,
(62)(n+1)T(r,h)≤T(r,Δcff)+T(r,Δcgg)+O(1)(n+1)T(r,h)≤N(r,Δcff)+N(r,Δcgg)(n+1)T(r,h)≤+S(r,f)+S(r,g)(n+1)T(r,h)≤T(r,f)+T(r,g)+S(r,f)+S(r,g).
Combining T(r,h)=T(r,f/g)=T(r,f)+T(r,g)+O(1), we obtain n(T(r,f)+T(r,g))≤S(r,f)+S(r,g), which is impossible.Therefore,h is a constant; then substituting f=gh into (60), we have hn+1≡1. Hence f(z)=tg(z), where t is a constant and tn+1=1.The proof of Theorem5 is complete.


### 3.3. Proof of Theorem6

Let(63)F(z)=[f(z)nΔcf](k),G(z)=[g(z)nΔcg](k),F1(z)=f(z)nΔcf,G1(z)=g(z)nΔcg.
Then F(z) and G(z) share 1 IM, and F1(k)=F, G1(k)=G. By Lemma 10, we have
(64)nT(r,f)+S(r,f)≤T(r,F1)≤(n+1)T(r,f)+S(r,f),(65)nT(r,g)+S(r,g)≤T(r,G1)≤(n+1)T(r,g)+S(r,g).
Since f is transcendental entire, by the definition of F, we have
(66)N2(r,1F)=N¯(r,1F)+N¯(2(r,1F)N2(r,1F)=N(r,1F)-[N(3(r,1F)-2N¯(3(r,1F)].
Using the argument in (35), we have
(67)N(3(r,1F)-2N¯(3(r,1F)≥(n-k-2)N(r,1f).
It follows from Lemma 12 and (66), (67), we have
(68)N2(r,1F)≤N(r,1F)-(n-k-2)N(r,1f)N2(r,1F)≤N(r,1fnΔcf)-(n-k-2)N(r,1f)N2(r,1F)=+S(r,f)≤nN(r,1f)+N(r,1Δcf)N2(r,1F)=-(n-k-2)N(r,1f)+S(r,f)N2(r,1F)≤(k+3)T(r,f)+S(r,f).
From Lemma 15, we have
(69)N¯(r,1F)≤Nk+1(r,1fnΔcf)+S(r,f)N¯(r,1F)≤(k+1)N¯(r,1f)+N(r,1Δcf)+S(r,f)N¯(r,1F)≤(k+2)T(r,f)+S(r,f).Similarly,(70)N2(r,1G)≤(k+3)T(r,g)+S(r,g),N¯(r,1G)≤(k+2)T(r,g)+S(r,f).
By Lemma 14, one of the following cases holds: (i)
T(r,G)≤N2(r,1/G)+N2(r,1/F)+N¯(r,1/F)+2N¯(r,1/G)+S(r,F)+S(r,G), the same inequality holding for T(r,F);(ii)
F≡(AG+B)/(CG+D).For case (i), we have
(71)T(r,G)≤N2(r,1G)+N2(r,1F)+N¯(r,1F)T(r,G)≤+2N¯(r,1G)+S(r,F)+S(r,G),T(r,F)≤N2(r,1F)+N2(r,1G)+N¯(r,1G)T(r,F)≤+2N¯(r,1F)+S(r,F)+S(r,G).
Therefore, we get
(72)T(r,F)+T(r,G)≤2[N2(r,1F)+N2(r,1G)]T(r,F)+T(r,G)=+3[N¯(r,1G)+N¯(r,1F)]T(r,F)+T(r,G)=+S(r,F)+S(r,G).
By (64) and Lemma 15, we have
(73)nT(r,f)≤T(r,F1)+S(r,f)≤T(r,F)-N2(r,1F)nT(r,f)=+Nk+2(r,1F1)+S(r,f)nT(r,f)≤T(r,F)-N2(r,1F)+(k+2)N¯(r,1f)nT(r,f)=+N(r,1Δcf)+S(r,f)nT(r,f)≤T(r,F)-N2(r,1F)nT(r,f)=+(k+3)T(r,f)+S(r,f).
Similarly,
(74)nT(r,g)≤T(r,G)-N2(r,1G)+(k+3)T(r,g)+S(r,g).
By (70), (72), (73), and (74), we obtain
(75)(n-5k-12){T(r,f)+T(r,g)}≤S(r,f)+S(r,g),
which is a contradiction since n≥5k+13.For case (ii), we have(76)F≡AG+BCG+D,
where A, B, C, and D are finite complex numbers satisfying AD≠BC. Therefore, by the first fundamental theorem, T(r,F)=T(r,G)+S(r,F).Next, we consider three cases.Case 1.
A
C
≠
0; from (76), we get
(77)F-AC=B-AD/CCG+D.
By the second fundamental theorem and (69), we have
(78)T(r,F)≤N¯(r,1F-A/C)+N¯(r,1F)+S(r,F)T(r,F)=N¯(r,G)+(k+2)T(r,f)+S(r,F)T(r,F)≤(k+2)T(r,f)+S(r,F).
From (73), we obtain (n-2k-5)T(r,f)≤S(r,f), contradicting to n≥5k+13.Case 2.
A
≠
0, and C=0. Then, F≡AG+B/D.
IfB≠0, by the second fundamental theorem and (69), (70), we have
(79)T(r,F)≤N¯(r,1F-B/D)+N¯(r,1F)+S(r,F)T(r,F)=N¯(r,1G)+N¯(r,1F)+S(r,F)T(r,F)≤(k+2)T(r,f)+(k+2)T(r,g)+S(r,F).
Similarly,
(80)T(r,G)≤(k+2)T(r,f)+(k+2)T(r,g)+S(r,G).
From (73), (74), (79), and (80), we get
(81)(n-3k-7)[T(r,f)+T(r,g)]≤S(r,F)+S(r,G),
which is a contradiction to n≥5k+13.
IfB=0, then F≡AG/D. If A/D=1, then F≡G; that is, [fnΔcf](k)≡[gnΔcg](k); using the argument in (54) and noting that n≥5k+13, we obtain f(z)=tg(z), where t is a constant and tn+1=1. If A/D≠1, by the condition that F and G share 1 IM, then F≠1 and G≠1. we obtain then F≠1 and F≠A/D. By the second fundamental theorem, we have
(82)T(r,F)≤N¯(1F-1)+N¯(1F-A/D)+S(r,F)≤S(r,F),
which is impossible.Case 3.
A
=
0, and C≠0. Then, F≡B/(CG+D).
IfD≠0, by the second fundamental theorem and (69), (70), we have
(83)T(r,F)≤N¯(r,1F-B/D)+N¯(r,1F)+S(r,F)T(r,F)=N¯(r,1G)+N¯(r,1F)+S(r,F)T(r,F)≤(k+2)T(r,f)+(k+2)T(r,g)+S(r,F).
Similarly,
(84)T(r,G)≤(k+2)T(r,f)+(k+2)T(r,g)+S(r,G).
From (73), (74), (83), and (84), we get
(85)(n-3k-7)[T(r,f)+T(r,g)]≤S(r,F)+S(r,G),
which is a contradiction to n≥5k+13.
IfD=0, then F≡B/CG. If B/C=1, then F·G≡1; using the argument in (48) in Theorem 5 and noting that n≥5k+13, we get a contradiction. If B/C≠1, by the condition that F and G share 1 IM, we obtain F≠1 and F≠B/C. By the second fundamental theorem, we have
(86)T(r,F)≤N¯(1F-1)+N¯(1F-B/C)+S(r,F)≤S(r,F),
which is impossible.
The proof of Theorem6 is complete.


## 3.1. Proof of Theorem4

LetG(z)=f(z)n(f(z+c)-1)Δcf. Since f is a transcendental entire function of finite order, from Lemma 7, we have
(27)(n+2)T(r,(r,f(z)))≤T(r,f(z)n+1(f(z+c)-1))+S(r,f)≤m(r,f(z)n+1(f(z+c)-1))+S(r,f)≤m(r,f(z)n+1(f(z+c)-1)G)+m(r,G)+S(r,f)≤T(r,G)+S(r,f).
By the second main theorem, we deduce that
(28)T(r,G)≤N¯(r,G)+N¯(r,1G)+N¯(r,1G-a)+S(r,G)T(r,G)≤N¯(r,1G-a)+N¯(r,1f)+N¯(r,1f(z+c)-1)T(r,G)=+N¯(r,1Δcf)+S(r,f)T(r,G)≤N¯(r,1G-a)+N¯(r,1f)+T(r,f(z+c)-1)T(r,G)=+T(r,Δcf)+S(r,f)T(r,G)≤N¯(r,1G-a)+N¯(r,1f)+T(r,f(z+c)-1)T(r,G)=+m(r,Δcff·f)+S(r,f)T(r,G)≤N¯(r,1G-a)+N¯(r,1f)+T(r,f(z+c)-1)T(r,G)=+m(r,Δcff)+m(r,f)+S(r,f)T(r,G)≤N¯(r,1G-a)+3T(r,f)+S(r,f).
According to (27) and (28), we have
(29)(n-1)T(r,f)≤N¯(r,1G-a)+S(r,f).
Noting that n≥2, we get that G-a has infinitely many zeros.This completes the proof of Theorem4.


## 3.2. Proof of Theorem5

Since[f(z)nΔcf](k) and [g(z)nΔcg](k) share 1 CM, we have
(30)[f(z)nΔcf](k)-1[g(z)nΔcg](k)-1=eh(z),
where h(z) is a polynomial. Set F=f(z)nΔcf, G=g(z)nΔcg,
(31)F1=F(k),F2=-eh(z)G(k),F3=eh(z),ssssssssssssssssssssthenF1+F2+F3=1,T(r)=max1≤j≤3T(r,Fj),S(r)=o(T(r)).Next, we will prove thatF1, F2, and F3 are linearly dependent and either F2 or F3 is a constant.Now, we suppose that neitherF2 nor F3 is a constant and F1, F2, and F3 are linearly independent; then by Lemma 9, we have
(32)T(r,F1)≤∑j=13‍N2(r,1Fj)+∑j=13‍N¯(r,Fj)+o(T(r)).
Since Fj(j=1,2,3) are entire functions, by the above inequality, we get
(33)T(r,F1)≤N2(r,1F(k))+N2(r,1G(k))+o(T(r)).
From (33) and the first main theorem, we have
(34)T(r,1F(k))=T(r,F(k))+O(1)=T(r,F1)+O(1)T(r,1F(k))≤N2(r,1F(k))+N2(r,1G(k))+o(T(r))T(r,1F(k))≤N(r,1F(k))T(r,1F(k))=-[N(3(r,1F(k))-2N¯(3(r,1F(k))]T(r,1F(k))=+N(r,1G(k))T(r,1F(k))=-[N(3(r,1G(k))-2N¯(3(r,1G(k))]T(r,1F(k))=+o(T(r)).
Assuming that z0 is zero of f(z) (or g(z)) with multiplicity p, if z0 is zero of f(z+c) (or g(z+c)) with multiplicity q(≥1), let m=min{p,q}, then z0 is a zero of F(k) (or G(k)) with multiplicity np+m-k≥np-k≥3, and if z0 is not zero of f(z+c) (or g(z+c)), then z0 is a zero of F(k) (or G(k)) with multiplicity np-k≥3. Therefore, we get that
(35)N(3(r,1F(k))-2N¯(3(r,1F(k))≥(n-k-2)N(r,1f),(36)N(3(r,1G(k))-2N¯(3(r,1G(k))≥(n-k-2)N(r,1g),
since
(37)nm(r,1f)=m(r,1fn)=m(r,ΔcfF)nm(r,1f)≤m(r,1F)+m(r,Δcff·f)nm(r,1f)≤m(r,F(k)F·1F(k))+m(r,Δcff)nm(r,1f)=+m(r,f)+S(r,f)nm(r,1f)≤m(r,1F(k))+T(r,f)+S(r,f)nm(r,1f)=T(r,1F(k))-N(r,1F(k))+T(r,f)+S(r,f).
Therefore, from (34), (35), (36), (37), and Lemma 12,
(38)(n-1)T(r,f)≤(k+2)N(r,1f)+(k+2)N(r,1g)(n-1)T(r,f)=+T(r,g)+o(T(r)).
On the other hand, from (30), we have G(k)+e-h-e-hF(k)=1. Obviously, according to our assumptions, neither e-h nor e-hF(k) is a constant and F1, F2, and F3 are linearly independent. Similarly, we have
(39)(n-1)T(r,g)≤(k+2)N(r,1g)+(k+2)N(r,1f)(n-1)T(r,g)≤+T(r,f)+o(T(r)).
From (38) and (39), we obtain that
(40)[n-2k-6](T(r,f)+T(r,g))≤o(T(r)),
which is a contradiction to n≥2k+7.Therefore,F1, F2, and F3 are linearly dependent, and there exist constants C1, C2, C3 which are not all equal to zero such that
(41)C1F1+C2F2+C3F3=0.Suppose thatC1=0; we have C2F2+C3F3=0. If C2≠0, we get F2=-(C3/C2)F3; that is, G(k)=C3/C2; thus g(z) is a polynomial; it is impossible. Similarly, if C2=0, we also deduce a contradiction.Suppose thatC1≠0, from (41); we know that (C2,C3)≠(0,0). If C2≠0, from (41), we have
(42)(1-C2C1)F2+(1-C3C1)F3=1
and C1≠C2, C1≠C3. That is,
(43)(1-C2C1)G(k)+1eh=1-C3C1.
From Lemma 11, we have
(44)T(r,G(k))≤N¯(r,1G(k))+N¯(r,G(k))+N¯(r,eh)+S(r,g)T(r,G(k))=N¯(r,1G(k))+S(r,g)≤N(r,1G(k))T(r,G(k))=-[N(2(r,1G(k))-N¯(2(r,1G(k))]+S(r,g).
By the similar argument in (37), we have
(45)nm(r,1g)≤T(r,1G(k))-N(r,1G(k))nm(r,1g)≤+T(r,g)+S(r,g).Fromn≥2k+7>k+2, if z0 is zero of g(z) with multiplicity p, then z0 is a zero of G(k) with multiplicity np-k≥2, and we get
(46)N(2(r,1G(k))-N¯(2(r,1G(k))≥(n-k-1)N(r,1g).
According to (44), (45), and (46), we have
(47)(n-1)T(r,g)≤(k+1)N(r,1g)+S(r,g),
which is a contradiction to n≥2k+7.Therefore,C2=0, C3≠0, which gives (1-C1/C3)F1+F2=1. Similarly, we derive a contradiction by calculation.Hence, we deduce that eitherF2 or F3 is a constant.SupposeF2=c≠1; from F1+F2+F3=1, we have F(k)+eh=1-c; in the same manner as above, we get a contradiction. Therefore, c=1; that is, F2=1. Suppose F3=c≠1; similarly as above, we get c=1; that is, F3=1.Therefore, we conclude thatF2=1 or F3=1.IfF2=1, since F1+F2+F3=1, we have F1=-F3=-eh(z). That is
(48)[fnΔcf](k)·[gnΔcg](k)≡1.
Since n≥2k+7 and f and g are transcendental entire functions with hyperorder less than one, we get that f and g have no zeros. Thus,
(49)f(z)=ea(z),g(z)=eb(z),
where a(z), b(z) are nonzero polynomials.Substitute (49) into (48); we have
(50)[ena(z)(ea(z+c)-ea(z))](k)·[enb(z)(eb(z+c)-eb(z))](k)≡1.
Let na(z)+a(z+c)=A1, na(z)+a(z)=A2, nb(z)+b(z+c)=B1, and nb(z)+b(z)=B2. If k=1, we have
(51)(A1′eA1-A2′eA2)·(B1′eB1-B2′eB2)≡1.
From (51), we know that A1′eA1-A2′eA2=eA2(A1′eA1-A2-A2′)≠0; If A1′≠0, then we have A2′=0; thus, A2 must be a constant. By Lemma 16, we have ρ(a(z))≥1; thus, ρ2(f)≥1, which is a contradiction. If A1′=0, then A1 must be a constant; similarly, we also deduce a contradiction.Ifk=2, by calculation, we have
(52)A1′′eA1+(A1′)2eA1-A2′′eA2-(A2′)2eA2=eA2[eA1-A2(A1′′+(A1′)2)-(A2′′+(A2′)2)]≠0.
If A1′′+(A1′)2≠0, then A2′′+(A2′)2=0. If A2 is transcendental entire, then we have
(53)m(r,A2′)=m(A2′′A2′)=S(r,A2′),
which is a contradiction to A2′ being transcendental entire. If A2 is a polynomial, from Lemma 16, which induces that ρ2(f)≥1, we get a contradiction. If A1′′+(A1′)2=0, similar as above, we get a contradiction. For k≥3, using the similar Method as above, we also deduce a contradiction. Therefore, There are not transcendental entire functions f(z) and g(z) satisfying (48).IfF3=1, that is, eh(z)=1, from (30), we get
(54)[fnΔcf](k)≡[gnΔcg](k).
From (54), we have
(55)fnΔcf≡gnΔcg+p(z),
where p(z) is a polynomial of degree at most k-1. Suppose p(z)≢0; then we get
(56)fnΔcfp(z)=gnΔcgp(z)+1.
Therefore, from the second main theorem, we have
(57)(n+1)T(r,f)≤T(fnΔcfp(z))+S(r,f)(n+1)T(r,f)≤N¯(fnΔcfp(z))+N¯(p(z)fnΔcf)(n+1)T(r,f)=+N¯(p(z)gnΔcg)+S(r,f)(n+1)T(r,f)≤N¯(1f)+N¯(1Δcf)+N¯(1g)(n+1)T(r,f)=+N¯(1Δcg)+S(r,f)(n+1)T(r,f)≤2T(r,f)+2T(r,g)+S(r,f).
Similarly, we have
(58)(n+1)T(r,g)≤2T(r,f)+2T(r,g)+S(r,f).
Therefore,
(59)(n+1)[T(r,f)+T(r,g)]≤4[T(r,f)+T(r,g)]+S(r,f)+S(r,g),
which is a contradiction to n≥2k+7. Thus, p(z)≡0, which implies that
(60)fnΔcf≡gnΔcg.Letf/g=h; if h is not a constant, then by (60), we have
(61)hn+1≡fΔcf·Δcgg.
Thus,
(62)(n+1)T(r,h)≤T(r,Δcff)+T(r,Δcgg)+O(1)(n+1)T(r,h)≤N(r,Δcff)+N(r,Δcgg)(n+1)T(r,h)≤+S(r,f)+S(r,g)(n+1)T(r,h)≤T(r,f)+T(r,g)+S(r,f)+S(r,g).
Combining T(r,h)=T(r,f/g)=T(r,f)+T(r,g)+O(1), we obtain n(T(r,f)+T(r,g))≤S(r,f)+S(r,g), which is impossible.Therefore,h is a constant; then substituting f=gh into (60), we have hn+1≡1. Hence f(z)=tg(z), where t is a constant and tn+1=1.The proof of Theorem5 is complete.


## 3.3. Proof of Theorem6

Let(63)F(z)=[f(z)nΔcf](k),G(z)=[g(z)nΔcg](k),F1(z)=f(z)nΔcf,G1(z)=g(z)nΔcg.
Then F(z) and G(z) share 1 IM, and F1(k)=F, G1(k)=G. By Lemma 10, we have
(64)nT(r,f)+S(r,f)≤T(r,F1)≤(n+1)T(r,f)+S(r,f),(65)nT(r,g)+S(r,g)≤T(r,G1)≤(n+1)T(r,g)+S(r,g).
Since f is transcendental entire, by the definition of F, we have
(66)N2(r,1F)=N¯(r,1F)+N¯(2(r,1F)N2(r,1F)=N(r,1F)-[N(3(r,1F)-2N¯(3(r,1F)].
Using the argument in (35), we have
(67)N(3(r,1F)-2N¯(3(r,1F)≥(n-k-2)N(r,1f).
It follows from Lemma 12 and (66), (67), we have
(68)N2(r,1F)≤N(r,1F)-(n-k-2)N(r,1f)N2(r,1F)≤N(r,1fnΔcf)-(n-k-2)N(r,1f)N2(r,1F)=+S(r,f)≤nN(r,1f)+N(r,1Δcf)N2(r,1F)=-(n-k-2)N(r,1f)+S(r,f)N2(r,1F)≤(k+3)T(r,f)+S(r,f).
From Lemma 15, we have
(69)N¯(r,1F)≤Nk+1(r,1fnΔcf)+S(r,f)N¯(r,1F)≤(k+1)N¯(r,1f)+N(r,1Δcf)+S(r,f)N¯(r,1F)≤(k+2)T(r,f)+S(r,f).Similarly,(70)N2(r,1G)≤(k+3)T(r,g)+S(r,g),N¯(r,1G)≤(k+2)T(r,g)+S(r,f).
By Lemma 14, one of the following cases holds: (i)
T(r,G)≤N2(r,1/G)+N2(r,1/F)+N¯(r,1/F)+2N¯(r,1/G)+S(r,F)+S(r,G), the same inequality holding for T(r,F);(ii)
F≡(AG+B)/(CG+D).For case (i), we have
(71)T(r,G)≤N2(r,1G)+N2(r,1F)+N¯(r,1F)T(r,G)≤+2N¯(r,1G)+S(r,F)+S(r,G),T(r,F)≤N2(r,1F)+N2(r,1G)+N¯(r,1G)T(r,F)≤+2N¯(r,1F)+S(r,F)+S(r,G).
Therefore, we get
(72)T(r,F)+T(r,G)≤2[N2(r,1F)+N2(r,1G)]T(r,F)+T(r,G)=+3[N¯(r,1G)+N¯(r,1F)]T(r,F)+T(r,G)=+S(r,F)+S(r,G).
By (64) and Lemma 15, we have
(73)nT(r,f)≤T(r,F1)+S(r,f)≤T(r,F)-N2(r,1F)nT(r,f)=+Nk+2(r,1F1)+S(r,f)nT(r,f)≤T(r,F)-N2(r,1F)+(k+2)N¯(r,1f)nT(r,f)=+N(r,1Δcf)+S(r,f)nT(r,f)≤T(r,F)-N2(r,1F)nT(r,f)=+(k+3)T(r,f)+S(r,f).
Similarly,
(74)nT(r,g)≤T(r,G)-N2(r,1G)+(k+3)T(r,g)+S(r,g).
By (70), (72), (73), and (74), we obtain
(75)(n-5k-12){T(r,f)+T(r,g)}≤S(r,f)+S(r,g),
which is a contradiction since n≥5k+13.For case (ii), we have(76)F≡AG+BCG+D,
where A, B, C, and D are finite complex numbers satisfying AD≠BC. Therefore, by the first fundamental theorem, T(r,F)=T(r,G)+S(r,F).Next, we consider three cases.Case 1.
A
C
≠
0; from (76), we get
(77)F-AC=B-AD/CCG+D.
By the second fundamental theorem and (69), we have
(78)T(r,F)≤N¯(r,1F-A/C)+N¯(r,1F)+S(r,F)T(r,F)=N¯(r,G)+(k+2)T(r,f)+S(r,F)T(r,F)≤(k+2)T(r,f)+S(r,F).
From (73), we obtain (n-2k-5)T(r,f)≤S(r,f), contradicting to n≥5k+13.Case 2.
A
≠
0, and C=0. Then, F≡AG+B/D.
IfB≠0, by the second fundamental theorem and (69), (70), we have
(79)T(r,F)≤N¯(r,1F-B/D)+N¯(r,1F)+S(r,F)T(r,F)=N¯(r,1G)+N¯(r,1F)+S(r,F)T(r,F)≤(k+2)T(r,f)+(k+2)T(r,g)+S(r,F).
Similarly,
(80)T(r,G)≤(k+2)T(r,f)+(k+2)T(r,g)+S(r,G).
From (73), (74), (79), and (80), we get
(81)(n-3k-7)[T(r,f)+T(r,g)]≤S(r,F)+S(r,G),
which is a contradiction to n≥5k+13.
IfB=0, then F≡AG/D. If A/D=1, then F≡G; that is, [fnΔcf](k)≡[gnΔcg](k); using the argument in (54) and noting that n≥5k+13, we obtain f(z)=tg(z), where t is a constant and tn+1=1. If A/D≠1, by the condition that F and G share 1 IM, then F≠1 and G≠1. we obtain then F≠1 and F≠A/D. By the second fundamental theorem, we have
(82)T(r,F)≤N¯(1F-1)+N¯(1F-A/D)+S(r,F)≤S(r,F),
which is impossible.Case 3.
A
=
0, and C≠0. Then, F≡B/(CG+D).
IfD≠0, by the second fundamental theorem and (69), (70), we have
(83)T(r,F)≤N¯(r,1F-B/D)+N¯(r,1F)+S(r,F)T(r,F)=N¯(r,1G)+N¯(r,1F)+S(r,F)T(r,F)≤(k+2)T(r,f)+(k+2)T(r,g)+S(r,F).
Similarly,
(84)T(r,G)≤(k+2)T(r,f)+(k+2)T(r,g)+S(r,G).
From (73), (74), (83), and (84), we get
(85)(n-3k-7)[T(r,f)+T(r,g)]≤S(r,F)+S(r,G),
which is a contradiction to n≥5k+13.
IfD=0, then F≡B/CG. If B/C=1, then F·G≡1; using the argument in (48) in Theorem 5 and noting that n≥5k+13, we get a contradiction. If B/C≠1, by the condition that F and G share 1 IM, we obtain F≠1 and F≠B/C. By the second fundamental theorem, we have
(86)T(r,F)≤N¯(1F-1)+N¯(1F-B/C)+S(r,F)≤S(r,F),
which is impossible.
The proof of Theorem6 is complete.



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*Source: 240369-2013-11-12.xml*