Uniqueness of meromorphic functions with their reduced linear c shift operators sharing two or more values or sets

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R E S E A R C H

Open Access

Uniqueness of meromorphic functions with

their reduced linear

c

-shift operators sharing

two or more values or sets

Abhijit Banerjee

1*

_{and Saikat Bhattacharyya}

2

*_{Correspondence:}

[email protected];

[email protected]

1_{Department of Mathematics,} University of Kalyani, West Bengal, India

Full list of author information is available at the end of the article

Abstract

In the paper, we introduce a new notion of reduced linearc-shift operatorLr cf, and

with the aid of this new operator, we study the uniqueness of meromorphic functions

f(z) andLr

cfsharing two or more values in the extended complex plane. The results

obtained in the paper signiﬁcantly improve a number of existing results. Further, using the notion of weighted sharing of sets, we deal the same problem. We exhibit a handful number of examples to justify certain statements relevant to the content of the paper. We are also able to determine the form of the function that coincides with its reduced linearc-shift operator. At the end of the paper, we pose an open question for future research.

MSC: 30D35

Keywords: Meromorphic functions; Uniqueness; Diﬀerence operator; Shared set; Weighted sharing

1 Introduction and results

At the outset, we assume that the readers are familiar with the standard notations and basic results of Nevanlinna’s value distribution theory (see [6,14]). Letf be a nonconstant meromorphic functions deﬁned in the open complex planeC.

We recall that T(r,f) denotes the Nevanlinna characteristic function of a noncon-stant meromorphic functionf andN(r,_f_–1_a) =N(r;a;f) (N(r,_f_–1_a) =N(r;a;f)) denotes the counting function (reduced counting function) ofa-points of a meromorphic functionf.

We need the following deﬁnitions and notations.

Deﬁnition 1.1 For a nonconstant meromorphic functionf andS⊂C∪ {∞}, let

Ef(S) =

a∈S

(z,p)∈C×N:f(z) =awith multiplicityp

and

Ef(S) =

a∈S

(z, 1)∈C×N:f(z) =a.

Then we say thatf andgshare the setSCM(IM)ifEf(S) =Eg(S) (Ef(S) =Eg(S)).

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WhenSis a singleton set, the deﬁnition coincides with the traditional deﬁnition of value sharing.

In 2001, Lahiri [8,9] introduced the deﬁnition of weighted sharing, which plays a key role in uniqueness theory as far as relaxation of sharing is concerned. In terms of Deﬁnition1.1, weighted sharing of sets can be expressed as follows.

Deﬁnition 1.2([9]) Letpbe a nonnegative integer or inﬁnity. Fora∈C∪ {∞}, we denote

byEp(a;f) the set of alla-points off, where ana-point of multiplicitymis countedmtimes ifm≤pandp+ 1 times ifm>p. ForS⊂C∪ {∞}, we deﬁneEf(S;k) =_a_∈_SEk(a;f), where

kis a nonnegative integer or inﬁnity. IfEf(S;k) =Eg(S;k), then we say thatf andgshare the setSwith weightkand write it as (S,k).

Clearly,Ef(S) =Ef(S,∞) andEf(S) =Ef(S, 0).

Deﬁnition 1.3([8]) We denote byN2(r,a;f) the sumN(r,a;f) +N(r,a;f| ≥2).

Letcbe a nonzero complex constant, and letf(z) be a meromorphic function. The shift operator is denoted byf(z+c). Also, we use the notationscf andkcf to denote the

difference andkth-order difference operators off(z), which are defined respectively by

cf(z) =f(z+c) –f(z), kcf(z) =c

k_c–1f(z), k∈N,k≥2.

Carefully observing the definitions, we see that all the variants of difference operators are nothing but linear combinations of different shift operators. So generalizingk_cf, it will be reasonable to introduce the linearc-shift operatorLcf =Lc(f)(z) as follows:

Lcf =Lc(f)(z) = k

j=0

ajf(z+jc),

where aj ∈C for j= 1, 2, . . . ,k with ak = 0. For convenience, putting ak =bk,ak–1 =

–bk–1, . . . ,a0= (–1)kb0, wherebi are nonzero complex constants with kj=0(–1)k–jbj= 0,

we get a special operator denoted byLr

cf =Lrc(f)(z) and call it the reduced linearc-shift

operator.

Putting bk =k_k, bk–1 =

k k–1

, bk–2 =

k k–2

, . . . ,b0 =

k 0

in Lr

cf, we easily verify that Lr

c(f)(z) =kcf.

The problem of uniqueness of meromorphic functions sharing two values were ﬁrst ini-tiated by Rubel and Yang [12] as follows.

Theorem A([12]) Let f be a nonconstant entire function.If f and f share two distinct values(a,∞)and(b,∞),then f ≡f .

In 1979, relaxing the nature of sharing, Mues and Steinmetz [11] improved TheoremA

as follows.

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As derivative and diﬀerence operators are two natural extensions of a function, there is an intimate relation between them. So it is natural to investigate an analogous result corresponding to TheoremBundercf.

In this direction, some progress has been made by Zhang and Liao [15].

Theorem C([15]) Let f(z)be a transcendental entire function of ﬁnite order,c be a nonzero constant,and a,b be two distinct ﬁnite constants.Ifcf(≡0)and f share(a,∞)and(b,∞), then f ≡cf.Furthermore,f(z)must be of the form f(z) = 2

z

c_h₍_z_),_{where h}₍_z₎_{is an entire}

function with period c.

Consideringf(z) =eπiz_{, Zhang and Liao [}₁₅_{] showed that though}_f₍_z_{) and}

1f share 0

CM, TheoremCceases to hold. From the next example we easily see that in TheoremC, two value sharing cannot even be replaced by one “nonzero” value sharing.

Example1.1 Letf(z) =e(i₄π+1₂log2)z_c _{+ 1 +}_i_{. Then}_f _and cf =ie(

iπ

4+12log2)zc share 1 CM.

Herecf=4cn+1f,n∈N.

In fact, we can easily form the following series of examples, rather to say counterexam-ples, which fortify the fact that in TheoremC, two value sharing cannot be replaced by one “nonzero” value sharing.

Example1.2 Letf(z) =e(74iπ+12log2)zc_{+ 1 –}_i_{. Then}_f _and_c_f _{= –}_ie(74iπ+12log2)zc _{share 1 CM,}

wherecf =4cn+1f,n∈N.

Example1.3 Letf(z) =eiπ3cz₊3 2+

√

3i

2 . Thenfandcf = (– 1 2+

√

3i 2 )e

iπz

3c _{share 1 CM, where}

cf =3cn+1f,n∈N.

Example1.4 Letf(z) =e2i3πcz_–1 2+

3√3i

2 . Thenf andcf= (– 3 2+

√

3i 2 )e

2iπz

3c _share√₃_i_CM.

Example1.5 Letf(z) =e4i3πcz₊1 2+

3√3i

2 . Thenf andcf = (– 3 2–

√

3i 2 )e

4iπz

3c _share√₃_i_CM.

Example1.6 Letf(z) =e5i3πcz₊3 2–

√

3i

2 . Thenf andcf= (– 1 2–

√

3i 2 )e

5iπz

3c _{share 1 CM, where}

cf =3cn+1f,n∈N.

We note that in TheoremsA,B, andC, researchers are engaged in finding the uniqueness of a function with its first difference operator, but all are practically tacit about higher-order difference operators.

Recently, for meromorphic functions, Jiang and Chen [7] obtained an analogous result corresponding to TheoremC.

Theorem D([7]) Let f(z)be a nonconstant meromorphic function of ﬁnite order such that N(r,f) =S(r,f),let c∈Cbe a nonzero constant such that f(z+c) –f(z)≡0,and let a and b be two nonzero distinct ﬁnite complex constants.If f andcf share(a,∞)and(b,∞),then f(z+c) = 2f(z).

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Theorem 1.1 Let f be a nonconstant meromorphic function of ﬁnite order,let c be a con-stant such that f(z+c) –f(z)≡0,and let a,b be two distinct nonzero ﬁnite constants.

Suppose Lr_cf and f(z)share(a,∞)and(b,∞).If(k+ 1)N(r,f) + 2(k+ 2)N(r,f) <T(r,f),

then f ≡Lr cf.

Earlier in 2013, Chen and Yi [3] took into account the sharing of inﬁnity in the direction of TheoremD. Their result is as follows.

Theorem E([3]) Let f(z)be a transcendental meromorphic function such that its order of growthσ(f)is not an integer or inﬁnite,and let c∈Cbe a constant such that f(z+c)≡f(z).

Ifcf(z)and f(z)share three distinct values(a,∞)and(b,∞), (∞,∞),then f(z+c)≡

2f(z).

Recently Lu and Lu [10] removed the restriction “σ(f) is not an integer” in the theorem and proved the following result.

Theorem F([10]) Let f(z)be a transcendental meromorphic function of ﬁnite order,and let c∈Cbe a constant such that f(z+c)≡f(z).Ifcf(z)and f(z)share three distinct values

(a,∞), (b,∞),and(∞,∞),then f(z+c)≡2f(z).

Considering TheoremsEandF, it will be pertinent to extend the theorems in the direc-tion of Theorem1.1. In this respect, in the following theorem, we see that at the expense of allowing the sharing of{∞}along with the two shared values in Theorem1.1, we have been able to remove the inequality and withdrawn the restriction over the sharing values to be nonzero.

Theorem 1.2 Let f be a nonconstant transcendental meromorphic function of ﬁnite order that is not of period c,and let a and b be two distinct ﬁnite constants.Suppose Lr

cf and f(z) share(a,∞), (b,∞),and(∞,∞).Then f ≡Lr

cf.

The following examples satisfy Theorem1.1for entire functions whenLr

cf =kcf and Lr

cf =kcf.

Example1.7 Letf(z) =enz_{. We choose}_c₌1

nlog2. Then for any even integerkandL r cf =

k_cf, it is easy to verify that the condition of Theorem1.1is satisﬁed andk_cf =f.

Example1.8 Letf(z) =sin(πz c )e

zlogω

c , where_cis a constant, and_ωk= 1.

Whenkis an odd integer, choosingbi= –1, 0≤i≤k, we have k_j₌₀(–1)k–j_bj_{= 0 and} Lr

cf =ωkf(z) +ωk–1f(z) +ωk–2f(z) +· · ·+f(z) =f(z).

Whenkis an even integer, choosingbi(1 +ω) + 1 = (b0+bk)(1 +ω), 2≤i≤k– 1, and b1=b0+bk, we have kj=0(–1)k–jbj= 0 and

Lr_cf =bkωkf(z) +bk–1ωk–1f(z) +· · ·+b0f(z)

= (b0+bk)f(z) +bk–1ωk–1f(z) +bk–2ωk–2f(z) +· · ·+b1ωf(z)

=b1(1 +ω)f(z) +

ωk–1+ωk–2+· · ·+ω2b1–

1 (1 +ω)

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The following examples satisfy Theorem1.1for meromorphic functions whenLr cf =

k

cf andLrcf=kcf.

Example1.9 Letf(z) =ezlogc2e 6(k+3)πzi

c

e2πczi_–1 , wherecis a constant. Then for any even integer k andLr

cf =kcf, it is easy to verify that the condition of Theorem1.1is satisﬁed and

k cf =f.

Example1.10 Letf(z) =e3(k+3)cπzi

e2πczi_–1 , wherecis a nonzero constant. Heref(z+sc) =f(z) when sis even, and = –f(z), whensis odd. Then for any even integerk, choosingb1=b0+bk, b2=b3=· · ·=bk–1, and 2b1+ (k– 2)bi– 1 = 0, 2≤i≤k– 1, we have kj=0(–1)k–jbj= 0 and Lr

cf =f(z).

With the observation as made in Sect.4, we can construct the following examples satis-fying Theorem1.2for entire functions whenLr

cf=kcf andLrcf =kcf.

Example1.11 Letf(z) ={α

z c 1 +α

z c

2 +· · ·+α z c k}exp(

2zπi

c ). Choosingαi, 1≤i≤k, as the roots

of the equation (1 –z)k_{– 1 = 0, it is easy to verify that}k cf=f.

z c 1 +α

z c

2 +· · ·+α z c k}sin(

2zπi c ).

Whenkis odd integer,αi, 1≤i≤k, are the roots of the equationzk– 2z+ 1 = 0 for

choosingbi=1₂, 0≤i≤k.

Whenkis even integer,αi, 1≤i≤k, are the roots of the equationzk– (3 –k)z+k– 2 = 0

for choosingb0= –k₂,bi= –₂1for oddi, andbi=1₂for eveni.

Clearly, in both the cases we have k_j₌₀(–1)k–j_b

j= 0 andLrcf =f(z).

In the same way, with the observation as made in Sect.4, for a meromorphic function, we can construct the following examples satisfying Theorem1.2when Lr_cf =k_cf and

Lr

cf =kcf.

z c 1 +α

z c

2 +· · ·+α z c

k}sin(12z_cπi). Here forL r

cf =kcf andαi, 1≤i≤k,

being the roots of the equation (1 –z)k– 1 = 0, it is easy to verify thatk_cf =f.

Remark1.1 We note that for a suitable choice ofk,αican be determined in Example1.11

and1.13asα1= 0 fork= 1,α1= 0 andα2= 2 fork= 2,α1= 0,α2=3+3₂i, andα2=3–3₂ifor k= 3, and so on. In fact, instead of the functions chosen in Examples1.11and1.13, any periodic function of periodccan be selected.

z c 1 +α

z c 2}e

6(k+3)πzi c

e2πczi–1 , wherecis a nonzero constant.

Whenkis an odd integer,αi,i= 1, 2, are the roots of the equationz2– 2z– 1 = 0 for

choosingb0=1₂,b1= 1,b2=1₂, andbi= 0 for 3≤i≤k.

Whenkis an even integer,αi,i= 1, 2, are the roots of the equationz2+z= 0 for choosing b0= 1,b1=1₂,b2= –1₂, andbi= 0 for 3≤i≤k.

Clearly, in both the cases, we have k_j₌₀(–1)k–j_bj_{= 0 and}_Lr cf =f(z).

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Example1.15 Letf(z) = Ae_Pdz₍_t₎/c+B, whereA,B,d, andcare nonzero constants, and letP(t)

is a polynomial in t=sin2(π_cz) orcos2₍πz

c). Then for eachk, we havekcf =

Aedz/c₍_ed_–1)k P(t) .

Now forα=₍B_ed(e_–1)d–1)k_–1k, it is easy to see thatf andkcf shareα, butf ≡kcf.

The following example shows that in Theorems1.1and1.2the relationf ≡cf does

not hold for a functionf of inﬁnite order.

Example1.16 Letf(z) =eφ(z)₊_A_{, where}_φ₍_z_{) =}_ezclog2_{, and}_c_,_A_{are nonzero constants.}

Thencf(z) =e2φ(z)–eφ(z). Here we see thatf andcf(z) share two nonzero values (A+

1) +√A+ 1 and (A+ 1) –√A+ 1 butf ≡cf.

The next counterexample is an extension of this example for anyk cf.

Example1.17 Letf(z) =esin(πcz)_{, where}_c_{is a nonzero constant. Then}

k_cf(z) = (–1)k–1₂k–1

1 –e2sin(π_cz) esin(πz

c)

.

Here we see thatf andk_cf share two nonzero values±

(–1)k–1₂k–1

1+(–1)k–1₂k–1 CM, butf ≡kcf.

In view of Theorems 1.1and1.2, it is interesting to investigate the conditions under which the conclusions of Theorems1.1and1.2hold when sharing of the valuesaandb

are relaxed from CM to IM. In our next two theorems, we deal in this regard.

Theorem 1.3 Let f be a nonconstant meromorphic function of ﬁnite order,let c be a con-stant such that f(z+c) –f(z)≡0,and let a and b be two nonzero distinct ﬁnite constants.

Suppose Lr_cf and f(z)share(a, 0), (b, 0).If3(k+ 1)N(r,f) + (k+ 4)N(r,f) + 3N(r,1_f) <T(r,f),

then f ≡Lr_cf.

Theorem 1.4 Let f be a nonconstant meromorphic function of ﬁnite order,let c be a con-stant such that f(z+c) –f(z)≡0, and let a and b be two nonzero distinct ﬁnite con-stants.Suppose Lr_cf and f(z)share(a, 0), (b, 0),and(∞,∞).If(k+ 1)N(r,f) + (k+ 4)N(r,f) + 3N(r,1_f) <T(r,f),then f ≡Lr_cf.

The following example shows that in Theorem1.1or Theorem1.2, two ﬁnite value shar-ing cannot be replaced by sharshar-ing a set containshar-ing two elements forLr_cf=k_cf.

Example1.18 Letf(z) =λexp{

z clog(1+λ)}

sin(2π_cz) , wherec,λare constants such thatλ

k+1_{= –1. Then}

k_cf(z) =λk+1exp{

z

clog(1 +λ)} sin(2π_cz) .

Clearly,fandk

cf share the set{ωp,ωp+2 n–1

}, whereω=e22πni_{, and}_n_and_p_{are two integers}

such that g.c.d.(p, 2n_{) = 1. Here}_f _≡k cf.

So it is also interesting to see whether the two value sharing results concerning the uniqueness off andk

cf can be extended up to two set sharing. In this respect, we have

the next two theorems, where the functionsf andk

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Theorem 1.5 Let S1={awi–1(z)|i= 1, 2, . . . ,n}and S2={bui–1(z)|i= 1, 2, . . . ,m},where w= exp(2_nπ)and u=exp(2_mπ),and a and b are two constants satisfying ab= 0and a2mn₌_b2mn_. Suppose f(z)is a nonconstant meromorphic function of ﬁnite order satisfying Ef(Si, 2) =

Ek_cf(Si, 2)for i= 1, 2,where n≥9and m≥9have no common factors.Then f ≡k_cf.

Theorem 1.6 Let Sifor i= 1, 2be deﬁned as inTheorem1.5.Suppose f(z)is a nonconstant entire function of ﬁnite order such that Ef(Si, 1) =Ek_cf(Si, 1)for i= 1, 2,where n≥5and m≥5have no common factors.Then f ≡k

cf.

The following examples show that Theorems1.5and1.6are not valid if the two functions

f andk_cf share only one set.

Example1.19 Letf(z) =φ(z), whereφ(z) = exp{zclog(1+λ)} sin(2πcz)

, wherecis a constant, and for

ζ =cos2π

n +isin 2π

n,λis a root of the equationζ

n–1_z2s_{– (}_ζ2n–2_{+ 1)}_zs₊_ζn–1_{= 0. Then clearly,}

s_cf(z) =ζn–1φ(z) orζ1–nφ(z). Therefore for a nonzero constantα,f andk_cf share the set{α,αζ,αζ2, . . . ,αζn–1}, butf ≡k_cf.

Example1.20 Letf(z) =φ(z), whereφ(z) =exp{z_clog(1 +λ)}, wherecandλare deﬁned as in the previous example. Then clearly,s_cf(z) =ζn–1φ(z). Therefore for a nonzero con-stantα,f andk_cf share the set{α,αζ,αζ2, . . . ,αζn–1}, butf ≡k_cf.

Note 1.1 In the previous examples,f andk

cf share the set{β,βζ,βζ2, . . . ,βζk–1},where

β=αis a constant,but f ≡k

cf.Note that here m and n have common factors.

Based on Theorems1.5and1.6, it is relevant to investigate whether analogous results can be established by replacing two shared set problem with one shared set together with one shared value. In this regard, we have the following results.

Theorem 1.7 Let S1be the set deﬁned as inTheorem1.5,and let S2={b},where a= 0, b= 0,and b2n₌_a2n_._{Suppose f}₍_z₎_{is a nonconstant entire function of ﬁnite order such that} Ef(Si, 1) =Ekcf(Si, 1)for i= 1, 2where n≥5.Then f≡kcf.

However, in the following theorem, we will show that to replace an entire function by a meromorphic one, the setS2must contain two elements.

Theorem 1.8 Let S1be the set deﬁned as inTheorem1.5,and let S2={b1,b2},where b1and b2are two distinct nonzero constants such that bn1=bn2,b2in=a2nfor i= 1, 2,and bn1bn2=a2n. Suppose f(z)is a nonconstant meromorphic function of ﬁnite order such that Ef(Si, 2) =

Ek

cf(Si, 2)for i= 1, 2where n≥9.Then f ≡ k cf.

Ascf andk

cf are nothing but the derivations from the shift operatorf(z+c) of the

functionf, it is interesting to explore whether analogous result corresponding to Theo-rem1.1can be obtained whenf(z) andf(z+c) share a set with two elements.

The following examples show that like for the diﬀerence operator, it is again not possible for meromorphic and entire functions.

Example1.21 Letf(z) = 2e πiz

c

1+e2πciz, wherecis a constant. Then we see that for a nonzero

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Example1.22 Letf(z) = 1+e πiz

c

1–eπciz, wherecis a constant. Then for a nonzero constant

α,f(z)

andf(z+c) share the set{α,1_α}, butf(z)≡f(z+c).

Example1.23 Letf(z) =sin(π₂_cz), wherecis a constant. Clearly,f(z) andf(z+c) share the set{√1

2, – 1

√

2}, butf(z)≡f(z+c).

So we have the following results analogous to Theorems1.7–1.8.

Theorem 1.9 Under the same situation as inTheorem1.7,ifk_cf is replaced by f(z+c),

then f(z)≡f(z+c).

Theorem 1.10 Under the same situation as inTheorem1.8,ifk

cf is replaced by f(z+c), then f(z)≡f(z+c).

2 Lemmas

In this section, we present some lemmas, which will be needed to proceed further.

Lemma 2.1([14]) Let f(z)be a nonconstant meromorphic function,and let dj(j= 1, . . . ,q)

be q distinct complex numbers.Then we have

m

r,

q

j=1

1

f –dj

=

q

j=1 m

r, 1

f–dj

+O(1).

Lemma 2.2([4,5]) Let f be a meromorphic function of ﬁnite order,and letηbe a nonzero complex constant.Then

m

r,f(z+η)

f(z)

+m

r, f(z)

f(z+η)

=S(r,f).

Lemma 2.3([5]) Let f(z)be a meromorphic function of ﬁnite order,and let c= 0be ﬁxed.

Then

Nr,f(z+c)≤Nr,f(z)+S(r,f),

Nr,f(z+c)≤Nr,f(z)+S(r,f).

Lemma 2.4([7]) Let f(z)be a nonconstant meromorphic function inC.Let d1,d2, . . . ,dn be n≥1distinct complex numbers.Then we have

n

j=1 m

r, 1

f –dj

≤m

r,1

f

+S(r,f), (2.1)

n

j=1 m

r, 1

f –dj

≤m

r, 1 cf

+S(r,f). (2.2)

Lemma 2.5 Let f(z)be a nonconstant meromorphic function inC.Let d1,d2, . . . ,dnbe n≥1distinct complex numbers.Then we have

n

j=1 m

r, 1

f –dj

≤m

r, 1

Lr cf

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Proof From the deﬁnition ofLr

By Lemmas2.1and2.2we have

n

and the lemma follows.

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Lemma 2.9([2]) Let F and G share(1, 2).Then one of the following cases occurs:

(i) T(r,F)≤N2(r, 0;F) +N2(r, 0;G) +N2(r,∞;F) +N2(r,∞;G)

+S(r,F) +S(r,G), the same inequality holds for T(r,G);

(ii) F≡G;

(iii) F.G≡1.

Lemma 2.10([1]) Let F and G be nonconstant meromorphic functions deﬁned inCsuch that they share(1, 1),and let

H=

3 Proofs of theorems

Proof of Theorem1.1 Assume thatf≡Lr cf.

Using Lemma2.7, we have

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≤T(r,f) +Nr,Lr_cf+S(r,f)

≤T(r,f) + (k+ 1)N(r,f) +S(r,f). (3.2)

Let

U= (L

r cf) Lr

cf–a

– f

f–a. (3.3)

Case 1.SupposeU≡0. AsLr

cf andf(z) share (a,∞), in view of Lemmas2.3and2.7, a simple calculation yields

N(r,∞;U)≤Nr,∞;Lr_cf+N(r,∞;f) +N∗r,a;L_crf,f+S(r,f)

≤(k+ 2)N(r,f) +S(r,f). (3.4)

Again using the logarithmic derivative theorem and Lemma2.7, we get

m(r,U)≤Sr,Lr_cf+S(r,f) =S(r,f). (3.5)

Note that

U f –b =

(Lr_cf)

Lr

cf(Lrcf–a) Lr_cf f–b–

f

(f–a)(f –b). (3.6)

So in view of the ﬁrst fundamental theorem, (3.4), (3.5), and Lemmas2.2and2.7, we get

m

r, 1

f –b

=m

r, 1

U

+m

r, U

f–b

≤T(r,U) +m

r, (L

r cf) Lr

cf(Lrcf–a)

+m

r, L

r cf f –b

+m

r, f

(f–a)(f –b)

+O(1)

≤T(r,U) +m

r,(L

r cf) a

1 (Lr

cf –a)

– 1

Lr cf

+m

r, L

r cf f–b

+m

r, f (a–b)

1

f–a)– 1 (f–b)

+S(r,f)

≤(k+ 2)N(r,f) +Sr,Lr_cf+S(r,f)≤(k+ 2)N(r,f) +S(r,f). (3.7)

In a similar manner, we can show that

m

r, 1

f –a

≤(k+ 2)N(r,f) +S(r,f). (3.8)

Now from (3.2), (3.7), and (3.8), applying the ﬁrst fundamental theorem, we obtain

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=m

r, 1

f–a

+m

r, 1

f–b

+N

r, 1

f–a

+N

r, 1

f–b

≤T(r,f) + (k+ 1)N(r,f) + 2(k+ 2)N(r,f) +S(r,f), (3.9)

which implies

T(r,f)≤(k+ 1)N(r,f) + 2(k+ 2)N(r,f) +S(r,f),

a contradiction.

Case 2.Next, supposeU≡0. Integrating we get

Lr_cf –a=C1(f –a),

whereC1is a nonzero constant. In a similar way, we can get

Lr_cf –b=C2(f –b),

whereC2is a nonzero constant. If either ofC1= 1 orC2= 1, then we are done. IfC1= 1

andC2= 1, then from the last two equations, after simple calculations, we get

(C1–C2)f(z) =C1a–C2b+b–a.

IfC1=C2, thenf is a constant, a contradiction. ThereforeC1=C2, and henceC1(a–b) =

(a–b). Asaandbare distinct, we haveC1=C2= 1, and sof ≡Lrcf.

Proof of Theorem1.2 Supposef≡Lr cf. Case 1. ab= 0.

SincefandLr

cf share (∞,∞), we haveN(r,f) =N(r,Lrcf), and hence any pole ofLrcf orf

of multiplicitypmust be a pole ofLr

cf–f of multiplicity≤p, that is,N(r,Lrcf–f)≤N(r,f).

So (3.1) reduces to

N

r, 1

Lr cf –a

+N

r, 1

Lr cf–b

≤N

r, 1

f –a

+N

r, 1

f –b

≤T(r,f) +S(r,f). (3.10)

LetUbe deﬁned as in (3.3).

Case 1.1First, supposeU≡0.

AsLr_cf andf(z) share (a,∞), (∞,∞), in view of Lemma2.7, we clearly have

N(r,∞;U)≤N∗r,∞;Lrcf,f

+N∗r,a;Lrcf,f

=S(r,f). (3.11)

Next, using (3.6), the ﬁrst fundamental theorem, (3.5), (3.11), Lemmas2.2and2.7and proceeding in the same way as in Theorem1.1, we see that (3.7) changes to

m

r, 1

f –b

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In a similar manner, we can show that

m

r, 1

f –a

=S(r,f). (3.13)

Now from (3.10), (3.12), and (3.13), applying the ﬁrst fundamental theorem, similarly to (3.9), we obtain

2T(r,f) +O(1)

=m

r, 1

f–a

+m

r, 1

f–b

+N

r, 1

f–a

+N

r, 1

f–b

≤T(r,f) +S(r,f),

which implies

T(r,f)≤S(r,f),

a contradiction.

Case 1.2.Next, supposeU≡0.

Here we can prove the theorem in the same way as in Theorem1.1. So we omit the details.

Case 2.Letab= 0.

Without loss of generality, we suppose thatb= 0. Let us consider the function

Φ=(L

r

cf)(Lrcf –f)

f(f–a) . (3.14)

Case 2.1.First, supposeΦ≡0.

Letz0be a zero off –aor that off of multiplicityp. AsLrcf andf(z) share (a,∞) and

(0,∞), clearly,a- or 0-points off will not be poles ofΦ. Noting thatf is a transcendental meromorphic function, each zero off of multiplicitypwill be a zero ofΦof multiplicity ≥p. Similarly, any pole offof multiplicityqwould be a pole ofLr

cf–fof multiplicitys≤q

and hence a pole ofΦof multiplicityq+s– 2q≤0. It follows thatΦhas no pole. Also, from (3.14), Lemma2.2, and the ﬁrst fundamental theorem we have

m(r,Φ)≤m

r, L

r cf f–a

+m

r,

Lr cf

f – 1

+S(r,f)≤S(r,f). (3.15)

So using the fact thatΦhas no pole, from (3.15) we see thatS(r,Φ) can be replaced by

S(r,f). Hence, in view of the ﬁrst fundamental theorem, we have

N

r,1

f

≤N

r, 1 Φ

+S(r,f)≤T

r, 1 Φ

–m

r, 1 Φ

+S(r,f)

≤T(r,Φ) –m

r, 1 Φ

+S(r,f)

≤–m

r, 1 Φ

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Again from Lemma2.2we note that

Using (3.17) in (3.16), from the ﬁrst fundamental theorem we get

T(r,f)≤S(r,f),

For two complex constantsaandband two nonconstant meromorphic functionsf and

g, byN(r,a;f |g=b) (N(r,a;f |g=b)) we mean the counting function of thosea-points of f that are (not) theb-points ofg, where an a-point off is counted according to its multiplicity. Again for a positive integers, byN(r,a;f |=s) we mean the reduced counting function of thosea-points off that are of multiplicity exactlys.

AsLr

cf andf share (a, 0) and (b, 0), we see that ana(b)-point off whose multiplicity is

greater than that ofLr

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≤N

Now from Lemma2.5we see that

m

which, in view of (3.18) and the ﬁrst fundamental theorem, yields

T(r,f)≤m mental theorem, we obtain

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+N

Proof of Theorem1.4 Here we proceed with the same argument as in Theorem1.3. So (3.2) reduces to

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Finally, by (3.25) and (3.27) we get

T(r,f)≤(k+ 1)N(r,f) + (k+ 4)N(r,f) + 3N

r,1

f

+S(r,f),

which is a contradiction, and sof≡Lr_cf.

Proof of Theorem1.5 LetF1= (f_a)n,G1= ( k cf

a )nandF2= ( f

b)m,G2= (

kcf

b )m. By the

state-ment of the theorem it follows thatFiandGishare (1, 2) fori= 1, 2.

First, suppose that (i) of Lemma2.9holds. Then we get

T(r,F1)≤N2(r, 0;F1) +N2(r, 0;G1) +N2(r,∞;F1) +N2(r,∞;G1) +S(r,F1) +S(r,G1),

that is,

nT(r,f)≤2N

r,1

f

+ 2N

r, 1 k cf

+ 2N(r,f) + 2Nr,k_cf

+S(r,f) +Sr,k_cf.

In a similar way, we can obtain

nTr,k_cf≤2N

r,1

f

+ 2N

r, 1 k cf

+ 2N(r,f) + 2Nr,k_cf

+S(r,f) +Sr,k_cf.

Thus

nT(r,f) +Tr,k_cf

≤4

N

r,1

f

+N

r, 1 k cf

+N(r,f) +Nr,k_cf+S(r,f) +Sr,k_cf

≤8T(r,f) +Tr,k_cf+S(r,f) +Sr,k_cf,

which is a contradiction forn≥9.

Next, considering cases (ii) and (iii) of Lemma2.9, we get

F1≡G1, i.e.,f ≡t1kcf (3.28)

or

F1.G1≡1, i.e.,f.kcf≡s1, (3.29)

wheretn

1= 1 andsn1=a2n.

Also,

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or

which is impossible.

Proof of Theorem1.6 LetF1= (f_a)n,G1= (

Now considering the functionsF1andG1, from Lemma2.9we get

T(r,F1)≤N2(r, 0;F1) +N2(r, 0;G1) +N2(r,∞;F1) +N2(r,∞;G1) +

In a similar way, we obtain

nTr,k_cf≤2N

which is a contradiction forn≥5.

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Next, suppose

F_i F_i –

2F_i Fi– 1

–

G _i G_i –

2G_i Gi– 1

≡0 fori= 1, 2.

Now considering the functionsF1andG1, by integration we get

F₁

(F1– 1)2 ≡

A G1

(G1– 1)2

, (3.32)

whereAis a nonzero constant.

From this we obtain thatT(r,F1) =T(r,G1) +O(1), that is,T(r,kcf) =T(r,f) +O(1).

Also, (3.32) implies thatF1andG1share (1,∞).

Again, asF1andG1are entire functions, we can say thatF1andG1share (∞,∞) and

N(r,∞;F1) =N(r,∞;G1) = 0.

Now it is obvious that

N2(r, 0;F1)≤2N

r,1

f

≤2T(r,f) +O(1).

Also, we obtain

N(r, 0;G1)≤2T

r,k_cf+O(1)≤2T(r,f) +S(r,f).

So

lim sup r−→∞

N2(r, 0;F) +N2(r, 0;G) + 2N(r,∞;F)

T(r) ≤

4

n< 1

forn≥5. Thus using Lemma2.8, we getF1=G1orF1.G1= 1.

Similarly,F2=G2orF2.G2= 1.

Now if we take

F1.G1≡1, i.e.,fn

k_cfn≡a2n,

then asf andk_cf are both entire functions, we getN(r,∞;kcf

f )≤N(r, 1 f).

Therefore

2nT(r,f) = 2T(r,F1) +O(1)≤T

r, 1

F2 1

+S(r,f)

≤T

r,

k_cf f

n

≤nT

r,

k cf f

+S(r,f)≤nT(r,f) +S(r,f),

which is a contradiction. So we haveF1≡G1, that is,f ≡t1(kcf), wheretn1= 1. Similarly,

we havef ≡t2kcf, wheret2m= 1.

Thus in the same way as in Theorem1.5, we getf ≡k

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Proof of Theorem1.7 SinceEf(S1, 1) =Ekcf(S1, 1), in the line of the proof of Theorem1.6,

we can see that

f ≡tk_cf, (3.33)

wheretn= 1.

Now ifbis a Picard value off, by the assumptionEf(S2, 1) =Ekcf(S2, 1) we know thatb

is a Picard value ofk

cf. Again from (3.33) we see that (tb) is a Picard value off. Sincef is

an entire function, we haveb=tb. Thust= 1, and hencef ≡k_cf.

Ifbis not a Picard value off, then there existsαsuch thatf(α) =k_cf(α) =b. By (3.33) we obtainb=tb. Thust= 1, and hencef≡k

cf.

Proof of Theorem1.8 SinceEf(S1, 2) =Ekcf(S1, 2), in the line of the proof of Theorem1.5,

we can see that

f ≡tk_cf, (3.34)

wheretn_{= 1, or}

f.k_cf ≡s, (3.35)

wheresn₌_a2n_{. Now we discuss the following cases.} Case 1.

Supposef andk

cf satisfy (3.34). We discuss the following subcases. Subcase 1.1.

Assume thatb1is not a Picard value off. Then there existsαsuch thatf(α) =b1. Since Ef(S2, 2) =Ekcf(S2, 2), we obtainkcf(α) =b1orkcf(α) =b2. Ifkcf(α) =b1, then by (3.34)

we haveb1=tb1. Thust= 1 andf ≡kcf. Ifkcf(α) =b2, then by (3.34) we see thatb1=tb2,

which contradicts the assumptionbn 1=bn2. Subcase 1.2.

Assume thatb2is not a Picard value off. Then in the similar way, we havef≡kcf. Subcase 1.3.

Assume thatb1 andb2are Picard values off. AsEf(S1, 2) =Ekcf(S1, 2),b1andb2 are

Picard valuesf andk_cf. Again by (3.34) we see that (tb1) and (tb2) are Picard values of f. Since a meromorphic function has at most two Picard values,b1=tb1or b1=tb2. If b1=tb1, thent= 1 andf ≡kcf. Ifb1=tb2, then it contradicts the assumptionbn1=bn2.

Case 2.

Supposef andk_cf satisfy (3.35). We discuss the following subcases.

Subcase 2.1.

Assume thatb1is not a Picard value off. Then there existsαsuch thatf(α) =b1. Since Ef(S2, 2) =Ekcf(S2, 2), we obtain thatkcf(α) =b1orkcf(α) =b2. Ifkcf(α) =b1, then by

(3.35) we haveb2₁=s, which contradicts the assumptionb₁2n=a2n_{. If}k

cf(α) =b2, then by

(3.35) we have (b1b2) =s, which again contradicts the assumptionbn1bn2=a2n. Subcase 2.2.

Assume thatb2is not a Picard value off. Then in a similar way, we arrive at a

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Subcase 2.3.

Assume thatb1andb2are Picard values off. SinceEf(S2, 2) =Ekcf(S2, 2),b1andb2are

Picard values off andk

cf. Again by (3.35) we see that bs1 and s

b2 are Picard values of f. Since a meromorphic function has at most two Picard values,b1=_bs₁ orb1=_bs₂, and

similarly in both situations, contradiction arises.

Proof of Theorem1.9 The proof can be carried out in the line of Theorem1.7.

Proof of Theorem1.10 The proof can be carried out in the line of Theorem1.8.

In this section, we have the following observation.

4 Observation

Consideringαj(= 1) forj= 1, 2, . . . ,kas the roots of kj=0(–1)k–jbjzj= 1, we claim that the

general solution of the relationLr_cf ≡f is of the form

f(z) =π1(z)α z c

1 +π2(z)α z c

2 +· · ·+πk(z)α z c k,

whereπi(z+c) =πi(z), 1≤i≤k,cbeing a nonzero constant.

Fork= 1, we see thatLr_cf =f impliesf(z+c) = (b0+1

b1 )f(z). Clearly, in this case the general

solution is

f(z) =π1(z)

b0+ 1 b1

z c

=π1(z)α z c 1,

whereα1is a root of the equationb1z–b0– 1 = 0.

Let us verify this fact:

Lr_cf =b1f(z+c) –b0f(z)

=b1

π1(z+c)α z+c

c 1

–b0

π1(z)α

z c 1

= (b1α1–b0)π1(z)α z c 1

=π1(z)α z c 1

=f(z).

Fork= 2, we see thatLr

cf =f impliesb2f(z+ 2c) –b1f(z+c) +b0f(z) =f(z). Letα1and

α2be the roots of the equation

b2z2–b1z+b0– 1 = 0.

In this case the general solution is of the form

f(z) =π1(z)α z c

1 +π2(z)α z c 2.

Let us verify this fact:

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=b2

In this case the general solution is of the form

f(z) =π1(z)α

So we conjecture that the general solution of the equationLr

cf =f for any integerkis

5 An open question

Due to suitable choice of the coeﬃcients of the linearc-shift operator, Theorems1.1and

1.2are valid forLr_cf. So the following question is inevitable: Do Theorems1.1and1.2hold ifLcf is considered instead ofLr

cf?

Acknowledgements

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Funding

Neither of the authors has support of funding agencies.

Availability of data and materials

The data and materials used in the paper are properly cited.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

Both authors contributed equally to the writing of this paper. Both authors read and approved the ﬁnal manuscript.

Author details

1_{Department of Mathematics, University of Kalyani, West Bengal, India.}2_{Department of Science and Humanities,} Jangipur Government Polytechnic, West Bengal, India.

Publisher’s Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional aﬃliations.

Received: 20 August 2019 Accepted: 18 November 2019

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