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

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:

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

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

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.

After that Liu [4] posed a significant question:

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].

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

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:

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.

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).

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.

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

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

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

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

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

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

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

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

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).

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

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

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

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

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,

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

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

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,

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

i.e.

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

i.e.

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,

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

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

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

which is a contradiction. Therefore

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,

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

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. □

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