On Entire and Meromorphic Functions That Share One Small Function with Their Differential Polynomial

Abstract

We study the uniqueness of meromorphic functions that share one small function with more general differential polynomial 𝑃[𝑓]. As corollaries, we obtain results which answer open questions posed by Yu (2003).

1. Introduction and Main Results

In this paper, a meromorphic functions mean meromorphic in the whole complex plane. We use the standard notations of Nevanlinna theory (see [1]). A meromorphic function 𝑎(𝑧) is called a small function with respect to 𝑓(𝑧) if 𝑇(𝑟,𝑎)=𝑆(𝑟,𝑓), that is, 𝑇(𝑟,𝑎)=𝑜(𝑇(𝑟,𝑓)) as 𝑟→∞ possibly outside a set of finite linear measure. If 𝑓(𝑧)−𝑎(𝑧) and 𝑔(𝑧)−𝑎(𝑧) have the same zeros with same multiplicities (ignoring multiplicities), then we say that 𝑓(𝑧) and 𝑔(𝑧) share 𝑎(𝑧) CM (IM).

For any constant 𝑎, we denote by 𝑁_𝑘)(𝑟,1/(𝑓−𝑎)) the counting function for zeros of 𝑓(𝑧)−𝑎 with multiplicity no more than 𝑘 and 𝑁_𝑘)(𝑟,1/(𝑓−𝑎)) the corresponding for which multiplicity is not counted. Let 𝑁_(𝑘(𝑟,1/(𝑓−𝑎)) be the counting function for zeros of 𝑓(𝑧)−𝑎 with multiplicity at least 𝑘 and 𝑁_(𝑘(𝑟,1/(𝑓−𝑎)) the corresponding for which the multiplicity is not counted.

Let 𝑓 and 𝑔 be two nonconstant meromorphic functions sharing value 1 IM. Let 𝑧₀ be common one point of 𝑓 and 𝑔 with multiplicity 𝑝 and 𝑞, respectively. We denote by𝑁_𝐿(𝑟,1/(𝑓−1))  (𝑁_𝐿(𝑟,1/(𝑓−1))) the counting (reduced) function of those 1 points of 𝑓 where 𝑝>𝑞; by 𝑁_𝐸¹⁾(𝑟,1/(𝑓−1)) the counting function of those 1-points of 𝑓 where 𝑝=𝑞=1; by 𝑁_𝐸⁽²(𝑟,1/(𝑓−1)) the counting function of those 1-points of 𝑓 where 𝑝=𝑞≥2. In the same way, we can define 𝑁_𝐿(𝑟,1/(𝑔−1)), 𝑁_𝐸¹⁾(𝑟,1/(𝑔−1)) and 𝑁_𝐸⁽²(𝑟,1/(𝑔−1)) (see [2]).

In 1996, Brück [3] posed the following conjecture.

Conjecture 1. Let 𝑓 be a nonconstant entire function such that the hyper-order 𝜎₂(𝑓) of 𝑓 is not a positive integer and 𝜎₂(𝑓)<∞. If 𝑓 and 𝑓^′ share a finite value 𝑎 CM, then (𝑓^′−𝑎)/(𝑓−𝑎)=𝑐, where 𝑐 is a nonzero constant.

In [3], under an additional hypothesis, Brück proved that the conjecture holds when 𝑎=1.

Theorem A. Let 𝑓 be a nonconstant entire function. If 𝑓 and 𝑓^′ share the value 1 CM and if 𝑁(𝑟,1/𝑓^′)=𝑆(𝑟,𝑓), then (𝑓^′−1)/(𝑓−1)=𝑐, for some constant 𝑐∈ℂ⧵{0}.
Many people extended this theorem and obtained many results. In 2003, Yu [4] proved the following theorem.

Theorem B. Let 𝑘≥1. Let 𝑓 be a nonconstant meromorphic function and 𝑎(𝑧) a meromorphic function such that 𝑎(𝑧)≢0,∞, 𝑓 and 𝑎 do not have any common pole and 𝑇(𝑟,𝑎)=𝑜(𝑇(𝑟,𝑓)) as 𝑟→∞. If 𝑓−𝑎 and 𝑓^(𝑘)−𝑎 share the value 0 CM and 4𝛿0,𝑓+2(8+𝑘)Θ∞,𝑓>19+2𝑘, (1) then 𝑓≡𝑓^(𝑘).

Theorem C. Let 𝑘≥1. Let 𝑓 be a nonconstant entire function and 𝑎(𝑧) be a meromorphic function such that 𝑎(𝑧)≢0,∞ and 𝑇(𝑟,𝑎)=𝑜(𝑇(𝑟,𝑓)) as 𝑟→∞. If 𝑓−𝑎 and 𝑓^(𝑘)−𝑎 share the value 0 CM and 𝛿>30,𝑓4, (2) then 𝑓≡𝑓^(𝑘).

In the same paper, the author posed the following questions.

Question 1. Can 𝑎 CM shared value be replaced by an IM shared value in Theorem C?

Question 2. Is the condition 𝛿(0,𝑓)>3/4 sharp in Theorem C?

Question 3. Is the condition 4𝛿(0,𝑓)+2(8+𝑘)Θ(∞,𝑓)>19+2𝑘 sharp in Theorem B?

In 2004, Liu and Gu [5] applied different method and obtained the following theorem which answers some questions posed in [4].

Theorem D. Let 𝑘≥1. Let 𝑓 be a nonconstant meromorphic function and 𝑎(𝑧) a meromorphic function such that 𝑎(𝑧)≢0,∞ and 𝑇(𝑟,𝑎)=𝑆(𝑟,𝑓) as 𝑟→∞. If 𝑓−𝑎 and 𝑓^(𝑘)−𝑎 share the value 0 CM and 𝑓^(𝑘) and 𝑎(𝑧) do not have any common poles of same multiplicity and 2𝛿0,𝑓+4Θ∞,𝑓>5, (3) then 𝑓≡𝑓^(𝑘).

Theorem E. Let 𝑘≥1. Let 𝑓 be a nonconstant entire function and 𝑎(𝑧) a meromorphic function such that 𝑎(𝑧)≢0,∞ and 𝑇(𝑟,𝑎)=𝑆(𝑟,𝑓) as 𝑟→∞. If 𝑓−𝑎 and 𝑓^(𝑘)−𝑎 share the value 0 CM and 𝛿>10,𝑓2, (4) then 𝑓≡𝑓^(𝑘).

Recently, Zhang and Lü [6] considered the problem of meromorphic functions sharing one small function with its 𝑘th derivative and proved the following theorem.

Theorem F. Let 𝑘(≥1), 𝑛(≥1) be integers and 𝑓 a nonconstant meromorphic function. Also let 𝑎(𝑧)≢0,∞ be a small meromorphic function with respect to 𝑓. If 𝑓^𝑛 and 𝑓^(𝑘) share the value 𝑎(𝑧) IM and (2𝑘+6)Θ∞,𝑓+4Θ0,𝑓+2𝛿_𝑘+20,𝑓>2𝑘+12−𝑛 (5) or 𝑓^𝑛 and 𝑓^(𝑘) share the value 𝑎(𝑧) CM and (𝑘+3)Θ∞,𝑓+2Θ0,𝑓+𝛿_𝑘+20,𝑓>𝑘+6−𝑛, (6) then 𝑓≡𝑓^(𝑘).

Regarding these results, a natural question is what can be said when a nonconstant meromorphic function 𝑓 shares one nonzero small meromorphic function 𝑎(𝑧) with 𝑃[𝑓], where 𝑃[𝑓] is a differential polynomial in 𝑓.

Definition 2. Any expression of the type 𝑃𝑓=^𝑛_𝑖=1𝛼_𝑖(𝑧)𝑓^𝑛𝑖0𝑓^′𝑛𝑖1𝑓^{′′𝑛𝑖2}⋯𝑓^{(𝑚)𝑛𝑖𝑚} (7) is called differential polynomial in 𝑓 of degree 𝑑(𝑃), lower degree 𝑑(𝑃), and weight Γ_𝑃, where 𝑛_𝑖0,𝑛_𝑖1,…,𝑛_𝑖𝑚 are nonnegative integers, 𝛼_𝑖=𝛼_𝑖(𝑧) are meromorphic functions satisfying 𝑇(𝑟,𝛼_𝑖)=𝑆(𝑟,𝑓) and 𝑑(𝑃)=max^𝑚_𝑗=0𝑛_𝑖𝑗,𝑑∶1≤𝑖≤𝑛(𝑃)=min^𝑚_𝑗=0𝑛_𝑖𝑗,Γ∶1≤𝑖≤𝑛_𝑃=max^𝑚_𝑗=0𝑗+1𝑛_𝑖𝑗∶1≤𝑖≤𝑛. (8)
Further, if 𝑑(𝑃)=𝑑(𝑃)=𝑛(say), then the differential polynomial 𝑃[𝑓] is called a homogeneous differential polynomial in 𝑓 of degree 𝑛.

Correspond to the above question, we obtain the following results, which extend and improve Theorems A–F and give answers to the questions possed by Yu [4] for more general differential polynomial.

Theorem 3. Let 𝑓 be a nonconstant meromorphic function and 𝑎(𝑧) be a small meromorphic function such that 𝑎(𝑧)≢0,∞. P[f] be a nonconstant differential polynomial in 𝑓 as defined in (7). If 𝑓 and 𝑃[𝑓] share the value 𝑎 IM and (2𝑄+6)Θ∞,𝑓+2+3𝑑(𝑃)𝛿0,𝑓>2𝑄+2𝑑(𝑃)+𝑑(𝑃)+7, (9) then 𝑓≡𝑃[𝑓].

Remark 4. Taking 𝑃[𝑓]=𝑓^(𝑘), that is, 𝑄=𝑘, 𝑑(𝑃)=𝑑(𝑃)=1 in (9), we get (2𝑘+6)Θ(∞,𝑓)+5𝛿(0,𝑓)>2𝑘+10, which improves (5) and extends the theorem to more general differential polynomial 𝑃[𝑓] as defined in (7).

Theorem 5. Let 𝑓 be a nonconstant meromorphic function and 𝑎(𝑧) be a small meromorphic function such that 𝑎(𝑧)≢0,∞. Let 𝑃[𝑓] a nonconstant differential polynomial in 𝑓 as defined in (7). If 𝑓 and 𝑃[𝑓] share the value 𝑎 CM and 3Θ∞,𝑓+𝑑(𝑃)+1𝛿0,𝑓>4, (10) then 𝑓≡𝑃[𝑓].

Remark 6. Taking 𝑃[𝑓]=𝑓^(𝑘), that is, 𝑄=𝑘, 𝑑(𝑃)=𝑑(𝑃)=1 in (10), we get 3Θ(∞,𝑓)+2𝛿(0,𝑓)>4, which improves (6) and extends the theorem to more general differential polynomial 𝑃[𝑓] as defined in (7).
Remark 6 gives answer to Question  3 of [4].

Theorem 7. Let 𝑓 be a nonconstant entire function and 𝑎(𝑧) a small meromorphic function such that 𝑎(𝑧)≢0,∞. Let 𝑃[𝑓] be a nonconstant differential polynomial in 𝑓 as defined in (7) If 𝑓 and 𝑃[𝑓] share the value 𝑎 IM and 3𝑑(𝑃)+2𝛿0,𝑓>2𝑑(𝑃)+2, (11) then 𝑓≡𝑃[𝑓].

Remark 8. Taking 𝑃[𝑓]=𝑓^(𝑘), that is, 𝑄=𝑘, 𝑑(𝑃)=𝑑(𝑃)=1 in (11), we get 𝛿>40,𝑓5. (12) Remark 8 gives answer to Question  1 of Yu [4].

Theorem 9. Let 𝑓 be a nonconstant entire function and 𝑎(𝑧) be a small meromorphic function such that 𝑎(𝑧)≢0,∞. 𝑃[𝑓] be a nonconstant differential polynomial in 𝑓 as defined in (7). If 𝑓 and 𝑃[𝑓] share the value 𝑎 CM and 𝑑(𝑃)+1𝛿0,𝑓>1, (13) then 𝑓≡𝑃[𝑓].

Remark 10. Taking 𝑃[𝑓]=𝑓^(𝑘), that is, 𝑄=𝑘, 𝑑(𝑃)=𝑑(𝑃)=1 in (13), we get 𝛿(0,𝑓)>1/2, which improves Theorem C and extends the theorem to more general differential polynomial 𝑃[𝑓] as defined in (7).
Remark 10 gives answer to Question  2 of Yu [4].

Remark 11. By proving Remarks 6, 8, and 10 we have answered Questions  3,  1, and 2 (of [4]), respectively, for the case 𝑓^(𝑘). Theorems 3–9 improve and generalize Theorems A–F for more general differential polynomial 𝑃[𝑓].

2. Lemma

Lemma 12 (see [7]). Let 𝑓 ba a meromorphic function and 𝑃[𝑓] be a differential polynomial in 𝑓. Then 𝑚𝑃𝑓𝑟,𝑓^𝑑(𝑃)≤𝑑(𝑃)−𝑑(𝑃𝑚1)𝑟,𝑓,𝑚𝑃𝑓+𝑆𝑟,𝑓𝑟,𝑓^𝑑(𝑃)≤𝑑(𝑃)−𝑑(𝑃)𝑚,𝑁𝑃𝑓𝑟,𝑓+𝑆𝑟,𝑓𝑟,𝑓^𝑑(𝑃)≤𝑑(𝑃)−𝑑(𝑃𝑁1)𝑟,𝑓+𝑄𝑁𝑟,𝑓+𝑁1𝑟,𝑓,𝑓≤+𝑆𝑟,𝑓𝑁(𝑟,𝑃𝑑(𝑃)𝑁𝑟,𝑓+𝑄𝑁,𝑇𝑓𝑟,𝑓+𝑆𝑟,𝑓𝑟,𝑃≤𝑄𝑁𝑟,𝑓+𝑑(𝑃)𝑇𝑟,𝑓+𝑆𝑟,𝑓, (14) where 𝑄=max{𝑛_𝑖1+2𝑛_𝑖2+3𝑛_𝑖3+⋯+𝑚𝑛_𝑖𝑚;1≤𝑖≤𝑛}.

Lemma 13 (see [8]). Let 𝑓 be a nonconstant meromorphic function, then 𝑁1𝑟,𝑓^(𝑘)≤𝑇𝑟,𝑓^(𝑘)1−𝑇𝑟,𝑓+𝑁𝑟,𝑓𝑁1+𝑆𝑟,𝑓𝑟,𝑓^(𝑘)1≤𝑁𝑟,𝑓+𝑘𝑁𝑟,𝑓+𝑆𝑟,𝑓. (15)(16)

Lemma 14 (see [9]). Let 𝐹𝐻=^′′𝐹^′−2𝐹^′−𝐺𝐹−1𝐺^′′′−2𝐺^′,𝐺−1 (17) where 𝐹 and 𝐺 are two nonconstant meromorphic functions. If 𝐹 and 𝐺 share 1 IM and 𝐻≢0, then 𝑁_𝐸¹⁾1𝑟,()()().𝐹−1≤𝑁𝑟,𝐻+𝑆𝑟,𝐹+𝑆𝑟,𝐺 (18)

Lemma 15. Let 𝑓 be a transcendental meromorphic function. Let 𝑃[𝑓] be defined as in (7). If 𝑃[𝑓]≢0, we have 𝑁1𝑟,𝑃𝑓≤𝑇𝑟,𝑃𝑓−𝑇𝑟,𝑓^𝑑(𝑃)+𝑑(𝑃)−𝑑(𝑃)𝑚1𝑟,𝑓1+𝑁𝑟,𝑓^𝑑(𝑃),𝑁1+𝑆𝑟,𝑓𝑟,𝑃𝑓≤𝑄𝑁𝑟,𝑓+𝑑(𝑃)−𝑑(𝑃)𝑚1𝑟,𝑓1+𝑁𝑟,𝑓^𝑑(𝑃)+𝑆𝑟,𝑓. (19)(20)

Proof. By the first fundamental theorem, we have 𝑁1𝑟,𝑃𝑓𝑓1=𝑇𝑟,𝑃−𝑚𝑟,𝑃𝑓+𝑂(1). (21) We have 𝑚1𝑟,𝑓^𝑑(𝑃)𝑃𝑓≤𝑚𝑟,𝑓^𝑑(𝑃)1+𝑚𝑟,𝑃𝑓𝑚1𝑟,𝑓^𝑑(𝑃)𝑃𝑓−𝑚𝑟,𝑓^𝑑(𝑃)1≤𝑚𝑟,𝑃𝑓1−𝑚𝑟,𝑓^𝑑(𝑃)𝑃𝑓+𝑚𝑟,𝑓^𝑑(𝑃)1≥−𝑚𝑟,𝑃𝑓 (22) or 1−𝑚𝑟,𝑃𝑓1≤−𝑚𝑟,𝑓^𝑑(𝑃)𝑃𝑓+𝑚𝑟,𝑓^𝑑(𝑃). (23) By (21), (23) and Lemma 12, we obtain (19).
Since 𝑇𝑓𝑓𝑓𝑃𝑓𝑟,𝑃=𝑚𝑟,𝑃+𝑁𝑟,𝑃≤𝑚𝑟,𝑓^𝑑(𝑃)+𝑚𝑟,𝑓^𝑑(𝑃)𝑓≤+𝑁𝑟,𝑃𝑑(𝑃)−𝑑(𝑃)𝑚𝑟,𝑓+𝑑(𝑃)𝑚+𝑟,𝑓𝑑(𝑃)𝑁𝑟,𝑓+𝑄𝑁≤𝑟,𝑓+𝑆𝑟,𝑓𝑑(𝑃)𝑚𝑟,𝑓+𝑑(𝑃)𝑁𝑟,𝑓+𝑄𝑁𝑟,𝑓+𝑆𝑟,𝑓, (24) we get 𝑇𝑟,𝑃𝑓≤𝑑(𝑃)𝑇𝑟,𝑓+𝑄𝑁𝑟,𝑓+𝑆𝑟,𝑓. (25) Substituting (25) in (19), we obtain (20).

Lemma 16 (see [10]). Let 𝑓 be a transcendental meromorphic function, 𝑃[𝑓] a differential polynomial in 𝑓 of degree 𝑑(𝑃) and weight Γ_𝑃. Then 𝑇(𝑟,𝑃)=𝑂(𝑇(𝑟,𝑓)), 𝑆(𝑟,𝑃)=𝑆(𝑟,𝑓).

3. Proof of Theorems

Proof of Theorem 3. Let 𝑃𝑓𝐹=𝑎𝑓,𝐺=𝑎. (26) From the conditions of Theorem 3, we know that 𝐹 and 𝐺 share 1 IM. From (26), we have 𝑇()𝑇(),𝑟,𝐹=𝑂𝑟,𝑓+𝑆𝑟,𝑓,𝑇𝑟,𝐺≤𝑇𝑟,𝑓+𝑆𝑟,𝑓𝑁(𝑟,𝐹)=𝑁(𝑟,𝐺)+𝑆𝑟,𝑓,𝑁(𝑟,𝐹)=𝑁𝑟,𝑓+𝑆𝑟,𝑓,𝑁()=𝑟,𝐺𝑁,𝑁𝑟,𝑓+𝑆𝑟,𝑓_𝐸1)1𝑟,𝐹−1=𝑁_𝐸¹⁾1𝑟,,𝐺−1+𝑆𝑟,𝑓𝑁_𝐸⁽²1𝑟,=𝐹−1𝑁_𝐸⁽²1𝑟,,𝐺−1+𝑆𝑟,𝑓𝑁_𝐿1𝑟,≤𝐹−1𝑁1𝑟,𝐹+𝑁(𝑟,𝐹)+𝑆𝑟,𝑓,𝑁1𝑟,=𝐹−1𝑁1𝑟,𝐺−1+𝑆𝑟,𝑓≤𝑁_𝐸¹⁾1𝑟,+𝐹−1𝑁_𝐸⁽²1𝑟,𝐹−1+𝑁_𝐿1𝑟,𝐹−1+𝑁_𝐿1𝑟,.𝐺−1+𝑆𝑟,𝑓 (27)(28)(29)(30)(31)(32)(33)
Let 𝐻 be defined by (17). Suppose that 𝐻≢0. By Lemma 14, (18) holds.
From (17) and (28), we have 𝑁(𝑟,𝐻)≤𝑁₍₂1𝑟,𝐹+𝑁₍₂1𝑟,𝐺+𝑁(𝑟,𝐺)+𝑁_𝐿1𝑟,𝐹−1+𝑁_𝐿1𝑟,𝐺−1+𝑁₀1𝑟,𝐹^′+𝑁₀1𝑟,𝐺^′, (34) where 𝑁₀(𝑟,1/𝐹^′) denotes the counting function corresponding to the zeros of 𝐹^′ which are not the zeros of 𝐹 and 𝐹−1. Similarly, 𝑁₀(𝑟,1/𝐺^′) is defined.
From the second fundamental theorem, we have 𝑇()()≤𝑟,𝐹+𝑇𝑟,𝐺𝑁(𝑟,𝐹)+𝑁1𝑟,𝐹+𝑁()+𝑟,𝐺𝑁1𝑟,𝐺+𝑁1𝑟,+𝐹−1𝑁1𝑟,𝐺−1−𝑁₀1𝑟,𝐹^′−𝑁₀1𝑟,𝐺^′+𝑆𝑟,𝑓. (35) Since 𝐹 and 𝐺 share 1 IM, we get from (33): 𝑁1𝑟,+𝐹−1𝑁1𝑟,𝐺−1=2𝑁_𝐸¹⁾1𝑟,𝐹−1+2𝑁_𝐿1𝑟,𝐹−1+2𝑁_𝐿1𝑟,𝐺−1+2𝑁_𝐸⁽²1𝑟,.𝐹−1 (36)
From this, (18), and (34), we have 𝑁1𝑟,+𝐹−1𝑁1𝑟,𝐺−1≤𝑁₍₂1𝑟,𝐹+𝑁₍₂1𝑟,𝐺+𝑁(𝑟,𝐺)+𝑁₀1𝑟,𝐹^′+3𝑁_𝐿1𝑟,𝐺−1+3𝑁_𝐿1𝑟,𝐹−1+𝑁_𝐸¹⁾1𝑟,𝐹−1+2𝑁_𝐸⁽²1𝑟,𝐺−1+𝑁₀1𝑟,𝐺^′+𝑆𝑟,𝑓. (37) It is clear that 𝑁_𝐿1𝑟,𝐹−1+2𝑁_𝐿1𝑟,𝐺−1+2𝑁_𝐸⁽²1𝑟,𝐺−1+𝑁_𝐸¹⁾1𝑟,1𝐹−1≤𝑁𝑟,()(1).𝐺−1≤𝑇𝑟,𝐺+𝑂 (38) Combining (37), and (38), we obtain 𝑁1𝑟,+𝐹−1𝑁1𝑟,𝐺−1≤𝑁₍₂1𝑟,𝐹+𝑁₍₂1𝑟,𝐺+𝑁(𝑟,𝐺)+2𝑁_𝐿1𝑟,𝐹−1+𝑁_𝐿1𝑟,()𝐺−1+𝑇𝑟,𝐺+𝑁₀1𝑟,𝐹^′+𝑁₀1𝑟,𝐺^′+𝑆𝑟,𝑓. (39) Substituting (39) in (35) and using (28), we obtain 𝑇(𝑟,𝐹)≤3𝑁()1𝑟,𝐺+𝑁𝑟,𝐹1+𝑁𝑟,𝐺+2𝑁_𝐿1𝑟,𝐹−1+𝑁_𝐿1𝑟,.𝐺−1+𝑆𝑟,𝑓 (40) Using (26) and (19), we get 𝑑(𝑃)𝑇𝑟,𝑓≤3𝑁(𝑟,𝐺)+𝑑(𝑃)−𝑑(𝑃)𝑚1𝑟,𝑓1+𝑁𝑟,𝑓1+𝑁𝑟,𝑓^𝑑(𝑃)+2𝑁_𝐿1𝑟,𝐹−1+𝑁_𝐿1𝑟,.𝐺−1+𝑆𝑟,𝑓 (41)
From (16), (20), and (26) we have 2𝑁_𝐿1𝑟,𝐹−1+𝑁_𝐿1𝑟,1𝐺−1≤2𝑁𝑟,𝐹^′1+𝑁𝑟,𝐺^′𝑁1≤2𝑟,𝐹+𝑁(1𝑟,𝐹)+𝑁𝑟,𝑓+𝑁1𝑟,𝑓+𝑆𝑟,𝑓≤2𝑁𝑟,𝑃𝑓+3𝑁1𝑟,𝑓+𝑁𝑟,𝑓+𝑆𝑟,𝑓≤2𝑄𝑁1𝑟,𝑓+2𝑁𝑟,𝑓^𝑑(𝑃)+2𝑑(𝑃)−𝑑(𝑃)𝑚1𝑟,𝑓+3𝑁1𝑟,𝑓+𝑁𝑟,𝑓≤()+𝑆𝑟,𝑓2𝑄+3𝑁𝑟,𝑓+2𝑑(𝑃)𝑁1+1𝑟,𝑓+2𝑑(𝑃)−𝑑(𝑃)𝑚1𝑟,𝑓+𝑆𝑟,𝑓. (42) From (41) and (42), we get 𝑑(𝑃)𝑇≤()𝑟,𝑓2𝑄+6𝑁𝑟,𝑓+2+3𝑑(𝑃)𝑁1𝑟,𝑓+3𝑑(𝑃)−𝑑(𝑃)𝑚1𝑟,𝑓≤()+𝑆𝑟,𝑓2𝑄+6𝑁𝑟,𝑓+2+3𝑑(𝑃𝑁1)𝑟,𝑓+3𝑑(𝑃)−𝑑(𝑃)𝑇1𝑟,𝑓+𝑆𝑟,𝑓3𝑑(𝑃)−2𝑑(𝑃)𝑇≤()𝑟,𝑓2𝑄+6𝑁𝑟,𝑓+2+3𝑑(𝑃)𝑁1𝑟,𝑓≤)++𝑆𝑟,𝑓(2𝑄+61−Θ∞,𝑓2+3𝑑(𝑃)≤1−𝛿0,𝑓×𝑇𝑟,𝑓+𝑆𝑟,𝑓2𝑄+3𝑑(𝑃)−)Θ++8(2𝑄+6∞,𝑓2+3𝑑(𝑃)𝛿0,𝑓𝑇𝑟,𝑓+𝑆𝑅,𝑓. (43) Therefore, we have (2𝑄+6)Θ∞,𝑓+2+3𝑑(𝑃𝛿−)0,𝑓2𝑄+2𝑑(𝑃)+8𝑇𝑟,𝑓≤𝑆𝑟,𝑓, (44) which is a contradiction to our hypothesis (9).
Thus 𝐻≡0. By integration, we get from (17) that 1=𝐴𝐺−1𝐹−1+𝐵, (45) where (𝐴≠0) and 𝐵 are constants. Thus ()()𝐺=𝐵+1𝐹+𝐴−𝐵−1(),()()𝐵𝐹+𝐴−𝐵𝐹=𝐵−𝐴𝐺+𝐴−𝐵−1().𝐵𝐺−𝐵+1 (46) We discuss the following three cases.
Case 1. Suppose that 𝐵≠0,−1. From (46), we have 𝑁1𝑟,()=𝐺−𝐵+1/𝐵𝑁(𝑟,𝐹). (47) From this and second fundamental theorem, we have 𝑇()≤𝑟,𝑓≤𝑇𝑟,𝐺+𝑆𝑟,𝑓𝑁(𝑟,𝐺)+𝑁1𝑟,𝐺+𝑁1𝑟,()≤𝐺−𝐵+1/𝐵+𝑆𝑟,𝑓𝑁(𝑟,𝐺)+𝑁1𝑟,𝐺+𝑁(𝑟,𝐹)+𝑆𝑟,𝑓≤2𝑁𝑟,𝑓+𝑁1𝑟,𝑓≤()+𝑆𝑟,𝑓2𝑄+6𝑁𝑟,𝑓+(2+3𝑑(𝑃)𝑁1𝑟,𝑓≤+𝑆𝑟,𝑓2𝑄+3𝑑(𝑃)−)Θ++8(2𝑄+6∞,𝑓2+3𝑑(𝑃)𝛿0,𝑓𝑇𝑟,𝑓+𝑆𝑅,𝑓. (48) Therefore, we have (2𝑄+6)Θ∞,𝑓+2+3𝑑(𝑃)𝛿−0,𝑓2𝑄+3𝑑(𝑃)+7𝑇𝑟,𝑓≤𝑆𝑟,𝑓, (49) which is a contradiction to our hypothesis (9).
Case 2. Suppose that 𝐵=0, From (46), we get ()𝐺=𝐹+𝐴−1𝐴,𝐹=𝐴𝐺−(𝐴−1), (50) we claim 𝐴=1.
If 𝐴≠1 from (50), we obtain 𝑁1𝑟,()1𝐺−𝐴−1/𝐴=𝑁𝑟,𝐹. (51) From this, second fundamental theorem, and (20), we have 𝑇()≤𝑟,𝑓≤𝑇𝑟,𝐺+𝑆𝑟,𝑓𝑁(𝑟,𝐺)+𝑁1𝑟,𝐺+𝑁1𝑟,()≤𝐺−𝐴−1/𝐴+𝑆𝑟,𝑓𝑁(𝑟,𝐺)+𝑁1𝑟,𝐺1+𝑁𝑟,𝐹≤+𝑆𝑟,𝑓𝑁1𝑟,𝑓+𝑁𝑟,𝑓+𝑄𝑁+𝑟,𝑓𝑑(𝑃)−𝑑(𝑃)𝑚1𝑟,𝑓1+𝑁𝑟,𝑓^𝑑(𝑃)≤()+𝑆𝑟,𝑓𝑄+1𝑁𝑟,𝑓+1+𝑑(𝑃)𝑁1𝑟,𝑓+𝑑(𝑃)−𝑑(𝑃)𝑇1𝑟,𝑓+𝑆𝑟,𝑓1−𝑑(𝑃)+𝑑(𝑃)𝑇≤()𝑟,𝑓2𝑄+6𝑁𝑟,𝑓+2+3𝑑(𝑃)𝑁1𝑟,𝑓≤+𝑆𝑟,𝑓2𝑄+3𝑑(𝑃)−)Θ++8(2𝑄+6∞,𝑓2+3𝑑(𝑃)𝛿0,𝑓𝑇𝑟,𝑓+𝑆𝑅,𝑓. (52) Hence, we have (2𝑄+6)Θ∞,𝑓+2+3𝑑(𝑃𝛿−)0,𝑓2𝑄+2𝑑(𝑃)+𝑑(𝑃)+7𝑇𝑟,𝑓≤𝑆𝑟,𝑓, (53) which is a contradiction to our hypothesis (9)
Thus, 𝐴=1.
From (50) we have 𝐹≡𝐺.
Therefore, we have 𝑓≡𝑃[𝑓].
Case 3. Suppose that 𝐵=−1, from (46) we have 𝐴𝐺=()−𝐹+𝐴+1,𝐹=1+𝐴𝐺−𝐴𝐺. (54) If 𝐴≠−1, we obtain from (54) that 𝑁1𝑟,()1𝐺−𝐴/𝐴+1=𝑁𝑟,𝐹. (55) By the same argument as in Case 2, we obtain a contradiction. Hence, 𝐴=−1.
From (54), we get 𝐹𝐺≡1, (56) that is, 𝑓⋅𝑃𝑓≡𝑎². (57) From (57), we have 𝑁1𝑟,𝑓+𝑁𝑟,𝑓=𝑆𝑟,𝑓. (58) Using (54), (57), Lemma 12, and first fundamental theorem, we get 𝑑(𝑃)𝑇1+1𝑟,𝑓=𝑇𝑟,𝑓^𝑑(𝑃)+11=𝑇𝑟,𝑓^𝑑(𝑃)𝑓𝑃𝑓=𝑇𝑟,𝑓^𝑑(𝑃)𝑎²𝑃𝑓+𝑆𝑟,𝑓=𝑚𝑟,𝑓^𝑑(𝑃)𝑃𝑓+𝑁𝑟,𝑓^𝑑(𝑃)≤+𝑆𝑟,𝑓𝑑(𝑃)−𝑑(𝑃)𝑚1𝑟,𝑓+𝑑(𝑃)−𝑑(𝑃)𝑁1𝑟,𝑓+𝑄𝑁𝑟,𝑓+𝑁1𝑟,𝑓≤+𝑆𝑟,𝑓𝑑(𝑃)−𝑑(𝑃)𝑚1𝑟,𝑓≤+𝑆𝑟,𝑓𝑑(𝑃)−𝑑(𝑃)𝑇1𝑟,𝑓+𝑆𝑟,𝑓. (59) From this, we have 𝑑(𝑃)+1)𝑇𝑟,𝑓≤𝑆𝑟,𝑓, (60) which is a contradiction. This completes the proof of Theorem 3.

Proof of Theorem 5. Let 𝐹 and 𝐺 be given by (26). From the assumption of Theorem 5, we know that 𝐹 and 𝐺 share 1 CM: 𝑁_𝐿1𝑟,=𝐹−1𝑁_𝐿1𝑟,𝐺−1=0. (61) Proceeding as in Theorem 3, we obtain (41).
Using (61) in (41), we get 𝑑(𝑃)𝑇𝑟,𝑓≤3𝑁(𝑟,𝐺)+𝑑(𝑃)−𝑑(𝑃)𝑚1𝑟,𝑓1+𝑁𝑟,𝑓^𝑑(𝑃)1+𝑁𝑟,𝑓+𝑆𝑟,𝑓≤3𝑁𝑟,𝑓+𝑑(𝑃)−𝑑(𝑃)𝑇1𝑟,𝑓−𝑁𝑟,𝑓+𝑑(𝑃)𝑁1+1𝑟,𝑓,𝑑+𝑆𝑟,𝑓(𝑃)𝑇𝑟,𝑓≤3𝑁𝑟,𝑓+𝑑(𝑃)𝑁1+1𝑟,𝑓≤𝑑+𝑆𝑟,𝑓(𝑃)−+𝑑+43Θ∞,𝑓(𝑃)+1𝛿0,𝑓𝑇𝑟,𝑓+𝑆𝑟,𝑓. (62) We have 3Θ∞,𝑓+𝑑(𝑃)+1𝛿0,𝑓−4𝑇𝑟,𝑓≤𝑆𝑟,𝑓, (63) which contradicts (10).
Thus, 𝐻≡0. Proceeding as in Theorem 3, we prove Theorem 5.

Proof of Theorem 7. 𝑓 is a nonconstant entire function. Taking 𝑁(𝑟,𝑓)=0 in proof of Theorem 3, we obtain Theorem 7.

Proof of Theorem 9. 𝑓 is a nonconstant entire function. Taking 𝑁(𝑟,𝑓)=0 in proof of Theorem 5, we obtain Theorem 9.

Acknowledgments

The authors thank the referee for his/her valuable suggestions. This research work is supported by the Department of Science and Technology Government of India, Ministry of Science and Technology, Technology Bhavan, New Delhi, India, under the sanction Letter no. (SR/S4/MS: 520/08).