Entire functions of restricted hyper-order sharing a set of two small functions IM with their linear c-shift operators

Abstract

Purpose

The paper aims to build the relationship between an entire function of restricted hyper-order with its linear c-shift operator.

Design/methodology/approach

Standard methodology for papers in difference and shift operators and value distribution theory have been used.

Findings

The relation between an entire function of restricted hyper-order with its linear c-shift operator was found under the periphery of sharing a set of two small functions IM (ignoring multiplicities) when exponent of convergence of zeros is strictly less than its order. This research work is an improvement and extension of two previous papers.

Originality/value

This is an original research work.

Citation

Banerjee, A. and Roy, A. (2022), "Entire functions of restricted hyper-order sharing a set of two small functions IM with their linear c-shift operators", Arab Journal of Mathematical Sciences, Vol. 28 No. 2, pp. 217-228. https://doi.org/10.1108/AJMS-10-2020-0099

1. Introduction

By a meromorphic function f, we always mean that it is defined on C. For such a meromorphic function, we recall some basic terminologies of value distribution theory such as the Nevanlinna characteristic function T(r, f), the proximity function m(r, f) and the counting function (reduced counting function) of a-points of fN(r,1f−a)=N(r,a;f) (N¯(r,1f−a)=N¯(r,a;f)). For a = ∞, we use N(r,f)=N(r,∞;f)N¯(r,f)=N¯(r,∞;f) to denote counting (reduced counting) function of poles of f (see [1]). With the help of the standard notations, we also would like to recall the following useful terms, namely exponent of convergence of zeros, order and hyper-order of f respectively defined as follows:

λ(f)=lim supr→∞log⁡Nr,1flog⁡r,ρ(f)=lim supr→∞log⁡T(r,f)log⁡randρ2(f)=lim supr→∞log⁡logT(r,f)log⁡r.

Usually, S(r, f) denotes any quantity satisfying S(r, f) = o(T(r, f)) for all r outside of a possible exceptional set of finite linear measure. We denote by S(f) the set of all meromorphic functions a(z) such that T(r, a(z)) = S(r, f) and a(z) is called small function compared to f(z). Let a(z) ∈ S and S be a subset of S(f) ∪{∞} and E_f(S) = ∪_a(z)∈S{z: f(z) − a(z) = 0}, where each zero is counted according to its multiplicity. If we do not count the multiplicity, then the set ∪_a(z)∈S{z: f(z) − a(z) = 0} is denoted by E¯f(S).

If E_f(S) = E_g(S) (E¯f(S)=E¯g(S)) we say that f and g share the set S CM or counting multiplicities (IM or ignoring multiplicities).

For a nonzero complex constant c, the shift operator of f(z) is denoted by f(z + c). The terms Δ_cf and Δckf will be used to denote the difference and k-th order difference operators of f(z), defined respectively as

Δcf(z)=f(z+c)−f(z),Δckf(z)=Δc(Δck−1f(z)),k∈N,k≥2.

We introduce the more generalized linear c-shift operator L_cf by

Lcf=Lc(f)(z)=∑j=0kajf(z+jc)(≢0),

where aj∈C for j = 0, 1, 2, …, k with a_k ≠ 0 (k ≥ 1).

The uniqueness problem of entire functions sharing set with their derivatives, shifts, different types of difference operators has been developed as an interesting direction of research in the realm of value distribution theory. In 1999, Li-Yang [2] made a pioneer work by considering the relation between an entire function and its derivative sharing a set with two elements. Following their footsteps, in 2005, Li [3] investigated the same type of problem for linear differential operator. Four years later, Liu [4] exhibited a similar result for an entire function f and its shift sharing a set with two small functions.

Let us start the discussion with another result of Liu [4] concerning difference operator.

Theorem A.

[4] Let f be a transcendental entire function of finite order, and let a be a nonzero finite constant. If f and Δ_cf share the set {a, −a} CM, then Δ_cf = f.

After that Liu [4] posed a significant question:

Question 1.1.

What happens if {a, −a} is replaced by {a(z), b(z)} in the above theorem, where a(z), b(z) ∈ S(f) are nonvanishing periodic entire functions with period c?

Being motivated by this question, Li [5] investigated the following theorem in a different direction that evolved as a new trend. Actually, Li [5] first diverted the attention of the research germinated from Question 1.1, in terms of relation between exponent of convergence of zero and order. We recall the theorem by Li [5].

Theorem B.

[5] Let f be a nonconstant entire function such that λ(f) < ρ(f) < ∞, ρ(f) ≠ 1, a, b be respectively two distinct entire functions such that ρ(a) < ρ(f) and ρ(b) < ρ(f). If f and Δ_cf share the set {a, b} CM, then Δ_cf = f for all z∈C.

By an example we can show that the restriction ρ(f) ≠ 1 in Theorem B can be removed.

Example 1.1.

Let f(z)=ez⁡log⁡2c. Then obviously Δ_cf = f. Clearly f and Δ_cf share the set {a, b} CM for two distinct entire functions a, b respectively such that ρ(a) < ρ(f) and ρ(b) < ρ(f) and also 0 = λ(f) < ρ(f) = 1.

After publication of Li’s [5] result, there was a long gap in research in this direction. Recently in 2019, concerning finite-order entire function, Qi-Wang-Gu [6] removed the restriction ρ(f) ≠ 1 in Theorem B. Not only that, they also ensured the particular form of f in the following manner:

Theorem C.

[6] Let f be a nonconstant entire function with λ(f) < ρ(f) < ∞, let a, b be respectively two distinct entire functions such that ρ(a) < ρ(f), ρ(b) < ρ(f). If f and Δ_cf share the set {a, b} CM, then f(z) = Ae^μz, where A, μ are two nonzero constants satisfying e^μc = 2. Furthermore, Δ_cf = f.

2. Main results

In our paper, we have extended and improved Theorem C in the following three directions:

1. We replace the difference operator by its linear c-shift operator to accommodate a larger class of operators, namely L_cf that includes difference operator.

2. We consider an entire function of ρ₂(f) < 1 instead of considering the same of finite order.

3. We relax the nature of the shared set {a, b} from CM to IM.

Thus, the following assertion extends and improves Theorem C in the way described above, and in fact it represents our main result in this paper.

Theorem 2.1.

Let f be a nonconstant entire function such that λ(f) < ρ(f) with ρ₂(f) < 1 and let a, b be two distinct entire functions such that ρ(a) < ρ(f) and ρ(b) < ρ(f). Let f and L_cf (≢0) share the set {a, b} IM, then L_cf = f. In addition, if b = −a, then L_cf = −f. In both cases f takes the form f(z) = Ah(z)e^μz, where A is a nonzero constant, h(z) is a polynomial and μ is a nonzero constant satisfying ∑j=0kajeμcj=1 and − 1 respectively. Furthermore,

1. when L_cf = f, then one of the following can occur:

   • If a₀ = 1, then k ≥ 2 and deg(h) ≤ (k − 2);

   • If a₀ ≠ 1, then deg(h) ≤ (k − 1).

2. When L_cf = −f, then one of the following occur:

   • if a₀ = −1, then k ≥ 2 and deg(h) ≤ (k − 2);

   • if a₀ ≠ − 1, then deg(h) ≤ (k − 1).

Remark 2.1.

In the above theorem, if we choose L_cf = Δ_cf, then k = 1, a₁ = 1 and a₀ = −1. Therefore conclusion (2) that means Δ_cf = −f is not possible. Thus from conclusion (1) we only have the form of the function as f(z) = Ae^μz, where A and μ are nonzero constants satisfying e^μc = 2 and also Δ_cf = f holds.

The following examples will successively show that in the above theorem, respectively for the cases k = 1, k = 2 and k = 3, all possible forms of the function exist.

First we consider the case k = 1.

Example 2.1.

Let f = Ae^μz, A ≠ 0. Choosing coefficients of L_cf for k = 1 as a1=1−a0eμc and a₀ ≠ 1 we have L_cf = f and ∑j=01ajeμcj=1. Next, choosing coefficients as a1=−1−a0eμc and a₀ ≠ − 1 we see that L_cf = −f and ∑j=01ajeμcj=−1.

Next we shall show that for k = 2, the forms of the function can be obtained.

Example 2.2.

Let f = (Az + B)e^μz, A ≠ 0. Choosing coefficients of L_cf for k = 2 as a2=−(1−a0)e2μc, a1=2(1−a0)eμc, one can easily check that L_cf = f and ∑j=02ajeμcj=1. On the other hand, choosing coefficients as a2=1+a0e2μc, a1=−2(1+a0)eμc, we easily can obtain L_cf = −f and ∑j=02ajeμcj=−1.

Example 2.3.

Consider the function f in Example 2.1. Choose coefficients of L_cf for k = 2 as a2=−4e2μc, a1=3eμc, a₀ = 2, then we have L_cf = f and ∑j=02ajeμcj=1. Next, choosing coefficients as a2=−6e2μc, a1=3eμc, a₀ = 2, we see that L_cf = −f and ∑j=02ajeμcj=−1.

For k = 3, all possible forms of the function are shown below.

Example 2.4.

Let f = (Az² + Bz + C)e^μz, A ≠ 0. Choosing coefficients of L_cf for k = 3 as a3=1−a0e3μc, a2=−3(1−a0)e2μc, a1=3(1−a0)eμc, one can easily check that L_cf = f and ∑j=03ajeμcj=1. Also, choosing coefficients as a3=−1−a0e3μc, a2=−3(−1−a0)e2μc, a1=3(−1−a0)eμc, one can easily check that L_cf = −f and ∑j=03ajeμcj=−1.

Example 2.5.

Consider the function f as in Example 2.2 and choosing the coefficients of L_cf as a3=5e3μc, a2=−9e2μc, a1=3eμc, a₀ = 2, we can have L_cf = f and ∑j=03ajeμcj=1. On the other hand, choosing coefficients as a3=9e3μc, a2=−15e2μc, a1=3eμc, a₀ = 2, we can get L_cf = −f and ∑j=03ajeμcj=−1.

Example 2.6.

Consider the function f in Example 2.1. Choosing coefficients of L_cf for k = 3 as a3=−2e3μc, a2=4e2μc, a1=−3eμc, a₀ = 2, clearly L_cf = f and ∑j=03ajeμcj=1. On the other hand, choosing coefficients as a3=−4e3μc, a2=4e2μc, a1=−3eμc, a₀ = 2, we easily get L_cf = −f and ∑j=03ajeμcj=−1.

Similar examples can be constructed for the case k ≥ 4 also.

Remark 2.2.

In Examples 2.4, 2.5, let us take Lcf=Δc3f. Choosing e^μc = 2 we see that though ∑j=03ajeμcj=(eμc−1)3=1 but Δc3f≠f. In a similar manner, for the function in Examples 2.4, 2.5, choosing eμc=3+3i2 we get ∑j=03ajeμcj=(eμc−1)3=−1 but Δc3f≠−f. But in Example 2.6, choosing e^μc such that ∑j=03ajeμcj=1 or − 1, we automatically have the respective conclusions Δc3f=f or Δc3f=−f. From this observation naturally one can infer that the case Lcf=Δckf needs special attention. In fact, we can conjecture that the degree of h could be zero that means h will be a nonzero constant.

In this respect, in Theorem 2.1 replacing L_cf by Δckf we can get the next corollary.

Corollary 2.1.

Under the same assumptions of Theorem 2.1 for the operator Δckf we have Δckf=f. In addition, if b = −a and k ≥ 2, then Δckf=−f. In both cases f takes the form f(z) = Be^μz, where B and μ are nonzero constants and μ satisfies ∑j=0k(−1)k−jkjeμcj=1 and − 1 respectively.

Next we provide two examples to show that ρ₂ < 1 is sharp.

Example 2.7.

Let f=eπizcee2πizc. Here λ(f) < ρ(f) and ρ₂(f) = 1. For a suitable choices of coefficients one can obtain Lcf=−eπizcee2πizc. For example, for even integer k, choose a_k + ⋯ + a₂ + a₀ = 0 and a_k−1 + ⋯ + a₃ + a₁ = 1. Clearly f and L_cf share the set {a, −a} CM, where a is an entire function such that ρ(a) < ρ(f). Though L_cf = −f, the form of f does not satisfy the conclusion of our theorem.

Example 2.8.

Let f=eeπizc. Here λ(f) < ρ(f) and ρ₂(f) = 1. For a suitable choices of coefficients, one can easily obtain Lcf=e−eπizc. For example, for even integer k, choose a_k + ⋯ + a₂ + a₀ = 0 and a_k−1 + ⋯ + a₃ + a₁ = 1. Clearly f and L_cf share the set {a+b,a−b} CM, where a, b are two complex constants such that a − b = 1. Then neither L_cf = ±f nor the form of f satisfies the conclusion of our theorem.

3. Preparatory lemmas

In this section, some useful lemmas are quoted from references [1, 7, 9–12], which will be needed in the sequel.

Lemma 3.1.

[9] Let T: [0, + ∞) → [0, + ∞) be a nondecreasing continuous function, and let s ∈ (0, + ∞). If the hyper-order of T is strictly less than 1,that is,

lim supr→∞log⁡logT(r)log⁡r=ρ2<1,

and δ ∈ (0, 1 − ρ₂), then

T(r+s)=T(r)+oT(r)rδ,

where r runs to infinity outside of a set of finite logarithmic measures.

Lemma 3.2.

[9] Let f(z) be a meromorphic function of ρ₂(f) < 1 and c∈C\{0}. Then

mr,f(z+c)f(z)=oT(r,f)r1−ρ2−ϵ.

Using the above two basic lemmas due to [9], we have the next lemma.

Lemma 3.3.

Let f(z) be a meromorphic function of ρ₂(f) < 1 and c∈C\{0}. Then for any ϵ > 0,

mr,f(z+ic)f(z+jc)=oT(r,f)r1−ρ2−ϵ.

Using Lemma 3.1, by a simple alteration of the result for finite-order meromorphic functions in [8], one can have the following lemma.

Lemma 3.4.

Let f(z) be a meromorphic function of ρ₂(f) < 1, then we have

N(r,f(z+c))=N(r,f)+S(r,f)

and

T(r,f(z+c))=T(r,f)+S(r,f).

Lemma 3.5.

[7] Let f be a transcendental meromorphic function in the plane of order less than 1. Let h > 0. Then there exists an ϵ-set E such that

g(z+c)g(z)→1,whenz→∞inC\E,

uniformly in c for |c| ≤ h.

Lemma 3.6.

[1, 10] Let f(z) be a transcendental meromorphic solution of equation

fnA(z,f)=B(z,f),

where A(z, f), B(z, f) are polynomials in f and its derivatives with meromorphic coefficients, say {a_λ: λ ∈ I}, such that m(r, a_λ) = S(r, f) for all λ ∈ I. If the total degree of B(z, f) is ≤ n, then

m(r,A(z,f))=S(r,f).

Lemma 3.7.

(see [12], Theorem 1.51) Suppose that f_i(z) (i = 1, 2, …, n) and g_i(z) (i = 1, 2, …, n) (n ≥ 2) are entire functions satisfying

1. ∑i=1nfi(z)egi(z)≡0,

2. g_j(z) − g_k(z) are not constants for 1 ≤ j < k < n,

3. for 1 ≤ i ≤ n, 1 ≤ k < l ≤ n,

T(r,fi)=o{T(r,egk−gl)}(r→∞,r∉E).

Then f_i(z) ≡ 0 (i = 1, 2, …, n).

Now we recall the following lemma due to Lu-Lu-Li-Xu (see [11], Corollary 3.2).

Lemma 3.8.

[11] Let g (≢0) be a nonconstant meromorphic solution of the linear difference equation

(3.1). ∑i=0kbig(z+ic)=R(z),

where R(z) is a polynomial and b_i′s for i = 0, 1, …k are complex constants with b_kb₀ ≠ 0, c∈C\{0} and k∈N. Then either ρ(g) ≥ 1 or g is a polynomial. In particular if b_k ≠ ± b₀, then ρ(g) ≥ 1.

4. Proof of the main theorem

Proof of Theorem 2.1.

According to our assumption λ(f) < ρ(f) and by Hadamard factorization theorem, let us assume that f(z) = h(z)e^η(z), where h(z) (≢0) is an entire function and η(z) is a nonconstant entire function satisfying

λ(f)=ρ(h)<ρ(f),ρ(eη)=ρ(f),ρ(η)=ρ2(eη)=ρ2(f)<1.

Therefore T(r, h) = S(r, f) and S(r, e^η) = S(r, f) = S(r). Here,

(4.1)T(r,f)≤T(r,h)+T(r,eη)≤T(r,eη)+S(r).

Let q(z)=Lcfeη. Clearly q ≢ 0. Then placing f(z) = h(z)e^η(z), in view of Lemma 3.4, we can deduce that

T(r,q)=m(r,q)=mr,Lc(h(z)eη(z))eη(z)=mr,∑j=0kajh(z+jc)eη(z+jc)−η(z)≤∑j=0kmr,eη(z+jc)eη(z)+S(r)=S(r,eη)=S(r).

Since q and h are not equivalent to zero, one can easily write

(4.2)(qeη−a)(qeη−b)(heη−a)(heη−b)=q2eη−aqeη−bqh2eη−aheη−bh.

Applying the Second Fundamental Theorem for small functions [1] on e^η and then applying the First Fundamental Theorem [1] on e^η − ω, we can obtain that

(4.3)T(r,eη)=N¯r,1eη−ω+S(r),

where ω (≢0) is a small function of e^η.

Here a≢b. Without loss of generality, let us assume that a≢0. Let z₀ be a zero of eη−ah but q(z₀) ≠ 0. Since f and L_cf that means he^η and qe^η share the set {a, b} IM, so in view of (4.2), z₀ is a zero of eη−aq or eη−bq. Let us denote by N¯(r,0;eη−ah,eη−aq) the reduced counting function of those common zeros of eη−ah and eη−aq, which are not zeros of q. Similarly, we denote by N¯(r,0;eη−ah,eη−bq) the reduced counting function of those common zeros of eη−ah and eη−bq, which are not zeros of q. Therefore from (4.3) we have,

T(r,eη)=N¯r,1eη−ah+S(r)=N¯r,0;eη−ah,eη−aq+N¯r,0;eη−ah,eη−bq+S(r),

which shows that either N¯r,0;eη−ah,eη−aq≠S(r) or N¯r,0;eη−ah,eη−bq≠S(r). Otherwise T(r, e^η) = S(r). This is not possible because in view of (4.1), we can draw a contradiction. Now, we consider two cases:

• Case 1. Suppose N¯r,0;eη−ah,eη−aq≠S(r). For sake of convenience, we resolve the case step by step.

• Step 1. In this step we will show that L_cf ≡ f.

Let z₁ is a zero of eη−ah and eη−aq. It is obvious that z₁ is a zero of ah−aq. If ah−aq≢0, then

N¯r,0;eη−ah,eη−aq≤N¯r,1ah−aq=S(r),

which is a contradiction. Therefore h = q that implies L_cf = f.

• Step 2. In this step we show that η(z) is a polynomial.

Expanding L_cf = f we can write

(4.4)∑j=0kajh(z+jc)eη(z+jc)=h(z)eη(z).

Choosing b_j = a_jh(z + jc) for j = 1, 2, …, k and b₀ = (a₀ − 1)h(z) we get ∑j=0kbjeη(z+jc)=0. Clearly ρ(b_l) = ρ(h(z + lc)) = ρ(h) < ρ(f) for all l = 0, 1, …, k. So, b_l′s are finite-order entire functions. We claim that ρ(e^{η(z + ic)−η(z + jc)}) < ∞, for at least one pair of i, j; such that 0 ≤ i < j ≤ k. On the contrary, let us suppose, for all 0 ≤ i < j ≤ k, ρ(e^{η(z + ic)−η(z + jc)}) = ∞. Then T(r, b_l) = o{T(r, e^{η(z + jc)−η(z + ic)})}, for 0 ≤ l ≤ k and 0 ≤ i < j ≤ k. Hence by Lemma 3.7, b_l ≡ 0 for all l = 0, 1, …, k, which is not possible. Thereby ρ(e^{η(z + ic)−η(z + jc)}) < ∞ implies that η(z + ic) − η(z + jc) is a polynomial. Let the degree of η(z + ic) − η(z + jc) be m. So, η^(m+1)(z + ic) − η^(m+1)(z + jc) = 0 that means η^(m+1)(z + ic) is periodic entire function of period (j − i)c. If η^(m+1)(z + ic) is nonconstant, then obviously ρ(η^(m+1)(z + ic)) ≥ 1, which yields ρ(η) = ρ(η(z + ic)) = ρ(η^(m+1)(z + ic)) ≥ 1. But ρ(η) < 1, a contradiction. Hence η^(m+1)(z + ic) is a constant and so η^(m+1)(z) is constant, which implies η(z) is polynomial.

• Step 3. In this step we wish to show that the degree of η(z) is 1.

On the contrary, suppose the deg(η(z)) = n(say) ≥ 2. Then for j = 1, 2, …, k,

eη(z+jc)−η(z)=ejcncnzn−1+Qj(z),

where Q_j(z) is a (n − 2)-th degree polynomial and c_n is the leading coefficient of η(z). Let g=ecncnzn−1. So, for j = 1, 2, …, k, eη(z+jc)−η(z)=gjeQj(z). Clearly T(r,eQj−Qk)=S(r,g) for all j = 1, 2, …, k − 1 and T(r,e−Qk)=S(r,g). Here we will draw a contradiction by deducing T(r, g) = S(r, g). Rewriting (4.4) we have

∑j=1kajh(z+jc)h(z)eQj(z)gj=1−a0.

i.e.

(4.5)gk−1g=1−a0akh(z)h(z+kc)e−Qk(z)−∑j=1k−1ajakh(z+jc)h(z+kc)eQj(z)−Qk(z)gj.

As η(z) is a polynomial, so ρ₂(h) ≤ ρ₂(e^η) = ρ(η) = 0, which implies ρ₂(h) = 0. Since ρ(h) − 1 < ρ(f) − 1 = ρ(e^η) − 1 = n − 1 = ρ(g), so by Lemma 3.3, for any ϵ > 0, mr,h(z+jc)h(z+kc)=oT(r,h)r1−ϵ=S(r,g), j = 0, 1, …, k − 1.

Let H(z,g)=∑j=0k−1Cjgj, where Cj=ajakh(z+jc)h(z+kc)eQj(z)−Qk(z) for j = 1, 2, …, k − 1 and C0=1−a0akh(z)h(z+kc)e−Qk(z). Thus, (4.5) can be written as g^k−1g = H(z, g). Clearly total degree of H(z, g) is at most k − 1 and m(r, C_j) = S(r, g) for j = 0, 1, … k − 1. Hence by Lemma 3.6, m(r, g) = S(r, g) that means T(r, g) = S(r, g), a contradiction. Therefore deg(η(z)) = 1. Let us assume that η(z) = μz + C, where μ and C be two nonzero constants.

• Step 4. In this step we deduce a necessary condition and actual form of the function.

Putting η(z) = μz + C in f(z) = h(z)e^η(z), we have f(z) = Ah(z)e^μz, where A = e^C is a nonzero constant. Now, applying this, (4.4) can be written as

(4.6)∑j=0kajh(z+jc)eμcj=h(z).

i.e.

∑j=0kajh(z+jc)h(z)eμcj=1.

Since ρ(h) < ρ(f) = ρ(e^μz) = 1, so by Lemma 3.5, there exist ϵ-set E, as z ∉ E and z → ∞, such that h(z+jc)h(z)→1. Thereby,

(4.7)∑j=0kajeμcj=1.

Since ρ(h) < 1, so by Lemma 3.8, we know that h(z) is a polynomial. Let us assume that h(z) = c_lz^l + c_l−1z^l−1 + ⋯ + c₁z + c₀. Putting it into (4.6) and then comparing coefficients and doing a simple calculation, we have (4.7) and

l∑j=1kjajeμcj=0,(l−1)∑j=1kj2ajeμcj=0,(l−2)∑j=1kj3ajeμcj=0,……,(l−(l−1))∑j=1kjlajeμcj=0.

Without loss of generality, we assume that all a_i′ s for i = 1, 2, …, k are nonzero. Now, the above system of equations can be written as

(4.8)A1X=B,

where A1=111…1123…k122232…k2………1l2l3l…kl(l+1)×k, X=a1eμca2e2μca3e3μc…akekμck×1 and B=1−a000…0(l+1)×1.

Let C be the corresponding augmented matrix. It is obvious that rank(A₁) = min{l + 1, k}. Clearly rank(C) = min{l + 1, k + 1}.

Suppose a₀ ≠ 1. So the nonhomogeneous system (4.8) has unique solution when l = k − 1, infinitely many solutions when l < k − 1 and no solutions when l > k − 1. Hence deg(h) ≤ (k − 1).

Next suppose a₀ = 1. Then for k = 1, L_cf = f and (4.7) both implies a₁ = 0, which is not possible. So in this case obviously k ≥ 2. Now, the homogeneous system A₁X = 0 has solutions when l ≤ k − 2. Thus, we have our desired Conclusion 1.

• Case 2. Suppose N¯r,0;eη−ah,eη−bq≠S(r). Let z₂ is a zero of eη−ah and eη−bq. It is obvious that z₂ is a zero of ah−bq. If ah−bq≢0, then

N¯r,0;eη−ah,eη−bq≤N¯r,1ah−bq=S(r),

which is a contradiction. Therefore

(4.9)ah=bq.

Since a≢0, therefore b ≢ 0. Let z₃ be a zero of eη−bh but q(z₀) ≠ 0. Since he^η and qe^η share the set {a, b} IM, so z₃ is a zero of eη−aq or eη−bq. Let us denote by N¯(r,0;eη−bh,eη−aq) the reduced counting function of those common zeros of eη−bh and eη−aq, which are not zeros of q. similarly, we denote by N¯(r,0;eη−bh,eη−bq) the reduced counting function of those common zeros of eη−bh and eη−bq, which are not zeros of q. Therefore from (4.3) we have,

T(r,eη)=N¯r,1eη−bh+S(r)=N¯r,0;eη−bh,eη−aq+N¯r,0;eη−bh,eη−bq+S(r),

which shows that either N¯r,0;eη−bh,eη−aq≠S(r) or N¯r,0;eη−bh,eη−bq≠S(r). Otherwise T(r, e^η) = S(r), which is not possible in view of (4.1). Now, we consider two subcases:

Subase 2.1. Suppose N¯r,0;eη−bh,eη−aq≠S(r). Then proceeding in a similar manner as used in starting portion of Case 2, we have bh=aq. In view of (4.9) we get, a² = b². As a≢b, so obviously b = −a. Therefore we must have q = −h that implies L_cf = −f. Further following the same steps as done in Case 1, we can have the form of the function as f(z) = Ah(z)e^μz satisfying

(4.10)∑j=0kajeμcj=−1.

Next adopting the similar calculations as done in Step 4, for a₀ ≠ − 1, we have deg(h) ≤ (k − 1) and for a₀ = −1, we have k ≥ 2 and deg(h) ≤ (k − 2). Thus, we have corresponding desired conclusion (2).

Subcase 2.2. Suppose N¯r,0;eη−bh,eη−bq≠S(r), which is similar to the Case 1 and so, we get the desired result.

Hence the proof is completed. □

Proof of Corollary 2.1.

To prove this corollary, it is sufficient to prove that deg(h) = 0, where h(z) = c_lz^l + c_l−1z^l−1 + ⋯ + c₁z + c₀ (l ≤ k − 1), c_l ≠ 0. We know for the operator Δckf, aj=(−1)k−jkj, where j = 0, 1, …, k.

From conclusion 1. of Theorem 2.1 we have, Δckf=f and f takes the form f = Ah(z)e^μz, where A is a nonzero constant, h(z) is a polynomial and μ is a nonzero constant satisfying ∑j=0k(−1)k−jkjeμcj=1, that is, (eμc−1)k=1. Now putting f = A(c_lz^l + c_l−1z^l−1 + ⋯ + c₁z + c₀)e^μz into Δckf=f and then comparing coefficient of z^k−1, we have

∑j=1k(lcljc+cl−1)(−1)k−jkjeμcj=cl−1⇒lclc∑j=1kj(−1)k−jkjeμcj+cl−1(eμc−1)k=cl−1.

As c ≠ 0, c_l ≠ 0 and (eμc−1)k=1 and also we know jkj=kk−1j−1, so we get

(4.11)l∑j=1k(−1)k−jkk−1j−1eμcj=0⇒l∑j=0k−1(−1)(k−1)−jk−1jeμcj=0⇒l(eμc−1)k−1=0.

Clearly in view of (eμc−1)k=1, l must be 0.

From conclusion 2. of Theorem 2.1 we have, b = −a, Δckf=−f and f takes the form f = Ah(z)e^μz, where A is a nonzero constant, h(z) is a polynomial and μ is a nonzero constant satisfying ∑j=0k(−1)k−jkjeμcj=−1, that is, (eμc−1)k=−1. Here obviously k ≥ 2. Proceeding in a similar manner as in above, again we can have (4.11) and in view of (eμc−1)k=−1, we can conclude l = 0. So in both cases l = 0 that means deg(h) = 0. Hence the corollary is proved. □

5. Observation and an open question

As we know λ(f) ≤ ρ(f) and since throughout the paper we have dealt with the case λ(f) < ρ(f), it will be interesting to inspect whether the same conclusions hold for the case λ(f) = ρ(f). In the next two examples we point out the fact that when λ(f) = ρ(f), the conclusion of Theorem 2.1 ceases to hold.

Example 5.1.

Let f = e^z(e^2z + 1). Choose c = πi and for even integer k, a_k + ⋯ + a₂ + a₀ = 0 and a_k−1 + ⋯ + a₃ + a₁ = 1. Then λ(f) = ρ(f) = 1 and L_cf = −e^z(e^2z + 1). Clearly L_cf and f share the set {a, −a} CM, where a is an entire function such that ρ(a) < ρ(f). Though L_cf = −f, the form of f does not satisfy the conclusion of Theorem 2.1.

Example 5.2.

Let f = −e^z + 3 and (ec−1)2=−1. Then Δc2f=ez and λ(f) = ρ(f) = 1. Clearly Δc2f and f share the set {1, 2} CM but Δc2f≠±f.

In view of the above two examples, we can conclude that in Theorem 2.1, λ(f) < ρ(f) is sharp, but the conclusion of the same theorem under the case λ(f) = ρ(f) is still an enigma. So we place it as an open question:

Question 5.1.

Under the hypothesis λ(f) = ρ(f), what will be the answer of the Question 1.1 concerning Δckf or even L_cf?