Uniqueness of Meromorphic Functions and Differential Polynomials

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Volume 2011, Article ID 514218,12pages doi:10.1155/2011/514218

Research Article

Uniqueness of Meromorphic Functions and

Differential Polynomials

Jin-Dong Li

Teaching Division of Mathematics, Chengdu University of Technology, Chengdu, Sichuan 610059, China

Correspondence should be addressed to Jin-Dong Li,[email protected]

Received 20 December 2010; Accepted 28 April 2011

Academic Editor: Rodica Costin

Copyrightq2011 Jin-Dong Li. 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 and diﬀerential polynomials sharing one value with weight and prove two main theorems which generalize and improve some results earlier given by M. L. Fang, S. S. Bhoosnurmath and R. S. Dyavanal, and so forth.

1. Introduction and Results

Letf be a nonconstant meromorphic function defined in the whole complex plane . It is assumed that the reader is familiar with the notations of the Nevanlinna theory such as Tr, f,mr, f,Nr, f, andSr, f, that can be found, for instance, in1–3.

Let f and g be two nonconstant meromorphic functions. Let abe a finite complex number. We say that f and g share the valueaCMcounting multiplicities if f −aand g−ahave the same zeros with the same multiplicities, and we say thatf and g share the valueaIMignoring multiplicitiesif we do not consider the multiplicities. Whenf andg share 1 IM, letz0 be a 1-point off of orderpand a 1-points ofg of orderq; we denote by

N11r,1/f −1the counting function of those 1-points off and g, wherep q 1 and

byN2_E r,1/f−1the counting function of those 1-points off and g, wherep q ≥ 2. NLr,1/f−1is the counting function of those 1-points of bothf andg, wherep > q. In

the same way, we can defineN11r,1/g−1,NE2r,1/g−1, andNLr,1/g−1. Iff

andgshare 1 IM, it is easy to see that

N

r, 1 f−1

N11

r, 1 f−1

NL

r, 1 f−1

NL

r, 1 g−1

N_E2

r, 1 f−1

N

r, 1 g−1

.

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Letfbe a nonconstant meromorphic function. Letabe a finite complex number andk a positive integer; we denote byNkr,1/f−a orNkr,1/f−athe counting function

for zeros off−awith multiplicity≤kignoring multiplicitiesand byNkr,1/f−a or

N_kr,1/f−athe counting function for zeros off−awith multiplicity at leastkignoring multiplicities. Set

Nk

r, 1

f−a

N

r, 1 f−a

N2

r, 1 f−a

· · ·Nk

r, 1

f−a

,

Θa, f1− lim

r→ ∞

Nr,1/f−a

Tr, f .

1.2

We further define

δka, f1−_rlim

→ ∞

Nkr,1/f−a

Tr, f . 1.3

In 2002, C. Y. Fang and M. L. Fang4proved the following result.

Theorem Asee4. Letfzandgzbe two nonconstant entire functions, and letn(≥8) be a

positive integer. Iffnzfz−1fzandgnzgz−1gzshare 1 CM, thenfz≡gz.

Fang5proved the following result.

Theorem Bsee5. Letfzandgzbe two nonconstant entire functions, and letn,kbe two

positive integers withn >2k8. Iffnzfz−1kandgnzgz−1kshare 1 CM, then

fz≡gz.

In6, for some general diﬀerential polynomials such as fnf−1mk, Liu proved the following result.

Theorem Csee6. Letfzandgzbe two nonconstant entire functions, and letn, m, kbe three

positive integers such thatn > 5k4m9. Iffnzfz−1mk andgnzgz−1mk

share 1 IM, then eitherfz ≡ gzor f and g satisfy the algebraic equationRf, g ≡ 0, where

Rω1, ω2 ωn₁ω1−1m−ω₂nω2−1m.

The following example shows that Theorem A is not valid when f and g are two meromorphic functions.

Example 1.1. Letf n2h−hn2/n11−hn2,g n21−hn1/n11−hn2, where

hez_{. Then}_fn_z_f_z₋₁_f_z_and_gn_z_g_z₋₁_g_z_{share 1 CM, but}_f_z_/_≡_g_z_.

Lin and Yi7and Bhoosnurmath and Dyavanal8generalized the above results and obtained the following results.

Theorem D see 7. Let fz and gz be two nonconstant meromorphic functions with

Θ∞, f > 2/n 1, and let n≥ 12 be a positive integer. If fnzfz − 1fz and

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Theorem E see 8. Let fz and gz be two nonconstant meromorphic functions satisfying

Θ∞, f>3/n1, and letn, kbe two positive integers withn >3k13. Iffnzfz−1k

andgnzgz−1kshare 1 CM, thenfz≡gz.

Naturally, one may ask the following question: is it really possible to relax in any way the nature of sharing 1 in the above results?

To state the next result, we require the following definition.

Definition 1.2see9. Letk be a nonnegative integer or infinity. Fora ∈ ∪ {∞}one

denotes byEka, fthe set of alla-points off, where ana-point of multiplicitymis counted

mtimes ifm≤kandk1 times ifm > k. IfEka, f Eka, g, one says thatf, gshare the

valueawith weightk.

We write thatf, gsharea, kto mean thatf, gshare the valueawith weightk; clearly iff, gsharea, k, thenf, gsharea, pfor all integerspwith 0 ≤ p≤ k. Also, we note that f, gshare a valueaIM or CM if and only if they sharea,0ora,∞, respectively.

Recently, with the notion of weighted sharing of values, Xu et al.10improved the above results and proved the following theorem.

Theorem Fsee10. Letfzandgzbe two nonconstant meromorphic functions, and letn, k

be two positive integers withn >5k11. IfΘ∞, f Θ∞, g>4/n,fnzfz−1k, and

gn_z_g_z₋₁k_{share 1}₁_,₂_{, then}_f_z_≡_g_z_.

Theorem Gsee10. Letfzandgzbe two nonconstant meromorphic functions, and letn, k

be two positive integers withn >7k23/2. IfΘ∞, f Θ∞, g>4/n,fnzfz−1k, and

gn_z_g_z₋₁k_{share 1}₁_,₁_{, then}_f_z_≡_g_z_.

Remark 1.3. The proof of Theorem E contains some mistakes: for example, one cannot get

formulas 6.9and 6.10in8. Therefore, the last inequality in page 1203 of8does not hold. So, Theorem E will not stand. Similarly, in10, the proof of Case 1 in Theorem F is incorrectsee page 63 of paper10. Hence, the conclusion of Theorems F and G will not stand.

Now one may ask the following questions which are the motivations of the paper.

iWhat happens if the conclusions of Theorems E, F, and G can not stand?

iiCan the valuenbe further reduced in the above results?

In the paper, we investigate the solutions of the above questions. We improve and gen-eralize the above related results by proving the following theorems.

Theorem 1.4. Letfz andgzbe two nonconstant meromorphic functions, and letn, k be two

positive integers withn >3k11. IfΘ∞, f>2/n,fnzfz−1k, andgnzgz−1k

share 11,2, thenfz≡gzorfnzfz−1k·gnzgz−1k≡1.

Theorem 1.5. Letfz andgzbe two nonconstant meromorphic functions, and letn, k be two

positive integers withn >5k14. IfΘ∞, f>2/n,fnzfz−1kandgnzgz−1k

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2. Some Lemmas

For the proof of our result, we need the following lemmas.

Lemma 2.1see1. Let f be nonconstant meromorphic function, and leta0, a1, . . . , an be finite

complex numbers such thatan/0. Then

Tr, anfnan−1fn−1· · ·a0

nTr, fSr, f. 2.1

Lemma 2.2see1. Letfzbe a nonconstant meromorphic function,ka positive integer, andc

a nonzero finite complex number. Then

Tr, f≤Nr, fN

r,1 f

N

r, 1

fk₋_c −N

r, 1

fk1 S

r, f

≤Nr, fN_k1

r, 1 f

N

r, 1

fk₋_c −N0

r, 1

fk1 S

r, f.

2.2

HereN0r,1/fk1is the counting function which only counts those points such thatfk10 but

ffk₋_c_/_0.

Lemma 2.3see11. Let f be a nonconstant meromorphic function, and letk, pbe two positive

integers. Then

Np

r 1

fk ≤Npk

r1 f

kNr, fSr, f. 2.3

ClearlyNr1/fk N1r1/fk.

Lemma 2.4see1. Letfzbe a transcendental meromorphic function, and leta1z,a2zbe

two meromorphic functions such thatTr, ai Sr, f,i1,2. Then

Tr, f≤Nr, fN

r, 1

f−a1

N

r, 1

f−a2

Sr, f. 2.4

Lemma 2.5. Letf andg be two nonconstant meromorphic functions, and letk(≥1),l(≥1) be two

positive integers. Suppose thatfkandgkshare1, l.

iIfl2 and

Δ12Θ

∞, f k2Θ∞, g Θ0, f Θ0, gδk10, fδk10, g> k7 2.5

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iiIfl1 and

Δ2 k3Θ

∞, f k2Θ∞, g Θ0, f Θ0, g2δk10, fδk10, g>2k9

2.6

then eitherfkgk≡1 orf≡g.

Proof. Let

hz fk2z fk1_z−2

fk1_z

fk_z₋₁ −

gk2_z

gk1_z2

gk1_z

gk_z₋₁. 2.7

Suppose thath /≡0.

Ifz0is a common simple 1-point offk andgk, substituting their Taylor series atz0

into2.7, we see thatz0is a zero ofhz. Thus, we have

N11

r, 1

fk₋₁ N11

r, 1

gk₋₁

≤N

r,1 h

≤Tr, h O1

≤Nr, h Sr, fSr, g.

2.8

By our assumptions,hzhave poles only at zeros of fk1 andgk1 and poles of f and g, and those 1-points of fk and gk whose multiplicities are distinct from the multiplicities of correspond to 1-points ofgkandfk, respectively. Thus, we deduce from

2.7that

Nr, h≤Nr, fNr, gN

r, 1 f

N

r,1 g

N0

r, 1

fk1

N0

r, 1

gk1 NL

r, 1

fk₋₁ NL

r, 1

gk₋₁ ,

2.9

hereN0r,1/fk1 has the same meaning as inLemma 2.2.

ByLemma 2.2, we have

Tr, f≤Nr, fNk1

r,1 f

N

r, 1

fk₋₁ −N0

r, 1

fk1 S

r, f,

Tr, g≤Nr, gNk1

r,1 g

N

r, 1

gk₋₁ −N0

r, 1

gk1 S

r, g.

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Sincefkandgkshare1,0, we get

N

r, 1

fk₋₁ N

r, 1

gk₋₁ 2N11

r, 1 f−1

2NL

r, 1

fk₋₁

2NL

r, 1

gk₋₁ 2N

2 E

r, 1

fk₋₁ .

2.11

We obtain from2.8–2.11that

Tr, fTr, g≤2Nr, f2Nr, gN

r,1 f N r,1 g N11 r, 1

fk₋₁

3NL

r, 1

fk₋₁ 3NL

r, 1

gk₋₁ 2N2E

r, 1

fk₋₁

N_k1 r, 1 f N_k1 r,1 g

Sr, fSr, g.

2.12

iIfl≥2, it is easy to see that

N11

r, 1

fk₋₁ 3NL

r, 1

fk₋₁ 3NL

r, 1

gk₋₁ 2N

2 E

r, 1

fk₋₁

≤N

r, 1

gk₋₁ S

r, fSr, g.

2.13

Since

N

r, 1

gk₋₁ ≤T

r, gk_S_{r, g}_m_{r, g}k_N_{r, g}k_S_{r, g}

≤mr, gm

r,gk

g N

r, gkNr, gSr, g

≤Tr, gkNr, gSr, g,

2.14

combining with2.12,2.13, and2.14, we obtain

Tr, f≤2Nr, f k2Nr, gN

r,1 f N r,1 g Nk1 r, 1 f Nk1 r, 1 g

Sr, fSr, g.

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Without loss of generality, we suppose that there exists a setIwith infinite measure such that Tr, g≤Tr, fforr∈I:

Tr, f≤k8−2Θ∞, f−k2Θ∞, g−Θ0, f−Θ0, g

−δk10, f−δk10, gεTr, fSr, F

2.16

forr∈Iand 0< ε <Δ1−k7, that is,

Δ1−k7−εT

r, f≤Sr, f, 2.17

that is,Δ1 ≤k7, which contradicts hypothesis2.5.

Therefore, we haveh≡0, that is,

fk2_z

fk1_z−2

fk1_z

fk_z₋₁

gk2_z

gk1_z −2

gk1_z

gk_z₋₁. 2.18

By solving this equation, we obtain

gk b1fk a−b−1

bfk _a₋_b , 2.19

wherea/0, bare two constants. Next, we consider three cases.

Case 1b /0,−1. For more details see the following subcases.

Subcase 1.1a−b−1/0. Then, by2.19, we haveNr,1/fk−a−b−1/b1

Nr,1/gk.

ByLemma 2.2andLemma 2.3, we get

r,1 f

N

r, 1

fk₋_a₋_b₋₁_/_b₁

−N0

r, 1

fk1 S

r, f

≤Nr, fNk1

r,1 f

N

r, 1

gk S

r, f

≤Nr, fN_k1

r,1 f

N_k1

r,1 g

kNr, gSr, f

≤2Nr, f k2Nr, gN

r, 1 f

N

r,1 g

Nk1

r, 1 f

Nk1

r, 1 g

Sr, fSr, g,

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that is,Tr, f≤k8−Δ1Tr, f Sr, f. Thus, by2.5we deduce thatTr, f ≤Sr, f

forr∈I, a contradiction.

Subcase 1.2a−b−10. Then, by2.19, we havegk b1fk/bfk1.

ThereforeNr,1/fk 1/b Nr, gk. ByLemma 2.2andLemma 2.3, we get

r,1 f

N

r, 1

fk ₁_/b −N0

r, 1

fk1 S

r, f

≤Nr, fNk1

r,1 f

Nr, gk_S_{r, f}

≤2Nr, f k2Nr, gN

r,1 f

N

r, 1 g

Nk1

r,1 f

Nk1

r,1 g

Sr, fSr, g,

2.21

that is,Tr, f≤k8−Δ1Tr, f Sr, f. Thus, by2.5we deduce thatTr, f ≤Sr, f

forr∈I, a contradiction.

Case 2b−1. For more details see the following subcases.

Subcase 2.1a1/0. Then by2.19, we geta/a1−fk gk. So, we haveNr,1/fk−

a1 Nr, gk. We can deduce a contradiction as in Case1.

Subcase 2.2a10. Then by2.19, we getfkgk≡1.

Case 3b0. For more details see the following subcases.

Subcase 3.1a−1/0. Then by2.19, we getfka−1/agk. So, we haveNr,1/fk−

1−a Nr, gk. We can deduce a contradiction as in Case1.

Subcase 3.2a−10. Then by2.19, we getfk ≡gk. From this, we obtainfgPz,

wherePzis a polynomial, and soTr, f Tr, g Sr, f. IfPz/≡0, then, byLemma 2.4, we get

Tr, f≤Nr, fN

r, 1 f

N

r, 1

f−P

Sr, f

≤Nr, fN

r, 1 f

N

r,1 g

Sr, f

≤2Nr, f k2Nr, gN

r,1 f

N

r,1 g

Nk1

r, 1 f

Nk1

r, 1 g

Sr, fSr, g.

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Thus, by2.5we deduce thatTr, f≤Sr, fforr ∈I, a contradiction. Therefore, we con-clude thatPz≡0, that is,f ≡g.

iiIfl1, it is easy to see that

N11

r, 1

fk₋₁ 2NL

r, 1

fk₋₁ 3NL

r, 1

gk₋₁ 2N

2 E

r, 1

fk₋₁

≤N

r, 1

gk₋₁ S

r, fSr, g.

2.23

ByLemma 2.3, we can get

NL

r, 1

fk₋₁ ≤N

r, 1

fk₋₁ N

r, 1

fk₋₁

≤N

r, fk

fk1 ≤N

r,fk1

fk S

r, f

≤N

r, 1

fk N

r, fSr, f

≤k1Nr, fNk1

r,1 f

Sr, f.

2.24

Combining with2.12,2.24, and2.14, we obtain

Tr, f≤k3Nr, f k2Nr, gN

r,1 f

N

r, 1 g

2Nk1

r,1 f

N_k1

r,1 g

Sr, fSr, g.

2.25

Without loss of generality, we suppose that there exists a setI with infinite measure such thatTr, g≤Tr, fforr∈I:

Tr, f≤2k10−k3Θ∞, f−k2Θ∞, g−Θ0, f

−Θ0, g−2δk10, f−δk10, gεTr, fSr, F

2.26

forr∈Iand 0< ε <Δ2−2k9, that is,

Δ2−2k9−εT

r, f≤Sr, f, 2.27

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Therefore, we haveh≡0, that is,

fk2_z

fk1_z−2

fk1_z

fk_z₋₁

gk2_z

gk1_z −2

gk1_z

gk_z₋₁. 2.28

By solving this equation, we obtain

gk b1fk a−b−1

bfk _a₋_b , 2.29

wherea/0, bare two constants. Using the argument ini, we can obtainfkgk ≡ 1 or f≡g. We here omit the details.

The proof ofLemma 2.5is completed.

3. Proof of

Theorem 1.4

Proof. LetFz fnf−1andGz gng−1. We have

Δ12Θ

∞, f k2Θ∞, g Θ0, f Θ0, gδk10, fδk10, g. 3.1

Since

Θ0, F 1− lim

r→ ∞

Nr,1/F

Tr, F 1−rlim→ ∞

Nr,1/fnf−1

n1Tr, f

1− lim

r→ ∞

Nr,1/fNr,1/f−1

n1Tr, f ≥1−rlim→ ∞

2Tr, f

n1Tr, f ≥ n−1 n1,

3.2

similarly,

Θ0, G≥ n−1 n1,

Θ∞, F≥ n

n1,

Θ∞, G≥ _nn

1.

3.3

Next, byLemma 2.1, we have

δ_k10, F 1−_rlim

→ ∞

N_k1r,1/F

Tr, F ≥1−rlim→ ∞

k1Nr,1/F Tr, F

≥1− lim

r→ ∞

k2Tr, f

n1Tr, f ≥1− k2

n1

n−k1

n1 .

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Similarly

δk10, G≥1−_nk2₁ n−_nk₁1. 3.5

From3.2–3.5, we get

Δ1≥2

n

n1 k2

n

n1

n−1

n1

n−1

n1

n−k1

n1

n−k1

n1 . 3.6

Sincen > 3k11, we getΔ1 > k7. Considering thatFk and Gk share 1,2, then, by

Lemma 2.5, we deduce that eitherFkGk≡1 orF≡G. Next, we consider the following two cases.

Case 1. FkGk≡1, that is,fnzfz−1k·gnzgz−1k≡1.

Case 2. F≡G, that is,

fn_f₋₁_gn_g₋₁_. _3.7

Suppose thatf /≡g; then we consider following two cases.

iLethf/gbe a constant. Then from3.7it follows thath /1, hn/1, hn1/1 and g 1−hn/1−hn1 constant, which leads to contradiction.

iiLethf/gbe not a constant. Sincef /≡g, we haveh /≡1, and hence we deduce that g 1−hn/1−hn1andg 1−hn/1−hn1h 1hh2_{· · ·}_hn−1_h/₁

hh2_{· · ·}_hn_{, where}_h_{is a nonconstant meromorphic function. It follows that}

Tr, fTr, ghnTr, h Sr, f. 3.8

On the other hand, by the second fundamental theorem, we get

Nr, fn

j1

N

r, 1

h−αj ≥n−2Tr, h S

r, f, 3.9

whereαj/1j1,2, . . . , nare distinct roots of the algebraic equationhn11.

So we have

Θ∞, f1− lim

r→ ∞

Nr, f

Tr, f ≤1−rlim→ ∞

n−2Tr, h Sr, f nTr, h Sr, f ≤1−

n−2

n

2

n, 3.10

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4. Proof of

Theorem 1.5

Proof. From3.2–3.5, we get

Δ2≥k3

n

n1 k2

n

n1

n−1

n1

n−1

n1 2

n−k1

n1

n−k1

n1 . 4.1

Sincen > 5k14, we getΔ2 > 2k9. Considering thatFk andGkshare1,2, then, by

Lemma 2.5, we deduce that eitherFkGk≡1 orF≡G.

Next, by using the argument inTheorem 1.4, we obtain the conclusion ofTheorem 1.5. We here omit the details.

Acknowledgments

This research was supported by the NSF of Key Lab of Geomathematics of Sichuan Province of China. The author wishes to thank the referee for his/her valuable comments and sugges-tions.

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