Unicity of Meromorphic Function Sharing One Small Function with Its Derivative

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Volume 2010, Article ID 507454,11pages doi:10.1155/2010/507454

Research Article

Unicity of Meromorphic Function Sharing One

Small Function with Its Derivative

Ang Chen,

1

_{Xiuwang Wang,}

2

_{and Guowei Zhang}

1

1_{Department of Mathematics, Shandong University, Jinan, Shandong 250100, China} 2_{Department of Mathematics, Xinxiang University, Xinxiang, Henan 453002, China}

Correspondence should be addressed to Ang Chen,[email protected]

Received 26 October 2009; Revised 2 February 2010; Accepted 3 February 2010

Academic Editor: Stanisława R. Kanas

Copyrightq2010 Ang Chen et al. 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 deal with the problem of uniqueness of a meromorphic function sharing one small function with its k’s derivative and obtain some results.

1. Introduction and Main Results

In this article, a meromorphic function means meromorphic in the open complex plane. We assume that the reader is familiar with the Nevanlinna theory of meromorphic functions and the standard notations such asTr, f, mr, f, Nr, f, Nr, f,and so on.

Let f and g be two nonconstant meromorphic functions; a meromorphic function

az/≡ ∞is called a small functions with respect tofprovided thatTr, a Sr, f.Note that the set of all small function offis a field. Letbzbe a small function with respect tof andg.We say thatfandgsharebzCMIMprovided thatf−bandg−bhave same zeros counting multiplicitiesignoring multiplicities.

Moreover, we use the following notations.

Letkbe a positive integer. We denote byNkr,1/f−athe counting function for the zeros off −awith multiplicity≤ kand byNkr,1/f−athe corresponding one for which the multiplicity is not counted. LetNkr,1/f−abe the counting function for the

zeros off−awith multiplicity≥ k, and letNkr,1/f−abe the corresponding one for

which the multiplicity is not counted. SetNkr,1/f−a Nr·1/f−a N2r,1/f−

a · · ·Nkr,1/f−a.And we define

δpa, f1−lim sup r−→∞

Npr,1/f−a

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Obviously, 1 ≥ Θa, f ≥ δpa, f ≥ δa, f ≥ 0. For more details, reader can see

1,2.

Br ¨ucksee3considered the uniqueness problems of an entire function sharing one value with its derivative and proved the following result.

Theorem A. Let fbe nonconstant entire function. If f and f share the value 1 CM and if

Nr,1/f Sr, f, thenf−1/f−1≡cfor some constantc∈C\ {0}.

Yang4, Zhang5, and Yu6extended Theorem A and obtained many excellent results.

Theorem Bsee5. Letfbe a nonconstant meromorphic function and, let kbe a positive integer. Suppose thatfand fkshare 1 CM and

2Nr, fN

r, 1 f

N

r, 1 fk

<λo1Tr, fk , 1.2

forr∈I,whereIis a set of infinite linear measure andλsatisfies 0< λ <1,thenfk−1/f−1≡c

for some constantc∈C\ {0}.

Theorem Csee6. Letfbe a nonconstant, nonentire meromorphic function andaz/≡0,∞be a small function with respect tof.If

1fandazhave no common poles,

2f−aandfk−ashare the value 0 CM,

34δ0, f 2k8Θ∞, f>2k19,thenf ≡fk,wherekis a positive integer.

In the same paper, Yu6posed four open questions. Lahiri and Sarkar7and Zhang

8studied the problem of a meromorphic or an entire function sharing one small function with its derivative with weighted shared method and obtained the following result, which answered the open questions posed by Yu6.

Theorem Dsee8. Letfbe a non-constant meromorphic function and, letkbe a positive integer. Also letaz/≡0, ∞be a meromorphic function such thatTr, a Sr, f.Suppose thatf−aand

fk₋_a_{share 0 IM and}

4Nr, f3N2

r, 1 fk

2N

r, 1 f/a

<λo1Tr, fk , 1.3

for 0< λ < 1, r ∈I,andI is a set of infinite linear measure. Thenfk−a\f−a≡cfor some constantc∈C\ {0}.

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Theorem 1.1. Letk≥1, n≥1be integers and letfbe a non-constant meromorphic function. Also letaz/≡0, ∞be a small function with respect tof. IffandfnkshareazIM and

4Nr, f2N

r, 1 f/a

2N2

r, 1 fnk

N

r, 1 fnk

≤λo1Tr,fnk ,

1.4

orfandfnkshareazCM and

2Nr, fN

r, 1 f/a

N2

r, 1 fnk

≤λo1Tr,fnk , 1.5

for 0< λ <1,r∈I, andIis a set of infinite linear measure, thenf−a\fnk−a≡c,for some constantc∈C\ {0}.

Theorem 1.2. Letk≥1, n≥1be integers andfbe a non-constant meromorphic function. Also let

az/≡0, ∞be a small function with respect tof. IffandfnkshareazIM and

2k6Θ∞, f3Θ0, f2δk20, f>2k10, 1.6

orfandfnkshareazCM and

k3Θ∞, fδ2

0, fδk20, f> k4, 1.7

thenf ≡fnk.

Clearly,Theorem 1.1improves and extends Theorems B and D, while1.2improves and extends Theorem C.

2. Some Lemmas

In this section, first of all, we give some definitions which will be used in the whole paper.

Definition 2.1. LetFandGbe two meromorphic functions defined inC; assume, thatFand

Gshare 1 IM; letz0be a zero ofF−1 with multiplicitypand a zero ofG−1 with multiplicity

q.We denote byN_E1r,1/F−1the counting function of the zeros of F−1 wherep q 1 and byN2_Er,1/F−1the counting function of zeros ofF−1 wherep q ≥ 2.We denotes byNLr,1/F−1the counting function of the zeros ofF−1 wherep > q≥ 1; each point is counted according to its multiplicity, andNLr,1/F−1denote its reduced form. In the same

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Definition 2.2. In this paperN0r,1/Fdenotes the counting function of the zeros ofFwhich are not the zeros ofFandF−1,and N0r,1/Fdenotes its reduced form. In the same way, we can defineN0r,1/GandN0r,1/G.

Next we present some lemmas which will be needed in the sequel. LetF, Gbe two non-constant meromorphic functions defined inC.We shall denote byHthe following function:

H

_F

F −2

F

F−1

−

_G

G −2

G

G−1

. 2.1

Lemma 2.3see2. LetF, Gbe two nonconstant meromorphic functions defined inC.IfF andG are sharing 1 IM, then

Nr, H≤Nr, F N2

r,1 F

N2

r, 1 G

NL

r, 1 F−1

NL

r, 1 G−1

N0

r, 1 F

N0

r, 1 G

Sr, f.

2.2

IfFandGare sharing 1 CM, then

Nr, H≤Nr, F N₂

r, 1 F

N₂

r, 1 G

N0

r, 1 F

N0

r, 1 G

Sr, f. 2.3

Lemma 2.4see1. Letfbe a meromorphic function andais a finite complex number. Then

iTr,1/f−a Tr, f O1,

iimr, fk/fl Sr, ffork > l≥0,

iiiTr, f≤Nr, f Nr,1/f−a1z Nr,1/f−a2z Sr, f,

wherea1z a2zare two meromorphic functions such thatTr, ai Sr, f,i1,2.

Lemma 2.5 see7. Let f be a non-constant meromorphic function, and k, p are two positive integers. Then

Np

r, 1 fk

≤Npk

r,1 f

kNr, fSr, f. 2.4

Lemma 2.6see9. Letfbe a non-constant meromorphic function and letnbe a positive integer.

Pf anfnan−1fn−1· · ·a1fwhereaiare meromorphic functions such thatTr, ai Sr, fi

1,2, . . . , n,andan/≡0.Then

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3. Proof of

Theorem 1.1

LetFfz/az, G fnzk/az,then

F−1 fz−az

az ,

G−1 f

n_zk₋_a_z

az .

3.1

From the definitions ofF, Gand recalling thatFandGshare value 1 IMCM, we get

N_E1

r, 1 F−1

N_E1

r, 1 G−1

Sr, f,

N2_E

r, 1 F−1

N2_E

r, 1 G−1

Sr, f,

3.2

NL

r, 1 F−1

≤N

r,1 F

Nr, F Sr, F, 3.3

N

r, 1 F−1

N

r, 1 G−1

Sr, F N1_E

r, 1 F−1

N2_E

r, 1 F−1

NL

r, 1 F−1

NL

r, 1 G−1

Sr, f.

3.4

We will distinguish two cases below.

Case 1H /≡0. From2.1it is easy to see thatmr, H Sr, f.

Subcase 1.1. Suppose thatfandfnkshareazIM. According to3.1,FandGshare 1 IM except the zeros and poles ofaz.By3.1, we have

Nr, F Nr, fSr, f, Nr, G Nr, fSr, f. 3.5

Letz0be a simple zero ofF−1 andG−1,butaz0/0,∞.Through a simple calculation we know thatz0is a zero ofH,so

N_E1

r, 1 F−1

≤N

r, 1 H

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From3.4–3.6andLemma 2.3, we have

Nr, 1 G−1

≤Nr, F 2NL

r, 1 F−1

2NL

r, 1 G−1

N₂r, 1 F N₂ r, 1 G

N2_E

r, 1 F−1

N0 r, 1 F N0 r, 1 G

Sr, f

≤Nr, f2N

r, 1 F 2NL r, 1 G−1

N2 r, 1 G N0 r, 1 G

Sr, f.

3.7

It follows by the second fundamental theorem,3.5, and3.7that

Tr, G≤Nr, G N

r, 1 G N r, 1 G−1

−N0

r, 1 G

Sr, G

≤2Nr, f2N

r, 1 F 2N r, 1 G N r, 1 G

Sr, f.

3.8

ByLemma 2.5, we have

Tr,fnk _≤₄_N_{r, f}₂_N

r, 1 f/a

2N2

r, 1 fnk

N

r, 1 fnk

Sr, f,

3.9

which contradicts1.4.

Subcase 1.2. Suppose thatfandfnkshareazCM.

Letz0be a simple zero ofF−1 andG−1, butaz0/0,∞.By a simple calculation, we can still getHz0 0.Therefore

N₁

r, 1 F−1

≤N r, 1 H

Sr, f≤Nr, H Sr, f. 3.10

Noting thatN1r,1/F−1N1r,1/G−1 Sr, f,by3.4andLemma 2.3, we can deduce

N

r, 1 G−1

≤Nr, F N₂

r,1 F N₂ r, 1 G N0 r, 1 F N0 r, 1 G N₂ r, 1 F−1

Sr, f.

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By the second fundamental theorem,3.5, and3.11, we have

Tr, G≤Nr, G N

r, 1 G

N

r, 1 G−1

−N0

r, 1 G

Sr, G

≤2Nr, fN2

r, 1 G

N

r, 1 F

Sr, f.

3.12

Taking into account3.1, we have

Tr,fnk _≤₂_N_{r, f}_N

r, 1

f/a

N2

r, 1

fnk

Sr, f. 3.13

This contradicts1.5.

Case 2H≡0. Integration yields

1

F−1 ≡

A

G−1 B, 3.14

whereA, Bare constants andA /0.It is easy to see thatFandGshare 1 CM. Now we claim thatB0.

IfNr, f/Sr, f,then by 3.14 we get B 0.So our claim holds. Hence we can assume that

Nr, fSr, f. 3.15

IfB /0,then we can rewrite3.14as

1

F−1 ≡

BG−1A/B

G−1 . 3.16

So

N

r, 1 G−1A/B

Nr, F Sr, f. 3.17

IfA /B,then byLemma 2.4and3.17we have

Tr, G≤Nr, G N

r, 1 G

N

r, 1 G−1A/B

Sr, f

≤N

r, 1 G

Sr, f≤Tr, G Sr, f.

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Hence

Tr, G N

r, 1 G

Sr, f, 3.19

that is,

Tr,fnk _N

r, 1

fnk

Sr, f. 3.20

This is a contradiction with1.4and 1.5. IfA B,then from3.14we get 1/F−1

AG/G−1.We rewrite it as

− a2

fn_Af₋_a₋_aA ≡

fnk

fn . 3.21

So by Lemmas2.4and2.6and3.15, we have

n1Tr, fT

r,

fnk fn

Sr, f

≤nN

r, 1 f

kNr, fSr, f≤nTr, fSr, f.

3.22

This implies thatTr, f Sr, f, sincen≥1. This is impossible. Hence our claim is right. So

G−1/F−1 A.Theorem 1.1is, thus, completely proved.

4. Proof of

Theorem 1.2

The proof is similar to the proof ofTheorem 1.1. LetF andGbe defined as inTheorem 1.1; hence, we have3.1–3.5. We still distinguish two cases.

Case 1. H /≡0

Subcase 1.1. Suppose thatf and fnk shareazIM, then we can still get3.6and3.7. Then by the second fundamental theorem,Lemma 2.3, and3.5we have

Tr, F≤Nr, F N

r,1 F

N

r, 1 F−1

−N0

r, 1 F

Sr, F

≤2Nr, f2N

r, 1 G

2N

r, 1 F

N

r, 1 F

Sr, f.

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ApplyingLemma 2.5to the above inequality and noticing the definition ofF, G,we get

Tr, f≤2k6Nr, f3N

r,1 f

2Nk2N

r,1 f

Sr, f

≤2k61−Θ∞, f3−3Θ0, f2−2δk20, fTr, fSr, f.

4.2

This implies that

2k6Θ∞, f3Θ0, f2δk20, f≤2k10. 4.3

Subcase 1.2. Suppose that f and fnk share az CM. Similarly as above, we can easily obtainN1r,1/F−1 N1r,1/G−1Sr, f; byLemma 2.3, we can deduce

N

r, 1 F−1

≤Nr, F N2

r, 1 F

N2

r, 1 G

N0

r, 1 F

N0

r, 1 G

N₂r, 1 G−1

Sr, f.

4.4

So by the second fundamental theorem,4.4, and usingLemma 2.5again, we have

Tr, F≤Nr, F N

r, 1 F

N

r, 1 F−1

−N0

r, 1 F

Sr, f

≤2Nr, fN2

r,1 f

N

r, 1 G

Sr, f

≤k5−k3Θ∞, f−δ2

0, f−δk20, fTr, fSr, f.

4.5

This implies that

k3Θ∞, fδ2

0, fδk20, f≤k4. 4.6

Case 2H≡0. Similarly, we can also get3.14. Next we claim thatB0. IfNr, f/Sr, f,

then it follows that B 0 from 3.14. Hence, we may assume that 3.15 holds. If

B /0 andB / −1,then

A G−1 ≡ −

BF−B1

F−1 , 4.7

and so

Nr, G N

r, 1 F−B1/B

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Again by second fundamental theorem and4.4we have

Tr, F N

r, 1 F

Sr, f, 4.9

that is,

Tr, f≤N

r, 1 f

Sr, f≤Tr, fSr, f. 4.10

Then we haveTr, f Nr,1/f,and it follows thatΘ0, f 0 and from3.15we have

Θ∞, f 1; then with1.6and1.7we may deduceδk20, f>1.It is impossible, and we can assume thatB−1; thus, we can get

fnk

a −A1≡ −A·a·

1

f. 4.11

It shows thatTr, f Tr,fnk.

IfA−1, by4.11, then we havef·fnk ≡a2_,_{which with the above equality may} lead toTr, f Sr, f,which is impossible. IfA /−1,then by second fundamental theorem,

Lemma 2.5,3.15, and4.11we have

Tr,fnk _≤_N

r, 1

fnk₋_a_A₁

N

r, 1 fnk

Sr, f,

≤kNr, fNk2

r,1 f

Sr, f≤Tr, fSr, f,

4.12

which with3.15may deduceNk2r,1/f Tr, f Sr, f; soδk2o, f 0,which with

Θ∞, f 1 and1.6may deduceΘ0, f>1,which is impossible. Hence our claim holds. Next we will prove thatA1.From3.17we haveG−1≡AF−1.Then

N

r, 1 G

N

r, 1 F1/A−1

. 4.13

IfA /1,then we have

Tr, F≤Nr, F N

r,1 F

N

r, 1 G

Sr, f. 4.14

ByLemma 2.5, we get

Tr, f≤k1Nr, fN

r,1 f

Nk2

r,1 f

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It implies that

k1Θ∞, f Θ0, fδk20, f≤k2. 4.16

Combining4.16with1.6yields

2k2 Θ0, f≥2k3Θ∞, f3Θ0, f2δ2k0, f−4Θ∞, f>2k6, 4.17

that is,Θ0, f>2.This is a contradiction. Combining4.16with1.7yields

k22Θ∞, f≥k3Θ∞, f Θ0, fδk20, f> k4, 4.18

that is, Θ∞, f > 1, which is also a contradiction. Hence A 1 and f ≡ fnk. Now

Theorem 1.2has been completely proved.

Acknowledgment

The authors would like to express their sincere thanks to the referee for helpful comments and suggestions.

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