Differential polynomials of L-functions with truncated shared values

Definition 1.1

Let f_i(i = 1, 2) be two nonconstant meromorphic functions, and let k ≥ 2 be a positive integer. If E_k)(1, f₁) = E_k)(1, f₂), we say that 1 is a truncated sharing value of f₁ and f₂.

Let f₁ be a nonconstant meromorphic function, let k ≥ 2 be a positive integer and a ∈ ℂ ⋃{∞}. We denote by N_(k r,1f1−a the counting function of those a-points of f₁ (counted with proper multiplicities) whose multiplicities are greater than k, denote by N_k) r,1f1−a the counting function of those a-points of f₁ (counted with proper multiplicities) whose multiplicities are less than k. Naturally, denote by N_(k r,1f1−a and N_k) r,1f1−a the reduced forms respectively.

Definition 1.2

Let f₁ be a nonconstant meromorphic function, and let k be a positive integer and b ∈ ℂ ⋃ {∞}. We define

Remark 1

From (1.1) and (1.2), we can get that 0 ≤ δ_k(b, f₁) ≤ δ_k–1(b, f₁) ≤ δ₁(b, f₁) ≤ Θ(b, f₁) ≤ 1.

Definition 1.3

Let f_i(i = 1, 2) be two nonconstant meromorphic functions such that E_k)(1, f₁) = E_k)(1, f₂), where k ≥ 2 is a positive integer. Let z₀ be a common zero of f₁ – 1 and f₂ – 1 with multiplicity p and q respectively, we denote by N¯Lr,1f1−1 the reduced counting function of those zeros of f₁ – 1 where p > q, by N¯2Er,1f1−1 the reduced counting function of those zeros f₁ – 1 where p = q ≥ 2. Especially, we denote by N¯L1r,1f1−1 the reduced counting function of those zeros of f₁ – 1 where p > q = 1, denote by N1Er,1f1−1 the reduced counting function of those zeros of f₁ – 1 where p = q = 1. In the same way, we can define N¯Lr,1f2−1,N¯2Er,1f2−1,N¯L1r,1f2−1,N1Er,1f2−1.

Definition 1.4

Let f_i(i = 1, 2) be two nonconstant meromorphic functions such that E_k)(1, f₁) = E_k)(1, f₂), where k ≥ 2 is a positive integer. We denote by N_(k+1(r, 1; f₁|f₂ ≠ 1) the reduced counting function of those zeros of f₁ – 1 where multiplicities are greater than k + 1, but not the zeros of f₂ – 1. In the same way, we can define N_(k+1(r, 1; f₂|f₁ ≠ 1).

Definition 1.5

Let f_i(i = 1, 2) be two nonconstant meromorphic functions. We denote by N₀ r,1f1′ the counting function of those zeros of f1′ which are not the zeros of f₁ and f₁ – 1, by N₀ r,1f1′ the corresponding reduced counting function. In the same way, we can define N₀ r,1f2′ and N₀ r,1f2′ .

Theorem 1.1

(see [3]) Let f and g be two nonconstant entire functions, and let n, k be two positive integers with n > 5k + 13. If (fⁿ(f – 1))^(k) and (gⁿ(g – 1))^(k) share 1 IM, then f ≡ g.

Theorem 1.2

(see [4]) Let f and g be two nonconstant entire functions, and let n, k and m ≥ 2 be three positive integers with n > k + (4k + 7)(1 – Θ(0)) + 4(1 – δ(1)). If E_m)(1, (fⁿ(f – 1))^(k)) = E_m)(1, (gⁿ(g – 1))^(k)), then f ≡ g.

Theorem 1.3

(see [5]) Let f and g be two nonconstant meromorphic functions such that λ(f) > 2, and let n, k be two positive integers with n > 9k + 18. If (fⁿ(f – 1))^(k) and (gⁿ(g – 1))^(k) share 1 IM, and Θ(∞, f) > 2/n, then f ≡ g.

Theorem 1.4

(see [6]) Let f and g be two nonconstant meromorphic functions such that λ(f) > 2, and let n, k and m ≥ 2 be three positive integers with n > 9k + 18. If E_m)(1, (fⁿ(f – 1))^(k)) = E_m)(1, (gⁿ(g – 1))^(k)), and Θ(∞, f) > 2/n, then f ≡ g.

Theorem 1.5

(see [7]) Let f be a transcendental meromorphic function, L be an L-function, n, k be two positive integers with n > 7k + 17 and k ≥ 2. If (fⁿ(f – 1))^(k) and (Lⁿ(L – 1))^(k) share 1 IM, then f ≡ L.

Theorem 1.6

Let f be a transcendental meromorphic function, L be an L-function, n, k, l ≥ 2 be three positive integers with n > 7k + 17 and k ≥ 2. If E_l)(1, (fⁿ(f – 1))^(k)) = E_l)(1, (Lⁿ(L – 1))^(k)), then f ≡ L.

Lemma 2.1

(see [8]) Let f₁ be a nonconstant meromorphic function, and let φ(≠ 0, ∞) be a small function of f₁. Then

Lemma 2.2

(see [9]) Let f₁ be a nonconstant meromorphic function, k(≥ 1), p(≥ 1) be two positive integers. Then

Lemma 2.3

(see [4]) Let f_i(i = 1, 2) be two nonconstant meromorphic functions, l(≥ 2) be positive integer. If f₁ and f₂ satisfy E_l)(1, f₁) = E_l)(1, f₂), then

Lemma 2.4

Let f_i(i = 1, 2) be two transcendental meromorphic functions such that E_l)(0, f1(k) – H) = E_l)(0, f2(k) – H), where k(≥ 1), l(≥ 2) are two positive integers, H is a nonzero polynomial. If

then f1(k)f2(k) ≡ H² or f₁ ≡ f₂.

Proof

Let

and

First of all, f₁, f1(k) are two transcendental meromorphic functions; therefore, T(r, H) = o{T(r, f₁)}. By the lemma of logarithmic derivative and Nevanlinna first fundamental theorem, we have

Therefore, S(r, f͠₁) ≤ S(r, f₁). In the same way, S(r, f͠₂) ≤ S(r, f₂).

Next, we claim F ≡ 0. Suppose that F ≢ 0. Let z₀ ∉ {z : H(z) = 0} be a common simple zero of f1(k) – H and f2(k) – H. From (2.3) we know that z₀ is a common simple zero of f͠₁ – 1 and f͠₂ – 1, which by calculating yields F(z₀) = 0. Thus, we have

Let z₁ ∉ {z : H(z) = 0} be a simple pole of f͠₁. Then by calculating we get that f1~″f1~′−2f1~′f1~−1 is analytic at z₁. Similarly, let z₂ ∉ {z : H(z) = 0} be a simple pole of f͠₂. Then by calculating we get that f2~″f2~′−2f2~′f2~−1 is analytic at z₂. In addition, each common zero of f͠₁ – 1 and f͠₂ – 1 with the same multiplicities is not the pole of F. From (2.4), we can get

We note that

It follows from the above two inequalities that

We deduce by the Nevanlinna first fundamental theorem that

From this, (2.3), and (2.5), we obtain

It follows from the lemma of logarithmic derivative and (2.6) that

Applying Lemma 2.1, we have

In the same way, we have

By Lemma 2.3, we get

Similarly,

Combining (2.8) with (2.9) we have

which together with (2.7) yields

By (2.10) and (2.11), we get

Applying Lemma 2.2, we can get

In the same way, we have

Suppose that there exists some subset I ⊆ R⁺, mesI = ∞ such that

as r ∈ I and r → ∞. Then by (2.12) and (2.14) we have

which contradicts the condition ι₁ > 4k + 13.

Similarly, if there exists some subset I ⊆ R⁺, mesI = ∞ such that

as r ∈ I and r → ∞, then by (2.13) and (2.15) we have

which contradicts the condition ι₂ > 4k + 13.

Therefore, we get F ≡ 0 and so by (2.4) it follows that

Integrating this equation, we obtain

where τ₁, τ₂ are two constants which are not equal to zero at the same time.

We discuss the following three cases.

1. If τ₂ = 0, by calculating, we can get from (2.3) and (2.16)

   f2=τ1f1+(1−τ1)H1, (2.17)

   where H₁ is a polynomial of degree deg(H₁) ≤ deg(H) + k. From (2.17) we get T(r, f₁) = T(r, f₂) + O(log r). If τ₁ ≠ 1, combining the Nevanlinna’s three small functions theorem with (2.17), we have

   T(r,f2)≤N¯(r,f2)+N¯r,1f2+N¯r,1f2−(1−τ1)H1+S(r,f2)=N¯(r,f2)+N¯r,1f2+N¯r,1f1+S(r,f2). (2.18)

   Therefore, we can obtain

   Θ(∞,f2)+Θ(0,f2)+Θ(0,f1)≤2. (2.19)

   By (2.1) we have

   ι1=(2k+4)Θ(∞,f1)+(2k+3)Θ(∞,f2)+3δk+1(0,f1)+2δk+1(0,f2)+Θ(0,f1)+Θ(0,f2)≤4k+13,

   which contradicts the condition ι₁ > 4k + 13. Hence, τ₁ = 1. Substituting this into (2.17) we get f₁ ≡ f₂. The second conclusion of Lemma 2.4 holds.

2. If τ₂ ≠ 0 and τ₁ = τ₂. We consider the following two subcases.

3. If τ₁ = τ₂ = –1, from (2.3) and (2.16) we get f1(k)f2(k) ≡ 1, the first conclusion of Lemma 2.4 holds.

4. If τ₁ = τ₂ ≠ –1, then, by (2.16) we have

   f1~≡(1+τ2)f2~−1τ2f2~,    f2~≡−1τ21f1~−1−1τ2. (2.20)

   From the right equality of (2.20) and (2.3) we get

   N¯r,1f1~−1−1τ2=N¯(r,f2)+O(log⁡r).

   From Lemma 2.1 and the above equality, we obtain

   T(r,f1)≤N¯(r,f1)+Nr,1f1+Nr,1f1~−1−1τ2−Nr,1f1~′+S(r,f1)≤N¯(r,f1)+Nk+1r,1f1+N¯r,1f1~−1−1τ2−N0r,1f1~′+O(log⁡r)+S(r,f1)≤N¯(r,f1)+Nk+1r,1f1+N¯(r,f2)+S(r,f1). (2.21)

   On the other hand, from the left equality of (2.20) and (2.3) we get

   N¯r,1f2~−1τ2+1=N¯r,1f1(k)+O(log⁡r).

   From Lemma 2.1, Lemma 2.2 and the above equality, we have

   T(r,f2)≤N¯(r,f2)+Nr,1f2+Nr,1f2~−1τ2+1−Nr,1f2~′+S(r,f2)≤N¯(r,f2)+Nk+1r,1f2+N¯r,1f2~−1τ2+1−N0r,1f2~′+O(log⁡r)+S(r,f2)≤N¯(r,f2)+Nk+1r,1f2+N¯r,1f1(k)+S(r,f2)≤N¯(r,f2)+Nk+1r,1f2+Nk+1r,1f1+kN¯(r,f1)+S(r,f1)+S(r,f2). (2.22)

   If (2.15) holds, from (2.22) we get

   Θ(∞,f2)+δk+1(0,f2)+δk+1(0,f1)+kΘ(∞,f1)≤k+2. (2.23)

   Substituting this into (2.2) we obtain

   ι2=(2k+4)Θ(∞,f2)+(2k+3)Θ(∞,f1)+3δk+1(0,f2)+2δk+1(0,f1)+Θ(0,f2)+Θ(0,f1)≤4k+13,

   which contradicts the condition ι₂ > 4k + 13.

   If (2.14) holds, from (2.21) we get

   Θ(∞,f1)+δk+1(0,f1)+Θ(∞,f2)≤2. (2.24)

   Substituting this into (2.1) we obtain

   ι1=(2k+4)Θ(∞,f1)+(2k+3)Θ(∞,f2)+3δk+1(0,f1)+2δk+1(0,f2)+Θ(0,f1)+Θ(0,f2)≤4k+13,

   which contradicts the condition ι₁ > 4k + 13.

5. If τ₂ ≠ 0 and τ₁ ≠ τ₂. We consider the following two subcases.

6. If τ₂ = –1, then τ₁ ≠ 0. Otherwise, from (2.16) we have 1f1~−1=−f2~+1f2~−1. Thus f͠₁ ≡ 0, which contradicts that f₁ is a transcendental meromorphic function.

   From (2.16) we have

   f1~≡τ1−f2~+τ1+1, (2.25)

   f2~≡(τ1+1)f1~−τ1f1~. (2.26)

   By (2.3) and (2.25) we obtain

   N¯r,τ1−f2~+τ1+1=N¯(r,f1)+O(log⁡r).

   From Lemma 2.1, we have

   T(r,f2)≤N¯(r,f2)+Nr,1f2+Nr,1f2~−(τ1+1)−Nr,1f2~′+S(r,f2)≤N¯(r,f2)+Nk+1r,1f2+N¯r,1f2~−(τ1+1)−N0r,1f2~′+O(log⁡r)+S(r,f2)≤N¯(r,f2)+Nk+1r,1f2+N¯(r,f1)+S(r,f2).

   If (2.15) holds, using the same argument as in Case 2, we have ι₂ ≤ 4k + 13, a contradiction.

   On the other hand, from (2.3) and (2.26) we have

   N¯r,1f1~−τ1(τ1+1)−1=N¯r,1f2~=N¯r,1f2(k)+O(log⁡r).

   Similarly, from Lemma 2.1, Lemma 2.2, we have

   T(r,f1)≤N¯(r,f1)+Nk+1r,1f1+Nk+1r,1f2+kN¯(r,f2)+S(r,f1)+S(r,f2).

   If (2.14) holds, using the same argument as in Case 2, we have ι₁ ≤ 4k + 13, a contradiction.

7. If τ₂ ≠ –1, then (2.16) can be rewritten as

   f1~−τ2+1τ2≡−τ1τ22⋅1f2~+(τ1−τ2)τ2−1,

   and

   f2~+τ1−τ2τ2≡−τ1τ22⋅1f1~−(τ2+1)τ2−1.

   Hence, combining these above equalities with (2.3), we have

   N¯r,1f2~+(τ1−τ2)τ2−1=N¯(r,f1)+O(log⁡r),

   and

   N¯r,1f1~−(τ2+1)τ2−1=N¯(r,f2)+O(log⁡r).

   Using the same argument as in Case 2, we have ι₂ ≤ 4k + 13 and ι₁ ≤ 4k + 13, a contradiction. Lemma 2.4 is proved.