ANALYSIS International Journal of Analysis 2314-4998 2314-498X Hindawi Publishing Corporation 926340 10.1155/2013/926340 926340 Research Article On Entire and Meromorphic Functions That Share One Small Function with Their Differential Polynomial 0000-0003-4268-5434 Bhoosnurmath Subhas S. Kabbur Smita R. Ehrnström Mats Department of Mathematics, Karnatak University, Dharwad 580003 India kud.ac.in 2013 6 2 2013 2013 24 09 2012 22 12 2012 2013 Copyright © 2013 Subhas S. Bhoosnurmath and Smita R. Kabbur. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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 ). 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 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.

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 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=q2. 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 ).

In 1996, Brück  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 , 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{0}.

Many people extended this theorem and obtained many results. In 2003, Yu  proved the following theorem.

Theorem B.

Let k1. 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 ff(k).

Theorem C.

Let k1. 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 ff(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  applied different method and obtained the following theorem which answers some questions posed in .

Theorem D.

Let k1. 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 ff(k).

Theorem E.

Let k1. 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 ff(k).

Recently, Zhang and Lü  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 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 ff(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 (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:1in},d_(P)=min{j=0mnij:1in},ΓP=max{j=0m(j+1)nij:1in}.

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  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 (9)(2Q+6)Θ(,f)+(2+3d_(P))δ(0,f)>2Q+2d_(P)+d¯(P)+7, then fP[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 (10)3Θ(,f)+(d_(P)+1)δ(0,f)>4, then fP[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 .

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 (11)  (3d_(P)+2)δ(0,f)>2d¯(P)+2, then fP[f].

Remark 8.

Taking P[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 .

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 (13)  (d_(P)+1)δ(0,f)>1, then fP[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 .

Remark 11.

By proving Remarks 6, 8, and 10 we have answered Questions  3,  1, and 2 (of ), respectively, for the case f(k). Theorems 39 improve and generalize Theorems A–F for more general differential polynomial P[f].

2. Lemma Lemma 12 (see [<xref ref-type="bibr" rid="B7">7</xref>]).

Let f 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;1in}.

Lemma 13 (see [<xref ref-type="bibr" rid="B8">8</xref>]).

Let f 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 [<xref ref-type="bibr" rid="B10">9</xref>]).

Let (17)H=(F′′F-2FF-1)-(G′′G-2GG-1), where F and G are two nonconstant meromorphic functions. If F and G share 1 IM and H0, then (18)NE1)(r,1F-1)N(r,H)+S(r,F)+S(r,G).

Lemma 15.

Let f 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 [<xref ref-type="bibr" rid="B9">10</xref>]).

Let f 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 Theorem <xref ref-type="statement" rid="thm1.1">3</xref>.

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).

Let H be defined by (17). Suppose that H0. 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).

Thus H0. By integration, we get from (17) that (45)1G-1=AF-1+B, where (A0) 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 B0,-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.

If A1 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 FG.

Therefore, we have fP[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)FG1, 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 Theorem <xref ref-type="statement" rid="thm1.2">5</xref>.

Let F 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, H0. Proceeding as in Theorem 3, we prove Theorem 5.

Proof of Theorem <xref ref-type="statement" rid="thm1.3">7</xref>.

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

Proof of Theorem <xref ref-type="statement" rid="thm1.4">9</xref>.

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).

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