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SOME RESULTS RELATED TO A CONJECTURE OF R. BRÜCK

JI-LONG ZHANG AND LIAN-ZHONG YANG

SHANDONGUNIVERSITY, SCHOOL OFMATHEMATICS& SYSTEMSCIENCES

JINAN, SHANDONG, 250100, P. R. CHINA

jilong_zhang@mail.sdu.edu.cn lzyang@sdu.edu.cn

Received 16 October, 2006; accepted 31 January, 2007 Communicated by N.K. Govil

ABSTRACT. In this paper, we investigate the uniqueness problems of meromorphic functions that share a small function with its differential polynomials, and give some results which are related to a conjecture of R. Brück and improve some results of Liu, Gu, Lahiri and Zhang, and also answer some questions of Kit-Wing Yu.

Key words and phrases: Meromorphic function; Shared value; Small function.

2000 Mathematics Subject Classification. 30D35.

1. INTRODUCTION ANDRESULTS

In this paper a meromorphic function will mean meromorphic in the whole complex plane.

We say that two meromorphic functionsf andgshare a finite value aIM (ignoring multiplici- ties) whenf −aandg −ahave the same zeros. Iff−aandg −ahave the same zeros with the same multiplicities, then we say thatf andgshare the valueaCM (counting multiplicities).

It is assumed that the reader is familiar with the standard symbols and fundamental results of Nevanlinna theory, as found in [5] and [15]. For any non-constant meromorphic functionf, we denote byS(r, f)any quantity satisfying

r→∞lim

S(r, f) T(r, f) = 0,

possibly outside of a set of finite linear measure in R. Suppose that a(z) is a meromorphic function, we say thata(z)is a small function off,ifT(r, a) =S(r, f).

Letlbe a non-negative integer or infinite. For anya ∈CS{∞}, we denote byEl(a, f)the set of all a-points offwhere an a-point of multiplicitymis countedmtimes ifm≤landl+ 1 times ifm > l. IfEl(a, f) =El(a, g), we say thatf andg share the valueawith weightl(see [6]).

This work was supported by the NNSF of China (No. 10671109) and the NSF of Shandong Province, China(No.Z2002A01).

The authors would like to thank the referee for valuable suggestions to the present paper.

259-06

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We say thatf andg share(a, l) iff and g share the valueawith weightl. It is easy to see thatf andg share(a, l)impliesf andg share(a, p)for0 ≤p ≤ l. Also we note thatf andg share a valueaIM or CM if and only iff andgshare(a,0)or(a,∞)respectively (see [6]).

L.A. Rubel and C.C. Yang [9], E. Mues and N. Steinmetz [8], G. Gundersen [3] and L.- Z. Yang [10], J.-H. Zheng and S.P. Wang [18], and many other authors have obtained elegant results on the uniqueness problems of entire functions that share values CM or IM with their first ork-th derivatives. In the aspect of only one CM value, R. Brück [1] posed the following conjecture.

Conjecture 1.1. Letf be a non-constant entire function. Suppose that ρ1(f)is not a positive integer or infinite, iff andf0 share one finite value a CM, then

f0−a f−a =c

for some non-zero constantc, whereρ1(f)is the first iterated order off which is defined by ρ1(f) = lim sup

r→∞

log log T(r, f) log r .

R. Brück also showed in the same paper that the conjecture is true ifa = 0orN r,f10

= S(r, f)(no growth condition in the later case). Furthermore in 1998, G.G. Gundersen and L.Z.

Yang [4] proved that the conjecture is true if f is of finite order, and in 1999, L. Z. Yang [11]

generalized their results to thek-th derivatives. In 2004, Z.-X. Chen and K. H. Shon [2] proved that the conjecture is true for entire functions of first iterated orderρ1 <1/2.In 2003, Kit-Wing Yu [16] considered the case thatais a small function, and obtained the following results.

Theorem A. Letf be a non-constant entire function, letk be a positive integer, and letabe a small meromorphic function off such thata(z)6≡ 0,∞. Iff−aandf(k)−ashare the value 0CM andδ(0, f)> 34, thenf ≡f(k).

Theorem B. Let f be a non-constant, non-entire meromorphic function, let k be a positive integer, and leta be a small meromorphic function off such thata(z) 6≡ 0,∞. Iff andado not have any common pole, and iff−aandf(k)−ashare the value0CM and4δ(0, f) + 2(8 + k)Θ(∞, f)>19 + 2k,thenf ≡f(k).

In the same paper, Kit-Wing Yu [16] posed the following questions.

Problem 1.1. Can a CM shared value be replaced by an IM shared value in Theorem A?

Problem 1.2. Is the conditionδ(0, f)> 34 sharp in Theorem A?

Problem 1.3. Is the condition4δ(0, f) + 2(8 +k)Θ(∞, f)>19 + 2ksharp in Theorem B?

Problem 1.4. Can the condition “f andado not have any common pole” be deleted in Theorem B?

In 2004, Liu and Gu [7] obtained the following results.

Theorem C. Letk ≥1and letf be a non-constant meromorphic function, and letabe a small meromorphic function off such thata(z)6≡0,∞. Iff−aandf(k)−ashare the value0CM, f(k) andado not have any common poles of the same multiplicities and

2δ(0, f) + 4Θ(∞, f)>5, thenf ≡f(k).

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Theorem D. Let k ≥ 1 and let f be a non-constant entire function, and let a be a small meromorphic function off such thata(z)6≡ 0,∞. Iff −aandf(k)−ashare the value0CM andδ(0, f)> 12,thenf ≡f(k).

Letpbe a positive integer anda∈ CS

{∞}. We denote byNp)

r,f−a1

the counting func- tion of the zeros off −awith the multiplicities less than or equal top, and byN(p+1

r,f−a1

the counting function of the zeros off −a with the multiplicities larger than p. And we use Np)

r,f−a1

andN(p+1

r,f−a1

to denote their corresponding reduced counting functions (ig- noring multiplicities) respectively. We also useNp

r, f−a1

to denote the counting function of the zeros of f −a where a p-folds zero is counted m times if m ≤ p andp times if m > p.

Define

δp(a, f) = 1−lim sup

r→∞

Np

r,f−a1 T(r, f) . It is obvious thatδp(a, f)≥δ(a, f)and

N1

r, 1 f −a

=N

r, 1 f−a

.

Lahiri [6] improved Theorem C with weighted shared values and obtained the following theorem.

Theorem E. Let f be a non-constant meromorphic function, k be a positive integer, and let a≡a(z)be a small meromorphic function off such thata(z)6≡0,∞. If

(i) a(z)has no zero (pole) which is also a zero (pole) off orf(k)with the same multiplicity, (ii) f −aandf(k)−ashare(0,2),

(iii) 2δ2+k(0, f) + (4 +k)Θ(∞, f)>5 +k, thenf ≡f(k).

In 2005, Zhang [17] obtained the following result which is an improvement and complement of Theorem D.

Theorem F. Letf be a non-constant meromorphic function,k (≥ 1)andl (≥ 0)be integers.

Also, leta ≡ a(z)be a small meromorphic function off such thata(z) 6≡ 0,∞. Suppose that f −aandf(k)−ashare(0, l). Thenf ≡f(k)if one of the following conditions is satisfied,

(i) l ≥2and

(3 +k)Θ(∞, f) + 2δ2+k(0, f)> k+ 4;

(ii) l = 1and

(4 +k)Θ(∞, f) + 3δ2+k(0, f)> k+ 6;

(iii) l = 0(i.e.f −aandfk−ashare the value0IM) and

(6 + 2k)Θ(∞, f) + 5δ2+k(0, f)>2k+ 10.

It is natural to ask what happens iff(k) is replaced by a differential polynomial (1.1) L(f) = f(k)+ak−1f(k−1)+· · ·+a0f

in Theorem E or F, where aj(j = 0,1, . . . , k − 1) are small meromorphic functions of f. Corresponding to this question, we obtain the following result which improves Theorems A∼ F and answers the four questions mentioned above.

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Theorem 1.2. Letf be a non-constant meromorphic function, k(≥ 1)andl(≥ 0)be integers.

Also, leta = a(z)be a small meromorphic function off such thata(z) 6≡ 0,∞. Suppose that f −aandL(f)−ashare(0, l). Thenf ≡L(f)if one of the following assumptions holds,

(i) l ≥2and

(1.2) δ2+k(0, f) +δ2(0, f) + 3Θ(∞, f) +δ(a, f)>4;

(ii) l = 1and

(1.3) δ2+k(0, f) +δ2(0, f) + 1

1+k(0, f) + k+ 7

2 Θ(∞, f) +δ(a, f)> k 2 + 5;

(iii) l = 0(i.e.f −aandL(f)−ashare the value0IM) and

(1.4) δ2+k(0, f) + 2δ1+k(0, f) +δ2(0, f) + Θ(0, f) + (6 + 2k)Θ(∞, f) +δ(a, f)>2k+ 10.

Since δ2(0, f) ≥ δ1+k(0, f) ≥ δ2+k(0, f) ≥ δ(0, f), we have the following corollary that improves Theorems A∼F.

Corollary 1.3. Letf be a non-constant meromorphic function,k(≥1)andl(≥0)be integers, and leta ≡ a(z)be a small meromorphic function off such thata(z) 6≡ 0,∞. Suppose that f −aandf(k)−ashare(0, l). Thenf ≡f(k)if one of the following three conditions holds,

(i) l ≥2and

2+k(0, f) + 3Θ(∞, f) +δ(a, f)>4;

(ii) l = 1and 5

2+k(0, f) + k+ 7

2 Θ(∞, f) +δ(a, f)> k 2 + 5;

(iii) l = 0(i.e.f −aandL(f)−ashare the value0IM) and

2+k(0, f) + (6 + 2k)Θ(∞, f) +δ(a, f)>2k+ 10.

2. SOME LEMMAS

Lemma 2.1 ([12]). Letf be a non-constant meromorphic function. Then

(2.1) N

r, 1

f(n)

≤T(r, f(n))−T(r, f) +N

r, 1 f

+S(r, f),

(2.2) N

r, 1

f(n)

≤N

r, 1 f

+nN(r, f) +S(r, f).

Suppose thatF andGare two non-constant meromorphic functions such thatF andGshare the value 1 IM. Let z0 be a 1-point of F of order p, a 1-point of G of order q. We denote by NL r,F1−1

the counting function of those 1-points of F where p > q, by NE1) r,F1−1 the counting function of those 1-points ofF wherep = q = 1, byNE(2 r,F−11

the counting function of those 1-points of F where p = q ≥ 2; each point in these counting functions is counted only once. In the same way, we can defineNL r,G−11

,NE1) r,G−11

andNE(2 r,G−11 (see [14]). In particular, ifF andGshare 1 CM, then

(2.3) NL

r, 1

F −1

=NL

r, 1 G−1

= 0.

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With these notations, ifF andGshare 1 IM, it is easy to see that N

r, 1

F −1 (2.4)

=NE1)

r, 1 F −1

+NL

r, 1

F −1

+NL

r, 1 G−1

+NE(2

r, 1

G−1

=N

r, 1 G−1

. Lemma 2.2 ([13]). Let

(2.5) H =

F00

F0 − 2F0 F −1

− G00

G0 − 2G0 G−1

,

whereF andGare two nonconstant meromorphic functions. IfF andGshare1IM andH 6≡0, then

(2.6) NE1)

r, 1

F −1

≤N(r, H) +S(r, F) +S(r, G).

Lemma 2.3. Let f be a transcendental meromorphic function, L(f) be defined by (1.1). If L(f)6≡0, we have

(2.7) N

r, 1

L

≤T(r, L)−T(r, f) +N

r, 1 f

+S(r, f),

(2.8) N

r, 1

L

≤kN(r, f) +N

r, 1 f

+S(r, f).

Proof. By the first fundamental theorem and the lemma of logarithmic derivatives, we have N

r, 1

L

=T(r, L)−m

r, 1 L

+O(1)

≤T(r, L)−

m

r, 1 f

−m

r,L(f) f

+O(1)

≤T(r, L)−

T(r, f)−N

r, 1 f

+S(r, f)

≤T(r, L)−T(r, f) +N

r, 1 f

+S(r, f).

This proves (2.7). Since

T(r, L) = m(r, L) +N(r, L)

≤m(r, f) +m

r,L f

+N(r, f) +kN(r, f)

=T(r, f) +kN(r, f) +S(r, f),

from this and (2.7), we obtain (2.8). Lemma 2.3 is thus proved.

Lemma 2.4. Letf be a non-constant meromorphic function,L(f)be defined by(1.1), and let pbe a positive integer. IfL(f)6≡0, we have

(2.9) Np

r, 1

L

≤T(r, L)−T(r, f) +Np+k

r, 1 f

+S(r, f),

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(2.10) Np

r, 1

L

≤kN(r, f) +Np+k

r, 1

f

+S(r, f).

Proof. From (2.8), we have Np

r, 1

L

+

X

j=p+1

N(j

r, 1 L

≤Np+k

r, 1 f

+

X

j=p+k+1

N(j

r, 1 f

+kN(r, f) +S(r, f), then

Np

r, 1 L

≤Np+k

r, 1 f

+

X

j=p+k+1

N(j

r, 1 f

X

j=p+1

N(j

r, 1 L

+kN(r, f) +S(r, f)

≤Np+k

r, 1 f

+kN(r, f) +S(r, f).

Thus (2.10) holds. By the same arguments as above, we obtain (2.9) from (2.7).

3. PROOF OFTHEOREM1.2 Let

(3.1) F = L(f)

a , G= f

a.

From the conditions of Theorem 1.2, we know that F andG share(1, l)except the zeros and poles ofa(z). From (3.1), we have

(3.2) T(r, F) =O T(r, f)

+S(r, f), T(r, G)≤T(r, f) +S(r, f),

(3.3) N(r, F) =N(r, G) +S(r, f).

It is obvious thatf is a transcendental meromorphic function. Let H be defined by (2.5). We discuss the following two cases.

Case 1. H 6≡0, by Lemma 2.2 we know that (2.6) holds. From (2.5) and (3.3), we have (3.4) N(r, H)≤N(2

r, 1

F

+N(2

r, 1 G

+N(r, G) +NL

r, 1

F −1

+NL

r, 1 G−1

+N0

r, 1

F0

+N0

r, 1 G0

, where N0 r,F10

denotes the counting function corresponding to the zeros of F0 which are not the zeros of F and F −1, N0 r,G10

denotes the counting function corresponding to the zeros ofG0 which are not the zeros ofGandG−1. From the second fundamental theorem in Nevanlinna’s Theory, we have

(3.5) T(r, F) +T(r, G)≤N

r, 1 F

+N(r, F) +N

r, 1 F −1

+N

r, 1

G

+N(r, G) +N

r, 1 G−1

−N0

r, 1 F0

−N0

r, 1 G0

+S(r, f).

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Noting thatF andGshare 1 IM except the zeros and poles ofa(z), we get from (2.4), N

r, 1

F −1

+N

r, 1 G−1

= 2NE1)

r, 1 F −1

+ 2NL

r, 1 F −1

+ 2NL

r, 1 G−1

+ 2NE(2

r, 1 G−1

+S(r, f).

Combining with (2.6) and (3.4), we obtain (3.6) N

r, 1

F −1

+N

r, 1 G−1

≤N(2

r, 1 F

+N(2

r, 1

G

+N(r, G) + 3NL

r, 1 F −1

+ 3NL

r, 1 G−1

+NE1)

r, 1 F −1

+ 2NE(2

r, 1 G−1

+N0

r, 1

F0

+N0

r, 1 G0

+S(r, f).

We discuss the following three subcases.

Subcase 1.1 l ≥2. It is easy to see that (3.7) 3NL

r, 1

F −1

+ 3NL

r, 1

G−1

+ 2NE(2

r, 1 G−1

+NE1)

r, 1

F −1

≤N

r, 1 G−1

+S(r, f).

From (3.6) and (3.7), we have (3.8) N

r, 1

F −1

+N

r, 1 G−1

≤N(2

r, 1 F

+N(2

r, 1

G

+N(r, G) +N

r, 1 G−1

+N0

r, 1 F0

+N0

r, 1

G0

+S(r, f).

Substituting (3.8) into (3.5) and by using (3.3), we have (3.9) T(r, F) +T(r, G)≤3N(r, G) +N2

r, 1

F

+N2

r, 1 G

+N

r, 1

G−1

+S(r, f).

Noting that

N2

r, 1 F

=N2 r, a

L

≤N2

r, 1 L

+S(r, f), we obtain from (2.9), (3.1) and (3.9) that

(3.10) T(r, f)≤3N(r, f) +N2+k

r, 1 f

+N2

r, 1

f

−m

r, 1 G−1

+S(r, f), which contradicts the assumption (1.2) of Theorem 1.2.

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Subcase 1.2 l = 1. Noting that 2NL

r, 1

F −1

+ 3NL

r, 1

G−1

+ 2NE(2

r, 1 G−1

+NE1)

r, 1

F −1

≤N

r, 1 G−1

+S(r, f),

NL

r, 1 F −1

≤ 1 2N

r, F

F0

≤ 1 2N

r,F0

F

+S(r, f)

≤ 1 2

N

r, 1

F

+N(r, F)

+S(r, f)

≤ 1 2

N1

r, 1

F

+N(r, f)

+S(r, f)

≤ 1 2

N1+k

r, 1

f

+ (k+ 1)N(r, f)

+S(r, f), and by the same reasoning as in Subcase 1.1, we get

T(r, f)≤ k+ 7

2 N(r, f) +N2+k

r, 1 f

+N2

r, 1

f

+1 2N1+k

r, 1

f

−m

r, 1 G−1

+S(r, f), which contradicts the assumption (1.3) of Theorem 1.2.

Subcase 1.3 l = 0. Noting that NL

r, 1

F −1

+ 2NL

r, 1 G−1

+ 2NE(2

r, 1 G−1

+NE1)

r, 1

F −1

≤N

r, 1 G−1

+S(r, f), 2NL

r, 1

F −1

+NL

r, 1 G−1

≤2N

r, 1 F0

+N

r, 1

G0

, and by the same reasoning as in the Subcase 1.2, we get a contradiction.

Case 2. H ≡0. By integration, we get from (2.5) that

(3.11) 1

G−1 = A

F −1 +B, whereA(6= 0)andB are constants. From (3.11) we have

(3.12) N(r, F) = N(r, G) = N(r, f) =S(r, f), Θ(∞, f) = 1, and

(3.13) 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 subcases.

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Subcase 2.1 Suppose thatB 6= 0,−1.From (3.13) we haveN r,1/ G− B+1B

=N(r, F).

From this and the second fundamental theorem, we have T(r, f)≤T(r, G) +S(r, f)

≤N(r, G) +N

r, 1 G

+N r, 1 G− B+1B

!

+S(r, f)

≤N

r, 1 G

+N(r, F) +N(r, G) +S(r, f)

≤N

r, 1 f

+S(r, f), which contradicts the assumptions of Theorem 1.2.

Subcase 2.2 Suppose thatB = 0. From (3.13) we have

(3.14) G= F + (A−1)

A , F =AG−(A−1).

If A 6= 1, from (3.14) we can obtain N r,1/ G− A−1A

= N(r,1/F). From this and the second fundamental theorem, we have

2T(r, f)≤2T(r, G) +S(r, f)

≤N(r, G) +N

r, 1 G

+N

r,1/

G−A−1 A

+N

r, 1 G−1

+S(r, f)

≤N

r, 1 G

+N

r, 1

F

+N

r, 1 G−1

+S(r, f),

which contradicts the assumptions of Theorem 1.2. ThusA = 1. From (3.14) we haveF ≡G, thenf ≡L.

Subcase 2.3 Suppose thatB =−1, from (3.13) we have

(3.15) G= A

−F + (A+ 1), F = (A+ 1)G−A

G .

If A 6= −1, we obtain from (3.15) that N r,1/ G−A+1A

= N(r,1/F). By the same reasoning discussed in Subcase 2.2, we obtain a contradiction. HenceA = −1. From (3.15), we getF ·G≡1, that is

(3.16) f·L≡a2.

From (3.16), we have

(3.17) N

r, 1

f

+N(r, f) = S(r, f),

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and soT

r,f(k)f

=S(r, f).From (3.17), we obtain 2T

r, f

a

=T

r,f2 a2

=T

r,a2 f2

+O(1)

=T

r,L f

+O(1) =S(r, f),

and soT(r, f) = S(r, f), this is impossible. This completes the proof of Theorem 1.2.

4. REMARKS

Letf andgbe non-constant meromorphic functions,a(z)be a small function off andg, and k be a positive integer or ∞. We denote byNk)E(r, a)the counting function of common zeros off−aandg −awith the same multiplicitiesp≤k,byN(k+10 (r, a)the counting function of common zeros off−aandg−awith the multiplicitiesp≥k+ 1, and denote byN0(r, a)the counting function of common zeros off−aandg −a; each point in these counting functions is counted only once.

Definition 4.1. Letfandgbe non-constant meromorphic functions,abe a small function off andg, andkbe a positive integer or∞. We say thatf andgshare“(a, k)”ifk= 0, and

N

r, 1 f−a

−N0(r, a) =S(r, f), N

r, 1

g−a

−N0(r, a) =S(r, g);

ork 6= 0, and

Nk)

r, 1 f −a

−Nk)E(r, a) =S(r, f), Nk)

r, 1

g−a

−Nk)E(r, a) =S(r, g),

N(k+1

r, 1 f −a

−N(k+10 (r, a) = S(r, f), N(k+1

r, 1

g−a

−N(k+10 (r, a) = S(r, g).

By the above definition and a similar argument to that used in the proof of Theorem 1.2, we conclude that Theorem 1.2 and Corollary 1.3 still hold if the condition thatf−aandL(f)−a (orf(k)−a) share(0, l)is replaced by the assumption that f −a andL(f)−a (orf(k) −a) share“(0, l)”.

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