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

Authors: Subhas S. Bhoosnurmath; Smita R. Kabbur
Journal: International Journal of Analysis
 (2013)

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## Abstract

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


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## Body

## 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 constanta, we denote by Nk)(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.Letf 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 byNL(r,1/(f-1))  (N¯L(r,1/(f-1))) the counting (reduced) function of those 1 points of f where p>q; by NE1)(r,1/(f-1)) the counting function of those 1-points of f where p=q=1; by NE(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 NL(r,1/(g-1)), NE1)(r,1/(g-1)) and NE(2(r,1/(g-1)) (see [2]).In 1996, Brück [3] posed the following conjecture.Conjecture 1.
Letf 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.
Letf 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.
Letk≥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
(1)4δ(0,f)+2(8+k)Θ(∞,f)>19+2k,
then f≡f(k).Theorem C.
Letk≥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
(2)δ(0,f)>34,
then f≡f(k).In the same paper, the author posed the following questions.Question 1.
Cana CM shared value be replaced by an IM shared value in Theorem C?Question 2.
Is the conditionδ(0,f)>3/4 sharp in Theorem C?Question 3.
Is the condition4δ(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.
Letk≥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
(3)2δ(0,f)+4Θ(∞,f)>5,
then f≡f(k).Theorem E.
Letk≥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
(4)δ(0,f)>12,
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.
Letk(≥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 fn and f(k) share the value a(z) IM and
(5)(2k+6)Θ(∞,f)+4Θ(0,f)+2δk+2(0,f)>2k+12-n
or fn and f(k) share the value a(z) CM and
(6)(k+3)Θ(∞,f)+2Θ(0,f)+δk+2(0,f)>k+6-n,
then f≡f(k).Regarding these results, a natural question is what can be said when a nonconstant meromorphic functionf 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(7)P[f]=∑i=1n‍αi(z)fni0(f′)ni1(f′′)ni2⋯(f(m))nim
is called differential polynomial in f of degree d¯(P), lower degree d_(P), and weight ΓP, where ni0,ni1,…,nim are nonnegative integers, αi=αi(z) are meromorphic functions satisfying T(r,αi)=S(r,f) and
(8)d¯(P)=max{∑j=0m‍nij:1≤i≤n},d_(P)=min{∑j=0m‍nij:1≤i≤n},ΓP=max{∑j=0m‍(j+1)nij:1≤i≤n}.
Further, ifd¯(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.
Letf 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
(9)(2Q+6)Θ(∞,f)+(2+3d_(P))δ(0,f)>2Q+2d_(P)+d¯(P)+7,
then f≡P[f].Remark 4.
TakingP[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.
Letf 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
(10)3Θ(∞,f)+(d_(P)+1)δ(0,f)>4,
then f≡P[f].Remark 6.
TakingP[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).
Remark6 gives answer to Question  3 of [4].Theorem 7.
Letf 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
(11)(3d_(P)+2)δ(0,f)>2d¯(P)+2,
then f≡P[f].Remark 8.
TakingP[f]=f(k), that is, Q=k, d¯(P)=d_(P)=1 in (11), we get
(12)δ(0,f)>45.
Remark 8 gives answer to Question  1 of Yu [4].Theorem 9.
Letf 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
(13)(d_(P)+1)δ(0,f)>1,
then f≡P[f].Remark 10.
TakingP[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).
Remark10 gives answer to Question  2 of Yu [4].Remark 11.
By proving Remarks6, 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]).
Letf ba a meromorphic function and P[f] be a differential polynomial in f. Then
(14)m(r,P[f]fd¯(P))≤(d-(P)-d_(P))m(r,1f)+S(r,f),m(r,P[f]fd_(P))≤(d-(P)-d_(P))m(r,f)+S(r,f),N(r,P[f]fd-(P))≤(d-(P)-d_(P))N(r,1f)+Q[N¯(r,f)+N¯(r,1f)]+S(r,f),N(r,P[f]≤d-(P)N(r,f)+QN-(r,f)+S(r,f),T(r,P(f))≤QN¯(r,f)+d¯(P)T(r,f)+S(r,f),
where Q=max{ni1+2ni2+3ni3+⋯+mnim;1≤i≤n}.Lemma 13 (see [8]).
Letf be a nonconstant meromorphic function, then
(15)N(r,1f(k))≤T(r,f(k))-T(r,f)+N(r,1f)+S(r,f)(16)N(r,1f(k))≤N(r,1f)+kN¯(r,f)+S(r,f).Lemma 14 (see [9]).
Let(17)H=(F′′F′-2F′F-1)-(G′′G′-2G′G-1),
where F and G are two nonconstant meromorphic functions. If F and G share 1 IM and H≢0, then
(18)NE1)(r,1F-1)≤N(r,H)+S(r,F)+S(r,G).Lemma 15.
Letf be a transcendental meromorphic function. Let P[f] be defined as in (7). If P[f]≢0, we have
(19)N(r,1P[f])≤T(r,P[f])-T(r,fd¯(P))+(d¯(P)-d_(P))m(r,1f)+N(r,1fd¯(P))+S(r,f),(20)N(r,1P[f])≤QN¯(r,f)+(d¯(P)-d_(P))m(r,1f)+N(r,1fd¯(P))+S(r,f).Proof.
By the first fundamental theorem, we have(21)N(r,1P[f])=T(r,P[f])-m(r,1P[f])+O(1).
We have
(22)m(r,1fd¯(P))≤m(r,P[f]fd¯(P))+m(r,1P[f])m(r,1fd¯(P))-m(r,P[f]fd¯(P))≤m(r,1P[f])-m(r,1fd¯(P))+m(r,P[f]fd¯(P))≥-m(r,1P[f])
or
(23)-m(r,1P[f])≤-m(r,1fd¯(P))+m(r,P[f]fd¯(P)).
By (21), (23) and Lemma 12, we obtain (19).
Since(24)T(r,P[f])=m(r,P[f])+N(r,P[f])≤m(r,P[f]fd_(P))+m(r,fd_(P))+N(r,P[f])≤(d-(P)-d_(P))m(r,f)+d_(P)m(r,f)+d-(P)N(r,f)+QN-(r,f)+S(r,f)≤d-(P)m(r,f)+d-(P)N(r,f)+QN-(r,f)+S(r,f),
we get
(25)T(r,P[f])≤d¯(P)T(r,f)+QN-(r,f)+S(r,f).
Substituting (25) in (19), we obtain (20).Lemma 16 (see [10]).
Letf be a transcendental meromorphic function, P[f] a differential polynomial in f of degree d-(P) and weight ΓP. Then T(r,P)=O(T(r,f)), S(r,P)=S(r,f).


## 3. Proof of Theorems

Proof of Theorem3.
Let(26)F=P[f]a,G=fa.
From the conditions of Theorem 3, we know that F and G share 1 IM. From (26), we have
(27)T(r,F)=O(T(r,f))+S(r,f),T(r,G)≤T(r,f)+S(r,f),(28)N¯(r,F)=N¯(r,G)+S(r,f),(29)N¯(r,F)=N¯(r,f)+S(r,f),N¯(r,G)=N¯(r,f)+S(r,f),(30)NE1)(r,1F-1)=NE1)(r,1G-1)+S(r,f),(31)N¯E(2(r,1F-1)=N¯E(2(r,1G-1)+S(r,f),(32)N¯L(r,1F-1)≤N¯(r,1F)+N¯(r,F)+S(r,f),(33)N¯(r,1F-1)=N¯(r,1G-1)+S(r,f)≤NE1)(r,1F-1)+N¯E(2(r,1F-1)+NL(r,1F-1)+NL(r,1G-1)+S(r,f).
LetH be defined by (17). Suppose that H≢0. By Lemma 14, (18) holds.
From (17) and (28), we have
(34)N(r,H)≤N(2(r,1F)+N(2(r,1G)+N¯(r,G)+NL(r,1F-1)+NL(r,1G-1)+N0(r,1F′)+N0(r,1G′),
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(35)T(r,F)+T(r,G)≤N¯(r,F)+N¯(r,1F)+N¯(r,G)+N¯(r,1G)+N¯(r,1F-1)+N¯(r,1G-1)-N0(r,1F′)-N0(r,1G′)+S(r,f).
Since F and G share 1 IM, we get from (33):
(36)N¯(r,1F-1)+N¯(r,1G-1)=2NE1)(r,1F-1)+2NL(r,1F-1)+2NL(r,1G-1)+2N¯E(2(r,1F-1).
From this, (18), and (34), we have
(37)N¯(r,1F-1)+N¯(r,1G-1)≤N(2(r,1F)+N(2(r,1G)+N¯(r,G)+N0(r,1F′)+3NL(r,1G-1)+3NL(r,1F-1)+NE1)(r,1F-1)+2N¯E(2(r,1G-1)+N0(r,1G′)+S(r,f).
It is clear that
(38)NL(r,1F-1)+2NL(r,1G-1)+2N¯E(2(r,1G-1)+NE1)(r,1F-1)≤N(r,1G-1)≤T(r,G)+O(1).
Combining (37), and (38), we obtain
(39)N¯(r,1F-1)+N¯(r,1G-1)≤N(2(r,1F)+N(2(r,1G)+N¯(r,G)+2NL(r,1F-1)+NL(r,1G-1)+T(r,G)+N0(r,1F′)+N0(r,1G′)+S(r,f).
Substituting (39) in (35) and using (28), we obtain
(40)T(r,F)≤3N¯(r,G)+N(r,1F)+N(r,1G)+2NL(r,1F-1)+NL(r,1G-1)+S(r,f).
Using (26) and (19), we get
(41)d¯(P)T(r,f)≤3N¯(r,G)+(d¯(P)-d_(P))m(r,1f)+N(r,1f)+N(r,1fd¯(P))+2NL(r,1F-1)+NL(r,1G-1)+S(r,f).
From (16), (20), and (26) we have
(42)2NL(r,1F-1)+NL(r,1G-1)≤2N(r,1F′)+N(r,1G′)≤2[N(r,1F)+N¯(r,F)]+N(r,1f)+N¯(r,f)+S(r,f)≤2N(r,1P[f])+3N¯(r,f)+N(r,1f)+S(r,f)≤2QN¯(r,f)+2N(r,1fd¯(P))+2(d¯(P)-d_(P))m(r,1f)+3N¯(r,f)+N(r,1f)+S(r,f)≤(2Q+3)N¯(r,f)+(2d¯(P)+1)N(r,1f)+2(d¯(P)-d_(P))m(r,1f)+S(r,f).
From (41) and (42), we get
(43)d¯(P)T(r,f)≤(2Q+6)N¯(r,f)+(2+3d¯(P))N(r,1f)+3(d¯(P)-d_(P))m(r,1f)+S(r,f)≤(2Q+6)N¯(r,f)+(2+3d_(P))N(r,1f)+3(d¯(P)-d_(P))T(r,1f)+S(r,f)(3d_(P)-2d¯(P))T(r,f)≤(2Q+6)N¯(r,f)+(2+3d_(P))N(r,1f)+S(r,f)≤{(2Q+6)(1-Θ(∞,f))+(2+3d_(P))(1-δ(0,f))}×T(r,f)+S(r,f)≤{(2Q+3d_(P)+8)-[(2Q+6)Θ(∞,f)+(2+3d_(P))δ(0,f)]}T(r,f)+S(R,f).
Therefore, we have
(44){(2Q+2d¯(P)+8)(2Q+6)Θ(∞,f)+(2+3d_(P))δ(0,f)-(2Q+2d¯(P)+8)}T(r,f)≤S(r,f),
which is a contradiction to our hypothesis (9).
ThusH≡0. By integration, we get from (17) that
(45)1G-1=AF-1+B,
where (A≠0) and B are constants. Thus
(46)G=(B+1)F+(A-B-1)BF+(A-B),F=(B-A)G+(A-B-1)BG-(B+1).
We discuss the following three cases.
Case 1. Suppose that B≠0,-1. From (46), we have
(47)N¯(r,1G-(B+1)/B)=N¯(r,F).
From this and second fundamental theorem, we have
(48)T(r,f)≤T(r,G)+S(r,f)≤N¯(r,G)+N¯(r,1G)+N¯(r,1G-(B+1)/B)+S(r,f)≤N¯(r,G)+N¯(r,1G)+N¯(r,F)+S(r,f)≤2N¯(r,f)+N¯(r,1f)+S(r,f)≤(2Q+6)N¯(r,f)+(2+3d_(P)N(r,1f)+S(r,f)≤{(2Q+3d_(P)+8)-[(2Q+6)Θ(∞,f)+(2+3d_(P))δ(0,f)]}T(r,f)+S(R,f).
Therefore, we have
(49){(2Q+6)Θ(∞,f)+(2+3d_(P))δ(0,f)-(2Q+3d_(P)+7)}T(r,f)≤S(r,f),
which is a contradiction to our hypothesis (9).
Case 2. Suppose that B=0, From (46), we get
(50)G=F+(A-1)A,F=AG-(A-1),
we claim A=1.
IfA≠1 from (50), we obtain
(51)N(r,1G-(A-1)/A)=N(r,1F).
From this, second fundamental theorem, and (20), we have
(52)T(r,f)≤T(r,G)+S(r,f)≤N¯(r,G)+N¯(r,1G)+N¯(r,1G-(A-1)/A)+S(r,f)≤N¯(r,G)+N¯(r,1G)+N(r,1F)+S(r,f)≤N¯(r,f)+N(r,1f)+QN¯(r,f)+(d¯(P)-d_(P))m(r,1f)+N(r,1fd¯(P))+S(r,f)≤(Q+1)N¯(r,f)+(1+d_(P))N(r,1f)+(d¯(P)-d_(P))T(r,1f)+S(r,f)(1-d¯(P)+d_(P))T(r,f)≤(2Q+6)N¯(r,f)+(2+3d_(P))N(r,1f)+S(r,f)≤{(2Q+3d_(P)+8)-[(2Q+6)Θ(∞,f)+(2+3d_(P))δ(0,f)]}T(r,f)+S(R,f).
Hence, we have
(53){2Q+2d_(P)+d¯(P)+7(2Q+6)Θ(∞,f)+(2+3d_(P))δ(0,f)-(2Q+2d_(P)+d¯(P)+7)}T(r,f)≤S(r,f),
which is a contradiction to our hypothesis (9)
Thus,A=1.
From (50) we have F≡G.
Therefore, we havef≡P[f].
Case 3. Suppose that B=-1, from (46) we have
(54)G=A-F+A+1,F=(1+A)G-AG.
If A≠-1, we obtain from (54) that
(55)N(r,1G-A/(A+1))=N(r,1F).
By the same argument as in Case 2, we obtain a contradiction. Hence, A=-1.
From (54), we get
(56)FG≡1,
that is,
(57)f·P[f]≡a2.
From (57), we have
(58)N(r,f)+N(r,1f)=S(r,f).
Using (54), (57), Lemma 12, and first fundamental theorem, we get
(59)(d¯(P)+1)T(r,f)=T(r,1fd¯(P)+1)=T(r,1fd¯(P)f)=T(r,P[f]fd¯(P)a2)+S(r,f)=m(r,P[f]fd¯(P))+N(r,P[f]fd¯(P))+S(r,f)≤(d¯(P)-d_(P))m(r,1f)+(d¯(P)-d_(P))N(r,1f)+Q[N¯(r,f)+N¯(r,1f)]+S(r,f)≤(d¯(P)-d_(P))m(r,1f)+S(r,f)≤(d¯(P)-d_(P))T(r,1f)+S(r,f).
From this, we have
(60)(d_(P))+1)T(r,f)≤S(r,f),
which is a contradiction. This completes the proof of Theorem 3.Proof of Theorem5.
LetF and G be given by (26). From the assumption of Theorem 5, we know that F and G share 1 CM:
(61)NL¯(r,1F-1)=NL¯(r,1G-1)=0.
Proceeding as in Theorem 3, we obtain (41).
Using (61) in (41), we get
(62)d¯(P)T(r,f)≤3N¯(r,G)+(d¯(P)-d_(P))m(r,1f)+N(r,1fd¯(P))+N(r,1f)+S(r,f)≤3N¯(r,f)+(d¯(P)-d_(P))(T(r,f)-N(r,1f))+(d¯(P)+1)N(r,1f)+S(r,f),d_(P)T(r,f)≤3N¯(r,f)+(d_(P)+1)N(r,1f)+S(r,f)≤{(d_(P)+4)-[3Θ(∞,f)+(d_(P)+1)δ(0,f)]}T(r,f)+S(r,f).
We have
(63){3Θ(∞,f)+(d_(P)+1)δ(0,f)-4}T(r,f)≤S(r,f),
which contradicts (10).
Thus,H≡0. Proceeding as in Theorem 3, we prove Theorem 5.Proof of Theorem7.
f is a nonconstant entire function. Taking N(r,f)=0 in proof of Theorem 3, we obtain Theorem 7.Proof of Theorem9.
f is a nonconstant entire function. Taking N(r,f)=0 in proof of Theorem 5, we obtain Theorem 9.



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