We assume that the reader is familiar with the elementary Nevanlinna theory, for detailed information, we refer the reader [15, 16, 22] and references therein. Meromorphic functions are non-constant unless otherwise specified. For such a function f and a∈ℂ¯=:ℂ∪{∞}, each z with f(z)=a will be called a-point of f. We will use here some standard definitions and basic notations from this theory. In particular, N(r,a;f) (N¯(r,a;f)) is the counting function (reduced counting function) of a-points of f, T(r,f) is the Nevanlinna characteristic function of f, and S(r,f) is any function of smaller order than T(r,f) when r→∞. We also denote ℂ*:=ℂ∖{0}.

For a meromorphic function f, the order ρ(f) and the hyper-order ρ2(f) of f are defined by

ρ(f)=limsupr→∞log+T(r,f)logr,  ρ2(f)=limsupr→∞log+log+T(r,f)logr. 

For a∈ℂ∪{∞}, we also define

Θ(a;f)=1−limsupr→+∞N¯r,1/(f−a)T(r,f).

We denote by S(f) the family of all meromorphic functions s for which T(r,s)=o(T(r,f)), where r→∞ outside of a possible exceptional set of finite logarithmic measure. Moreover, we also include all constant functions in S(f), and let S^(f)=S(f)∪{∞}. For s∈S^(f), we say that two meromorphic functions f and g share s CM when f-s and g-s have the same zeros with the same multiplicities. If multiplicities are not taking into account, then we say that f and g share s IM.

The five-point and four-point uniqueness theorems of Nevanlinna [30] are classical uniqueness results in the theory of meromorphic functions. The five-point theorems states that if two meromorphic functions f, g share five distinct values in the extended complex plane IM, then f≡g. The beauty of this result lies in the fact that there is no counterpart of this result in case of real valued functions. On the other hand, four-point theorem states that if two meromorphic functions f, g share four distinct values in the extended complex plane CM, then f≡T∘g, where T is a Möbius transformation.

These results initiated the study of uniqueness of two meromorphic functions f and g. The study of such uniqueness becomes more interesting if the function g has some expressions in terms of f.

The following definition will be used later.

Definition 1.1. f and g be two meromorphic functions such that f and g share the value a with weight k where a∈ℂ∪{∞}. We denote by N¯E(k+1r,1/(f−a) the counting function of those a-points of f and g where p=q≥ k+1, each point in this counting function is counted only once.

In what follows, let c be a non-zero constant. For a meromorphic function f, let us now denote its shift I_cf and difference operators Δcf, respectively, by Icf(z)=f(z+c) and Δcf(z)=(Ic−1)f(z)=f(z+c)−f(z).

For finite ordered meromorphic functions, Halburd and Korhonen [17], and independently Chiang and Feng [13], developed parallel difference versions of the famous Nevanlinna theory. As applications of this theory, we refer the reader to such articles as [3, 4, 7, 6, 9, 36]) for set sharing problems, as [1, 27, 32] for finding solutions to the Fermat-type difference equations, as [14] for Nevanlinna theory of the Askey–Wilson divided difference operators, and as [28] and references therein for meromorphic solutions to the difference equations of Malmquist type.

Regarding periodicity of meromorphic functions, Heittokangas et al. [20, 21] have considered the problem of value sharing between meromorphic functions and their shifts and obtained the following results.

Theorem A.([20]) Let f be a meromorphic function of finite order, and let c∈ℂ*. If f(z) and f(z+c) share three distinct periodic functions s1,s2,s3∈S^(f) with period cCM, then f(z)≡f(z+c) for all z∈ℂ.

In 2009, Heittokangas et al. [21] improved Theorem A by replacing "sharing three small functions CM" by "2 CM+1 IM" as follows.

Theorem B.([21]) Let f be a meromorphic function of finite order, and let c∈ℂ*. Let s1,s2,s3∈S^(f) be three distinct periodic function with period c. If f(z) and f(z+c) share s1,s2∈S^(f) CM and s₃ IM , then f(z)≡f(z+c) for all z∈ℂ.

In 2014, Halburd et al. [19] extended some results in this direction to meromorphic functions f whose hyper-order ρ2(f) is less than one. One may get much more information (see [1, 4, 11, 12, 13, 18, 19, 20, 21, 25, 26] and the references therein) about the relationship between a meromorphic function f(z) and it shift f(z+c).

In 2016, in this direction, Li and Yi [23] obtained a uniquenessresult of meromorphic functions f sharing four values with their shifts f(z+c).

Theorem C.([23]) Let f be a non-constant meromorphic function of hyper-order ρ2(f)<1and c∈ℂ*. Suppose that f and f(z+c) share 0, 1, η IM, and share ∞ CM, where η is a finite value such that η≠0,1. Then f(z)≡f(z+c) for all z∈ℂ.

We now recall here the definition of partially shared values by two meromorphic functions f and g.

Definition 1.2.([10]) Let f and G be non-constant meromorphic functions and s∈ℂ∪{∞}. Denote the set of all zeros of f-s by E(s,f) , where a zero of multiplicity m is counted m times. If E(s,f)⊂E(s,g), then we say that f and g partially share the value s CM. Note that E(s,f)=E(s,g) is equivalent to f and g share the value s CM. Therefore, it is easy to see that the condition "partially shared values CM" is more general than the condition"shared value CM".

In addition, let E¯(s,f) denote the set of zeros of f-s, where a zero is counted only once in the set, and E¯k)(s,f) denote the set of zeros of f-s with multiplicity l≤k, where a zero with multiplicity l is counted only once in the set. The reduced counting function corresponding to to E¯k)(s,f) are denoted by N¯k)(r,1/(f−s)).

Charak et al. [8] gave the following definition of partial sharing.

Definition 1.3.([8]) We say that a meromorphic function f share s∈S^ partially with a meromorphic function g if E¯(s,f)⊆E¯(s,g), where E¯(s,f) is the set of zeros of f(z)-s(z), where each zero is counted only once.

Let f and g be two non-constant meromorphic functions and s(z)∈S^(f)∩S^(g). We denote by N¯0(r,s;f,g) the counting function of common solutions of f(z)-s(z)=0 and g(z)−s(z)=0, each counted only once. Put

N¯12(r,s;f,g)=N¯r,1f−s+N¯r,1g−s−2N¯0(r,s;f,g).

It is easy to see that N¯12(r,s;f,g) denoted the counting function of distinct solutions of the simultaneous equations f(z)-s(z)=0 and g(z)−s(z)=0.

In 2016, Charak et al. [8] introduced the above notion of partial sharing of values and applying this notion of sharing, they have obtained the following interesting result.

Theorem D.([8]) Let f be a non-constant meromorphic function of hyper order ρ2(f)<1, and c∈ℂ*. Let s1,s2,s3,s4∈S^(f) be four distinct periodic functions with period c. If δ(s,f)>0 for some s∈S^(f) and

E¯(sj,f)⊆E¯(sj,f(z+c)),   j=1, 2, 3, 4,

then f(z)=f(z+c) for all z∈ℂ.

In 2018, Lin et al. [24] investigated further on the result of Charak et al. [8] replacing the condition "partially shared value E¯(s,f)⊆E¯(s,f(z+c))" by the condition "truncated partially shared value E¯k)(s,f)⊆E¯k)(s,f(z+c))", k is a positive integer. By the following example, Lin et. al. [24] have shown thatthe result of Charak et. al. [8] is not be true for k=1 if truncated partially shared values is considered.

Example 1.4.([24]) Let f(z)=2ez/(e2z+1) and c=π i, s₁=1, s₂=-1, s₃=0, s4=∞ and k=1. It is easy to see that f(z+πi)=−2ez/(e2z+1) and f(z) satisfies all the otherconditions of Theorem D, but f(z)≡f(z+c).

However, after a careful investigation, we find that Theorem D is not valid in fact for each positive integer k although f(z) and f(z+c) share value s∈{s1,s2,s3,s4} CM. We give here only two examples to show that the result of Charak et al. [8] is not true for k=2 and k=3.

Example 1.5. Let f(z)=aez(e2z+3)/3e2z+1, c=πi and s₁=a, s₂=-a, where a∈ℂ*, s₃=0, s4=∞ and k₁=2=k₂ . It i easy to see that f(z+πi)=−aez(e2z+3)/3e2z+1 and f(z) satisfies all the conditions of Theorem D, but f(z)≡f(z+c).

Example 1.6. Let f(z)=4aez(e2z+1)/e4z+6e2z+1 and c=πi, s₁=a, s₂=-a, where a∈ℂ*, s₃=0, s4=∞ and k₁=3=k₂. Then clearly f(z+πi)=−4aez(e2z+1)/e4z+6e2z+1 and f(z) satisfies all the conditions of Theorem D, but f(z)≡f(z+c).

In 2018, Lin et al. [24] established thefollowing result considering partially sharing values.

Theorem E.([24]) Let f be a non-constant meromorphic function of hyper-order ρ2(f)<1 and c∈ℂ*. Let k₁, k₂ be two positive integers, and let s1,s2∈S(f)∪{0}, and s3,s4∈S^(f) be four distinct periodic functions with period c such that f and f(z+c) share s3,s4 CM and

E¯kj)(sj,f)⊆E¯kj)(sj,f(z+c)),  j=1,2.

If Θ(0,f)+Θ(∞;f)>2/(k+1), where k=min{k1,k2}, then f(z)≡f(z+c) for all z∈ℂ.

As a consequence of Theorem E, Lin et al. [24] have obtained the following result.

Theorem F.([24]) Let f be a non-constant meromorphic function of hyper order ρ2(f)<1, Θ(∞,f)=1 and c∈ℂ*. Let s1,s2,s3∈S(f) be three distinct periodic functions with period c such that f(z) and f(z+c) share s₃ CM and

E¯k)(sj,f)⊆E¯k)(sj,f(z+c)),  j=1,2.

If k≥2, then f(z)≡f(z+c) for all z∈ℂ.

The authors have showed that number "k= 2" is sharp for the function f(z)=sinz and c=π. It is easy to see that f(z+c) and f(z) share the value 0 CM and E¯1)(1,f(z))=E¯1)(1,f(z+c))=ϕ and E¯1)(−1,f(z))=E¯1)(−1,f(z+c))=ϕ but f(z+c)≡f(z). Since Theorem F is true for k≥2, hence Lin et al. [24] investigated further to explore the situation when k=1 and obtained the result.

Theorem G([24]) Let f be a non-constant meromorphic function of hyper order ρ2(f)<1, Θ(∞,f)=1 and c∈ℂ*. Let s1,s2,s3∈S(f) be three distinct periodic functions with period c such that f(z) and f(z+c) share s₃ CM and

E¯1)(sj,f)⊆E¯1)(sj,f(z+c)),  j=1,2.

Then f(z)≡f(z+c) or f(z)≡−f(z+c) for all z∈ℂ. Moreover, the later occurs only if s1+s2=2s3.}

Remark 1.7. We find that in the proof of [24, Theorem 1.6], the authors Lin et al. made a mistake. In Theorem 1.6, they have obtained f(z+c)≡−f(z) as one of the conclusion when s1+s2=2s3, where originally it will be f(z+c)≡−f(z)+2s3. One can easily understand it from the following explanation. In [24, Proof of Theorem 1.6, page - 476] the authors have obtained α=−1, where α, the way they have defined, finally will be numerically equal with f(z+c)−s3/f(z)−s3=α, when s1+s2=2s3.Hence after combining, it is easy to see that f(z+c)−s3f(z)−s3=−1 and this implies that f(z+c)≡−f(z)+2s3.

In this paper, our aim is to take care of these points. We also want to extend the above results with certain suitable setting. Henceforth, for a meromorphic function f and c∈ℂ*, we recall here (see [2]) Lc(f):=c1f(z+c)+c0f(z), where c1(≠0),c0∈ℂ. Clearly, Lc(f) is a generalization of shift f(z+c) as well as the difference operator Δcf.

In this paper, to give a correct version of the result of Lin et al. with a general setting, we are mainly interested to find the affirmative answers of the following questions.

Question 1.8. Is it possible to extend f(z+c) upto Lc(f), in all the above mentioned results?

Question 1.9. Can we obtained a similar result of Theorem E, replacing the condition Θ(0;f)+Θ(∞;f)>2/(k+1), where k=min{k1,k2} by a more general one?

If answers of the above questions are found to be affirmative, then it is natural to raise the following questions.

Question 1.10. Is the new general condition, so far obtained, sharp?

Question 1.11. Can we find the class of all the meromorphic function which satisfies the difference equation Lc(f)≡f?

Answering the above questions is the main objective of this paper. We organize the paper as follows: In Section 2, we state the main results of this paper and exhibit several examples pertinent with the different issues regarding the main results. In Section 3, key lemmas have been stated and proved some of them. Section 4 is devoted specially to prove the main results of this paper. In Section 5, some questions have raised for further investigations on the main results of this paper.

We prove the following result generalizing that of Lin et al. [24].

Theorem 2.1. Let f be a non-constant meromorphic function of hyper order ρ2(f)<1 and c,c1∈ℂ*. Let k₁, k₂ be two positive integers, and s₁, s2∈S∖{0}, s₃, s4∈S^(f) be four distinct periodic functions with period c such that f and Lc(f) share s₃, s₄ CM and

E¯kj)(sj,f)⊆E¯kj)(sj,Lc(f)),  j=1,2. 

If

Θ(0;f)+Θ(∞;f)>1k1+1+1k2+1, 

then Lc(f)≡f. Furthermore, f assumes the following form

f(z)=1−c0c1zcg(z),

where g(z) is a meromorphic function such that g(z+c)=g(z), for all z∈ℂ.

Remark 2.2. The following examples show that the condition

Θ(0;f)+Θ(∞;f)>1k1+1+1k2+1

in Theorem 2.1 is sharp.

Example 2.3. Let f(z)=aez(e2z+3)/3e2z+1, c=πi and s₁=a, s₂=-a, where a∈ℂ*, s₃=0, s4=∞ and k₁=2=k₂. It is easy to see that Lπi(f)=−aez(e2z+3)/3e2z+1, where c1=c0+1, c0,c1∈ℂ*, and f(z) satisfies all the conditions of Theorem 2.1 and

Θ(0;f)+Θ(∞;f)=23=1k1+1+1k2+1,

where Θ(0,f)=1/3=Θ(∞,f), but Lπi(f)≡f.

Example 2.4. Let f(z)=4aez(e2z+1)/e4z+6e2z+1, c=πi, s₁=a, s₂=-a, where a∈ℂ*, s₃=0 , s4=∞ and k₁=3=k₂ . Then clearly Lπi(f)=−4aez(e2z+1)/e4z+6e2z+1, where c1=c0+1, c0,c1∈ℂ*, and f(z) satisfies all the conditions of Theorem 2.1 and

Θ(0;f)+Θ(∞;f)=12=1k1+1+1k2+1,

where Θ(0,f)=1/2, Θ(∞,f)=0 but we see that Lπi(f)≡f.

As the consequence of Theorem 2.1, we obtain the following result.

Theorem 2.5. Let f be a non-constant meromorphic function of hyper-order ρ2(f)<1, Θ(∞,f)=1 and c,c1∈ℂ*. Let s1,s2,s3∈S^(f) be three distinct periodic functions with period c such that f and Lc(f) share s₃ CM and

E¯kj)(sj,f)⊆E¯kj)(sj,Lc(f)),  j=1,2.

If k1,k2≥2, then Lc(f)≡f. Furthermore, f assumes the following form

f(z)=1−c0c1zcg(z),

where g(z) is a meromorphic function such that g(z+c)=g(z), for all z∈ℂ.

The following example shows that the number k₁=2=k₂ is sharp in Theorem 2.5.

Example 2.6. We consider f(z)=acosz, where a∈ℂ*, s1=a,s2=−a and s3=0. We choose Lπ(f)=c1f(z+π)+c0f(z), where c1, c0∈ℂ* with c1=c0+1. Clearly f and Lπ(f) share s₃ CM, Θ(∞,f)=1, E¯1)(a,f)=ϕ=E¯1)(a,Lπ(f)) and E¯1)(−a,f)=ϕ=E¯1)(−a,Lπ(f)), but f(z)≡Lπ(f).

Naturally, we are interested to explore the situation when k₁=1=k₂ and hence, we obtain the following result.

Theorem 2.7. Let f be a non-constant meromorphic function of hyper-order ρ2(f)<1 with Θ(∞,f)=1 and c,c1∈ℂ*. Let s1,s2,s3∈S^(f) be three distinct periodic functions with period c such that f and Lc(f) share s₃ CM and

E¯1)(sj,f)⊆E¯1)(sj,Lc(f)),  j=1,2.

Then Lc(f)≡f or Lc(f)≡−f+2s3. Furthermore,

• (i) If Lc(f)≡f, then

  f(z)=1−c0c1zcg(z).

• (ii) If Lc(f)≡−f+2s3, then

  f(z)=−1−c0c1zcg(z)+2s3,  forall z∈ℂ,

where g(z) is a meromorphic function such that g(z+c)=g(z). Moreover, Lc(f)≡−f+2s3 occurs only if s1+s2=2s3.

Remark 2.8. We see that Theorems 2.1, 2.2 and 2.3 directly improved, respectively, Theorems E, F and G.

Remark 2.9. We see from Example 2.3 that, in Theorem 2.3, the possibility Lc(f)≡−f+2s3 could be occurred.

The following example shows that the restrictions on the growth of f in our above results are necessary and sharp.

Example 2.10. Let f(z)=eg(z), where g(z) is an entire function with ρ(g)=1, and hence for c1=1/2=c0, it is easy to see that Lπ(z)=e2g(z)+1/2eg(z). We choose s1=1, s2=−1 and s3=∞. Clearly Θ(∞,f)=1, ρ2(f)=1, f and Lπ(f) share s₃ CM and E¯1)(1,f)⊆E¯1)(1,Lπ(z)) and E¯1)(−1,f)⊆E¯1)(−1,Lπ(z)) but we see that neither Lπ(z)≡f nor Lπ(z)≡−f+2s3. Also the function has not the specific form.

Remark 2.11. The next example shows that the condition Θ(∞,f)=1 in Theorem 2.3 can not be omitted.

Example 2.12. Let f(z)=1/cosz, c₁=1, c₀=0, s₁=1, s₂=-1 and s₃=0. Clearly Θ(∞,f)=0, f and L3π/2(f) share s₃ CM, E¯1)(1,f)⊆E¯1)(1,L3π/2(z)) and E¯1)(−1,f)⊆E¯1)(−1,L3π/2(z)).However, one may observe that neither L3π/2(z)≡f nor L3π/2(z)≡−f+2s3. Also the function has not the specific form.

In this section, we present some necessary lemmas which will play a key role in proving the main results. Henceforth, for a non-zero complex number c and for integers n≥ 1, we define the higher order difference operators Δcnf:=Δcn−1(Δcf).

Lemma 3.1.([34]) Let c∈ℂ, n∈ℕ, let f be a meromorphic function of finite order. Then any small periodic function a≡a(z)∈S(f)

mr,Δcnff(z)−a(z)=S(r,f), 

where the exponential set associated with S(r,f) is of at most finite logarithmic measure.

Lemma 3.2. ([29, 31]) If R(f) is rational in f and has small meromorphic coefficients, then

T(r,R(f))=degf(R)T(r,f)+S(r,f).

Lemma 3.3.([35]) Suppose that h is a non-constant entire function such that f(z)=eh(z), then ρ2(f)=ρ(h).

In [13, 17], the first difference analogue of the lemma on the logarithmic derivative was proved and for the hyper-order ρ2(f)<1, the following is the extension, see [19].

Lemma 3.4.([19]) Let f be a non-constant finite order meromorphic function and c∈ℂ. If c is of finite order, then

mr,f(z+c)f(z)=OlogrrT(r,f)

for all r outside of a set E with zero logarithmic density. If the hyper order ρ2(f)<1, then for each ϵ>0, we have

mr,f(z+c)f(z)=0T(r,f)r1−ρ2−ϵ

for all r outside of a set of finite logarithmic measure.

Lemma 3.5.([33]) Let f be a non-constant meromorphic function, sj∈S^(f), j=1,2,...,q, (q≥3). Then for any positive real number ϵ, we have

(q−2−ϵ)T(r,f)≤∑ j=1qN¯r,1f−sj, r∈E,

where E⊂[0,∞) and satisfies ∫Edloglogr<∞.

We now prove the following lemma, a similar proof of this lemma can also be found in [2].

Lemma 3.6. Let f be a non-constant meromorphic function such that

E¯(sj,f)⊆E¯(sj,c1f(z+c)+c0f(z)),  j=1,2,

where s1,s2∈S(f), c, c0, c1(≠0)∈ℂ*, then f is not a rational.

Proof We wish to prove this lemma by the method of contradiction. Let f be a rational function. Then f(z)=P(z)/Q(z) where P and Q are two polynomials relatively prime to each other and P(z)Q(z)≡0. Hence

(3.1)E(0,P)∩E(0,Q)=ϕ

It is easy to see that

c1f(z+c)+c0f(z)=c1P(z+c)Q(z+c)+c0P(z)Q(z)        =c1P(z+c)Q(z)+c0P(z)Q(z+c)Q(z+c)Q(z)        =P1(z)Q1(z),(say)

where P₁ and Q₁ are two relatively prime polynomials and P1(z)Q1(z)≡0.

Since E¯(s1,f)⊆E¯(s1,c1f(z+c)+c0f(z)) and f is a rational function, there must exists a polynomial h(z) such that

c1f(z+c)+c0f(z)−s1=(f−s1)h(z)

which can be re-written as

(3.2)c1P(z+c)Q(z)+c0P(z)Q(z+c)Q(z+c)Q(z)−s1≡P(z)Q(z)−s1h(z).

We now discuss the following cases:

Case 1. Let P(z) is non-constant.

Then by the Fundamental Theorem of Algebra, there exists z0∈ℂ such that P(z₀)=0. Then it follows from (3.2) that

(3.3)c1P(z0+c)Q(z0+c)≡(1−h(z0))s10, 

where s10=s1(z0).

Subcase 1.1. Let z0∈ℂ be such that s1(z0)=0.

Then from (3.3), it is easy to see that P(z₀+c)=0. Then we can deduce from (3.1) that P(z₀+mc)=0 for all positive integer m. However, this is impossible, and hence we conclude that the polynomial P(z) is a non-zero constant.

Subcase 1.2. Let z0∈ℂ be such that s1(z0)≠0.

Then from (3.3), we obtain

P(z0+c)≡s10c1(1−h(z0))Q(z0+c).

Next proceeding exactly same way as done in above, we obtain

(3.4)P(z0+mc)≡s10c1(1−h(z0))Q(z0+mc).

In view of (3.3) and (3.4), a simple computation shows that

P(z0+c)Q(z0+c)=P(z0+mc)Q(z0+mc)  for  all  positive  integers  m, 

which contradicts the fact that E(0,P)∩E(0,Q)=ϕ.

Therefore, it is easy to see that f(z) takes the form f(z)=η/Q(z), where P(z)=η=constant (≠0).

Case 2. Let Q(z) be non-zero constant.

Now

(3.5)c1f(z+c)+c0f(z)=c1η Q(z)+c0η Q(z+c)Q(z+c)Q(z).

Since E(s2,f)=E(s2,c1f(z+c)+c0f(z)), hence there must exists a polynomial h₁(z) such that c1f(z+c)+c0f(z)−s2=(f−s2)h1(z), which can be written as

(3.6)c1 Q(z)+c0Q(z+c)≡η−s2Q(z)dh1(z)Q(z+c).

Since Q(z), and hence Q(z+c) are a non-constant polynomials, hence by the Fundamental Theorem of Algebra, there must exist z₀ and z₁ such that Q(z0)=0=Q(z1+c).

Subcase 2.1. When Q(z₀)=0, then from (3.6), we see that h1(z0)=−c0/η, which is absurd.

Subcase 2.2. When Q(z1+c)=0, then from (3.6), we get Q(z₁)=0 , which is not possible.

Therefore, we conclude that Q(z) is a non-zero constant, say η2. Thus we have f(z)=η/η2, a constant, which is a contradiction. This completes the proof.

Lemma 3.7.([19]) Let T:[0,+∞]→[0,+∞] be a non-decreasing continuous function, and let s∈(0,+∞). If the hyper-order of T is strictly less than one, i.e.,

limsupr→+∞log+log+T(r)logr=ρ2<1,

then

T(r+s)=T(r)+oT(r)r1−ρ2−ϵ,

where ϵ>0 and r→∞, outside of a set of finite logarithmic measure.

In this section, we give the proofs of our main results.

Proof of Theorem 1.1. First of all we suppose that sj∈ℂ, j=1,2,3,4. By the assumption of the theorem, f(z) and Lc(f)=c1f(z+c)+c0f(z) share s₃, s₄ CM, hence we must have

(4.1)f−s3Lc(f)−s4f−s4Lc(f)−s3=eh(z),

where h(z) is an entire function with ρ(h)<1 by Lemma 3.3. In view of Lemma 3.4, we obtain

Tr,eh=S(r,f). 

Next we suppose that z0∈E¯k)(s1,f)∪E¯k)(s2,f). Then from (3.1), one may easily deduce that eh(z0)=1. For the sake of convenience, we setγ:=eh(z) and

S(r):=S(r,L(f))=S(r,f).

We now split the problem into two cases.

Case 1. Let eh(z)≠1.

A simple computation shows that that

(4.2)N¯k1)r,1f−s1≤Nr,1γ−1≤T(r,γ)+O(1)≤S(r)

and

(4.3)N¯k2)r,1f−s2≤Nr,1γ−1≤T(r,γ)+O(1)≤S(r). 

Without loss of generality, we may assume that s₃, s4∈S(f)∖{0}. By Lemma 3.5, for

ϵ∈0,13Θ(0;f)+Θ(∞;f)−1k1+1−1k2+1,

we obtain

(4.4)(4−ϵ)T(r,f)≤N¯(r,f)+N¯r,1f+∑ j=14 N ¯ r,1f−sj+S(r,f).

With the help of (4.2) and (4.3), it follows from (4.4) that

(2−ϵ)T(r,f)≤N¯(r,f)+N¯r,1f+∑ j=12 N ¯(kj+1r,1f−sj+S(r,f) 

which gives

Θ(0;f)+Θ(∞;f)≤1k1+1+1k2+1

and this contradicts

Θ(0;f)+Θ(∞;f)>1k1+1+1k2+1.

Case 2. Therefore, we have eh(z)≡1 and hence

(f−s3)(Lc(f)−s4)(f−s4)(Lc(f)−s3)=1.

On simplification, it is easy to obtain Lc(f)≡f(z), for all z∈ℂ.

We are now to find the class of all the meromorphic functions satisfying the difference equation Lc(f)≡f. By assumption of the result, and using Lemma 3.6, it is easy to see that f is not a rational function. Therefore f(z) must be a transcendentalmeromorphic function.

We also see that f(z) and f(z+c) are related by

(4.5)f(z+c)=1−c0c1f(z). 

Let f₁(z) and f₂(z) be two solutions of (4.5) (see [2] for more details). Then it is easy to see that

(4.6)f1(z+c)=1−c0 c1 f1(z) (4.7)f2(z+c)=1−c0 c1 f2(z).

We set h(z):=f1(z)/f2(z). Then in view of (4.6) and (4.7), we obtain

h(z+c)=f1(z+c)f2(z+c)=1−c0c1f1(z)1−c0c1f2(z)=f1(z)f2(z)=h(z),

for all z∈ℂ. Therefore, it is easy to verify that

f2(z)=1−c0 c1 zcg2(z),

where g₂(z) is a meromorphic function with g2(z+c)=g2(z), is a solution of (4.5). Hence, it is also easy to verify that f1(z)=f2(z)h(z), a solution of (4.5). Thus the linear combination

a1f1(z)+a2f2(z)=1−c0c 1zca1h(z)+a2g2(z)        =1−c0c 1zcσ(z),

where σ(z)=a1h(z)+a2g2(z) is such that σ(z+c)=σ(z), for all z∈ℂ, is the general solution of (4.5). Hence, the precise form of the function f is the following

f(z)=1−c0c1zcg(z),

where g(z) is a meromorphic function with g(z+c)=g(z), for all z∈ℂ.

This completes the proof.

Proof of Theorem 2.3 Let us suppose that g(z) is the canonical product of the poles of f. Then by Lemma 3.4, we obtain

(4.8)mr,g(z+c)g(z)=S(r,f). 

Since Θ(∞;f)=1, hence it is easy to see that

limsupr→+∞N¯(r,f)T(r,f)=0.

Therefore, it follows from (4.8) that

(4.9)Tr,g(z+c)g(z)=S(r,f).

Since f and Lc(f) share s₃ CM, by Lemma 3.3, we obtain

(4.10)Lc(f)−s3f−s3=eH(z)g(z)g(z+c),

where H(z) is an entire function with ρ(H)<1. By Lemma 3.4, we also obtain

(4.11)Tr,eH(z)g(z)g(z+c)=S(r,f).

Therefore, by Lemma 3.2 and (4.11), a simple computation shows that T(r,Lc(f))=T(r,f)+S(r,f). For the sake convenience, we set

β:=eH(z)g(z)g(z+c)  and  S(r):=S(r,Lc(f))=S(r,f)

If Lc(f)≡f(z). i.e., if β≠1, then with the help of (4.10) and from the assumption, we obtain

(4.12)N¯1)r,1f−s1≤Nr,1β−1≤T(r,β)+O(1)=S(r).

and

(4.13)N¯1)r,1f−s2≤Nr,1β−1≤T(r,β)+O(1)=S(r).

By Lemma 3.7, and using (4.12) and (4.13), by a simple computation, we obtain

(4.14)N¯1)r,1Lc (f)−s1 ≤N¯1)r,1f−s1 +S(r)=S(r).

and

(4.15)N¯1)r,1Lc (f)−s2 ≤N¯1)r,1f−s1 +S(r)=S(r).

On the other hand, it follows from (4.10) that

(4.16)Lc(f)−s1=(s3−s1)+β (f−s3)    =β f−s1+(β−1)s3β

Similarly, we obtain

(4.17)Lc(f)−s2=β f−s2 +(β−1)s3 β.

It is easy to see that

(4.18)Nr,1Lc(f)−s1=Nr,1f−s1+(β−1)s3β+S(r). 

and

(4.19)Nr,1Lc(f)−s2=Nr,1f−s2+(β−1)s3β+S(r).

Now our aim is to deal with the following three cases.

Case 1. Suppose that (β−1)s3+s1/β≠s2.

Since (β−1)s3+s1/β≠s1 and Θ(∞;f)=1, hence by Lemma 3.5 for ϵ∈0,1/2, it follows from (4.10), (4.12), (4.13), (4.14) and (4.18) that

(4.20)(2−ϵ)T(r,f)≤N¯(r,f)+N¯r,1f+N¯r,1f−s2+N¯r,1f−(β−1)s3+s1β≤ N ¯ (2r,1f−s1+ N ¯ (2r,1f−s2+ N ¯ (2r,1Lc(f)−s1≤12T(r,f)+12T(r,f)+12T(r,f)+S(r)=32T(r,f)+S(r,f),

which is a contradiction.

Case 2. Suppose that (β−1)s3+s2/β≠s1c.

Since (β−1)s3+s2/β≠s2 and Θ(∞;f)=1, hence by applying the same argument as in Case 1, we arrive at a contradiction.

Therefore, we must have Lc(f)≡f, and hence following the proof of Theorem 2.1, we obtain the precise form of the function.

Case 3. Suppose that

(β−1)s3+s2β=s1

and

(β−1)s3+s1β=s2.

An elementary calculation shows that β =-1, so that 2s3=s1+s2. Therefore, from (4.10), we have Lc(f)≡−f(z)+2s3 and by the same argument used in the previous cases, it is not hard to show that f(z) will take the form

f(z)=−1−c0c1zcg(z)+2s3,  for all z∈ℂ,

where g(z) is a meromorphic function with period c. This completes the proof.