On a uniqueness theorem of Tohge

Arch. Math. 84 (2005) 461–4690003–889X/05/050461–09DOI 10.1007/s00013-004-1217-6© Birkhauser Verlag, Basel, 2005 Archiv der Mathematik

On a uniqueness theorem of Tohge

By

Indrajit Lahiri and Arindam Sarkar

Abstract. We prove a uniqueness theorem for meromorphic functions which improves a resultof Tohge and answers an open question posed by him.

1. Introduction, definitions and results. Let f , g be nonconstant meromorphicfunctions defined in the open complex plane C. For a ∈ C ∪ {∞} we say that f , gshare the value a CM (counting multiplicities) if f , g have the same a-points with the samemultiplicity, and we say that f , g share the value a IM (ignoring multiplicities) if f , g havethe same a-points and the multiplicities are not taken into account.

Let k be a positive integer or infinity. We denote byEk)(a; f ) the set of distinct a-pointsof f with multiplicities not exceeding k. We shall say that a value a ∈ C ∪ {∞} is a Picardexceptional value (in short evP) of f if f does not assume the value a.

K. Tohge [8] proved the following result.

Theorem A [8]. Let f , g be nonconstant meromorphic functions sharing 0, 1,∞ CMand f ′, g′ share 0 CM. Then f and g satisfy one of the following: (i) f ≡ g,(ii) fg ≡ 1, (iii) (f − 1)(g − 1) ≡ 1, (iv) f + g ≡ 1, (v) f ≡ cg, (vi) f − 1 ≡ c(g − 1),(vii) [(c − 1)f + 1][(c − 1)g − c] + c ≡ 0, where c(�= 0, 1) is a constant.

Tohge [8] left the problem of weakening the restriction of CM sharing as open. RecentlyA. H. H. Al-Khaladi [1] dealt with this open problem and proved the following result whichimproved Theorem A.

Theorem B [1]. Let f , g be two nonconstant meromorphic functions sharing 0,∞ CMand f ′, g′ share 0 IM. If Ek)(1; f ) = Ek)(1; g), where k is a positive integer or infinity,then the conclusion of Theorem A holds.

In the paper we concentrate our attention on the problem of Tohge and Theorem A. Wesee in the following example that it is not possible in Theorem A to relax the nature ofsharing of ∞ from CM to IM.

Mathematics Subject Classification (2000): 30D35.

462 I. Lahiri and A. Sarkar arch. math.

E x a m p l e 1. Let f = (ez−1)−2 and g = (ez−1)−1. Then f , g share 0, 1 CM, ∞ IMand f ′, g′ share 0 CM but f , g do not satisfy any one of the relations given in Theorem A.

The purpose of the paper is to investigate the possibility of improving Theorem A byrelaxing the nature of sharing of the values 0, 1,∞ and thereby provide an answer to theabove problem of Tohge. To this end we employ the notion of weighted sharing, whichmeasures how close a shared value is to being shared IM or to being shared CM.

D e f i n i t i o n [3, 4]. Let k be a nonnegative integer or infinity. For a ∈ C ∪ {∞}we denote by Ek(a; f ) the set of all a-points of f where an a-point of multiplicity m iscounted m times if m� k and k + 1 times if m > k. If Ek(a; f ) = Ek(a; g), we say thatf , g share the value a with weight k.

The definition implies that if f , g share a value a with weight k then zo is an a-pointof f with multiplicity m(� k) if and only if it is an a-point of g with multiplicity m(� k)and zo is an a-point of f with multiplicity m(> k) if and only if it is an a-point of g withmultiplicity n(> k) where m is not necessarily equal to n.

We write f , g share (a, k) to mean that f , g share the value a with weight k. Clearlyif f , g share (a, k) then f , g share (a, p) for all integers p, 0 �p < k. Also we note thatf , g share a value a IM or CM if and only if f , g share (a, 0) or (a, ∞) respectively.

We now state the main results of the paper.

Theorem 1. Let f , g be two nonconstant meromorphic functions sharing (0, k), (1,m),(∞, 1), where (m − 1)(mk − 1) > (1 + m)2. If E1)(0; f ′) ⊂ E∞)(0; g′) and E1)(0; g′)⊂ E∞)(0; f ′) then one of the following holds: (i) f ≡ g, (ii) fg ≡ 1, (iii) (f − 1)(g − 1) ≡ 1, (iv) f + g ≡ 1, (v) f ≡ cg, (vi) f − 1 ≡ c(g − 1), (vii) [(c − 1)f + 1][(c − 1)g − c] + c ≡ 0, where c(�= 0, 1) is a constant.

Theorem 2. Let f , g be two nonconstant meromorphic functions sharing (0, 1), (1,m),(∞, k), where (m − 1)(mk − 1) > (1 + m)2. If E1)(0; f ′) ⊂ E∞)(0; g′) and E1)(0; g′)⊂ E∞)(0; f ′) then one of the following holds : (i) f ≡ g, (ii) fg ≡ 1, (iii) (f − 1)(g − 1) ≡ 1, (iv) f + g ≡ 1, (v) f ≡ cg, (vi) f − 1 ≡ c(g − 1), (vii) [(c − 1)f + 1][(c − 1)g − c] + c ≡ 0, where c(�= 0, 1) is a constant.

We note that the condition (m−1)(mk−1) > (1+m)2 in Theorems 1 and 2 is equivalentto (m − 1)(k − 1) > 4 and so is symmetric in k and m. Also we see that Theorems1 and 2 hold for the following pairs of least values ofm and k: (i)m = 3, k = 4; (ii)m = 4,k = 3; (iii) m = 2, k = 6; (iv) m = 6, k = 2.

R e m a r k 1. It is clear that the case (iv) in Theorems 1 and 2 is equivalent to (1 − 1f)

(1 − 1g) ≡ 1 and the case (vii) is equivalent to 1 − 1

f≡ c(1 − 1

g). So under the hypotheses

of Theorems 1 and 2 we see that one of the pairs (f, g), (1 − f, 1 − g), (1 − 1f, 1 − 1

g) is

either proportional or reciprocal.

Vol. 84, 2005 On a uniqueness theorem of Tohge 463

The following definitions will be required in the sequel.

D e f i n i t i o n 2 [5]. Let p be a positive integer and b ∈ C ∪ {∞}. Then by N(r, b;f | �p) we denote the counting function of those b-points of f (counted with propermultiplicities) whose multiplicities are not greater than p. By N(r, b; f | �p) we denotethe corresponding truncated counting function.

In an analogous manner we define N(r, b; f | �p) and N(r, b; f | �p).

D e f i n i t i o n 3. Let f and g share a value a IM. Let z be an a-point of f and g withmultiplicities pf (z) and pg(z) respectively.

We put

νf (z) = 1 if pf (z) > pg(z)

= 0 if pf (z)�pg(z)

and

µf (z) = 1 if pf (z) < pg(z)

= 0 if pf (z)�pg(z).

Let n(r, a; f > g) = ∑|z|�r

νf (z) and n(r, a; f < g) = ∑|z|�r

µf (z). We now denote

by N(r, a; f > g) and N(r, a; f < g) the integrated counting functions obtained fromn(r, a; f > g) and n(r, a; f < g) respectively.

Finally we put N∗(r, a; f, g) = N(r, a; f > g)+N(r, a; f < g).Throughout the paper we denote by f and g two nonconstant meromorphic functions

defined in C. We do not explain the standard definitions and notations of the value distri-bution theory as these are available in [2].

2. Lemmas. In this section we present some lemmas which will be needed inthe sequel.

Lemma 1 [3]. If f , g share (0, 0), (1, 0), (∞, 0) then (i) T (r, f )� 3T (r, g)+ S(r, f ),(ii) T (r, g)� 3T (r, f )+ S(r, g).

This shows that S(r, f ) = S(r, g) and we denote them by S(r).

Lemma 2 [6]. Letf , g share (0, 1), (1,m), (∞, k) andf �≡ g, where (m−1)(mk−1) >(1 +m)2. Then N(r, a; f | � 2) = S(r) and N(r, a; g | � 2) = S(r) for a = 0, 1,∞.

Lemma 3. Let f , g share (0, k), (1,m), (∞, 1) and f �≡ g, where (m− 1)(mk − 1) >(1 +m)2. Then N(r, a; f | � 2) = S(r) and N(r, a; g | � 2) = S(r) for a = 0, 1,∞.

P r o o f. If we apply Lemma 2 to 1/f and 1/g, we obtain Lemma 3. �


The following lemma can be proved in the same way as the statements (iii) and (iv) ofLemma 2.3 in [7].

Lemma 4. Let f, g share (0, 0), (1, 0), (∞, 0) and f �≡ g. If α = (f − 1)/(g − 1)and h = g/f then

(i) N(r, 0;α) = N(r,∞; f < g)+N(r, 1; f > g),(ii) N(r,∞;α) = N(r,∞; f > g)+N(r, 1; f < g),

(iii) N(r, 0;h) = N(r, 0; f < g)+N(r,∞; f > g),(iv) N(r,∞;h) = N(r, 0; f > g)+N(r,∞; f < g).

Lemma 5. Let f , g share (0, 1), (1,m), (∞, k) or (0, k), (1,m), (∞, 1) and f �≡ g,where (m−1)(mk−1) > (1+m)2. If α and h are defined as in Lemma 4 thenN(r, a;α) =S(r) and N(r, a;h) = S(r) for a = 0,∞.

P r o o f. The lemma follows from Lemmas 2, 3 and 4 becauseN∗(r, a; f, g)�N(r, a; f| � 2) for a = 0, 1,∞. �

Lemma 6. Let f and g share (0, 0), (1, 0), (∞, 0). If f is a bilinear transformation ofg then f and g satisfy one of the following: (i) f ≡ g, (ii) fg ≡ 1, (iii) (f −1)(g−1) ≡ 1,(iv) f +g ≡ 1, (v) f ≡ cg, (vi) f −1 ≡ c(g−1), (vii) [(c−1)f +1][(c−1)g−c]+c ≡ 0,where c(�= 0, 1) is a constant.

P r o o f. Let

f ≡ Ag + B

Cg +D,(1)

where A,B,C,D are constants and AD − BC �= 0.If 0, 1,∞ are not evP of f and g then obviously f ≡ g and the case (i) holds. So we

now suppose that f �≡ g and consider the following possibilities.

C a s e (i). Let 0, 1 be evP of f and g. Since ∞ is not an evP of f , g and f , g share(∞, 0), we get from (1) C = 0 and so f ≡ A

Dg + B

D, where D �= 0.

Since 0, 1 are evP of f and g, it follows that 0, 1, BD, A+B

Dare evP of f . Since f �≡ g,

it then follows that BD

= 1 and A+BD

= 0. So f + g ≡ 1, which is (iv).

C a s e (ii). Let 0,∞ be evP of f and g. Then from (1) we see that 0,∞, BD, AC

are evPof f . Since f �≡ g, this implies that A = 0 and D = 0. So from (1) we get fg ≡ B

C.

Since 1 is not an evP of f , g and f , g share (1, 0), it follows that BC

= 1 and so fg ≡ 1,which is (ii).

C a s e (iii). Let 1,∞ be evP of f and g. Then 1,∞, AC

are evP of f and 1,∞,−DC

areevP of g. So we see that A = C and C = −D, Again since 0 is not an evP of f and g, weget from (1) B = 0. Hence we obtain from (1) f ≡ Ag

Ag−A, where A �= 0. This implies(f − 1)(g − 1) ≡ 1, which is (iii).


C a s e (iv). Let 0 be an evP of f and g but 1,∞ be not. Then we get from (1) C = 0and A + B = D and so f ≡ cg + 1 − c, where c = A

A+B �= 0. Since f �≡ g, c �= 1 andf − 1 ≡ c(g − 1), which is (vi).

C a s e (v). Let 1 be an evP of f and g but 0,∞ be not. Then from (1) we get B = 0and C = 0. Letting c = A

D(�= 0) we obtain from (1) f ≡ cg, where c �= 1 because f �≡ g,

which is (v).

C a s e (vi). Let ∞ be an evP of f and g but 0, 1 be not. Then from (1) we getB = 0,A =C+D and so f ≡ (C+D)g

Cg+D . SinceB = 0 andAD−BC �= 0, it follows thatA = C+D �= 0.

So from above we get (c − 1)fg − cf + g ≡ 0, where c = DC+D �= 0. Since f �≡ g, it

follows that c �= 1. Now [(c−1)f +1][(c−1)g−c]+c ≡ (c−1)[(c−1)fg−cf +g] ≡ 0,which is (vii). This proves the lemma. �

3. Proof of the main results. We prove Theorem 1 only as the proof of Theorem 2is similar.

P r o o f o f T h e o r e m 1. If possible, we suppose that f is not a bilinear transformationofg. We see thatf = (1−α)/(1−αh) andg = (1−α)h/(1−αh), whereα andh are definedas in Lemma 4. Since f is not a bilinear transformation of g, it follows that α, h and αh arenonconstant. Let b = α′h/(αh′ +α′h). Then (f −b)(1−αh) = (1−α)−b(1−αh) = F ,say. Also (f − g)(1 − αh) = (1 − α)(1 − h) and (g − 1)(1 − αh) = h − 1 so thatf − g = (α − 1)(g − 1). Again

g′

g= h′(1 − αh)+ (h− 1)(α′h+ αh′)

h(1 − α)(1 − αh).

Therefore

g′(g − f )

g(g − 1)= h′(1 − αh)+ (h− 1)(α′h+ αh′)

h(1 − αh)

= (1 − α)(αh′ + α′h)− α′h(1 − αh)

αh(1 − αh).(2)

Again

f − b = (1 − α)− b(1 − αh)

1 − αh= (1 − α)(αh′ + α′h)− α′h(1 − αh)

(1 − αh)(αh′ + α′h)and so

(f − b)

(h′

h+ α′

α

)= (1 − α)(αh′ + α′h)− α′h(1 − αh)

αh(1 − αh).(3)

From (2) and (3) we get

g′(g − f )

g(g − 1)= (f − b)

(h′

h+ α′

α

).(4)


Since F ′ = −α′ − b′(1 − αh)+ b(α′h+ αh′) = −α′ − b′(1 − αh)+ α′h, we get

F ′

F− α′

α= −α′ − b′(1 − αh)+ α′h− α′

αF

F= 1

f − b

[α′

α(b − 1)− b′

]

and so

1

f − b=

F ′F

− α′α

α′α(b − 1)− b′ .(5)

From the definitions of α and h we see that T (r, α)� T (r, f ) + T (r, g) + O(1) andT (r, h)� T (r, f )+ T (r, g)+O(1). So by Lemma l we note that m(r, α′/α) = S(r, α) =S(r) and m(r, h′/h) = S(r, h) = S(r). Since the poles of α′/α are all simple and occur atthe zeros and poles of α, it follows that N(r, α′/α) = N(r, 0;α)+ N(r,∞;α). SimilarlyN(r, h′/h) = N(r, 0;h)+N(r,∞;h). Therefore in view of Lemms 5 we obtain

T

(r,α′

α

)= m

(r,α′

α

)+N

(r,α′

α

)

= N(r, 0;α)+N(r,∞;α)+ S(r, α) = S(r),

T

(r,h′

h

)= m

(r,h′

h

)+N

(r,h′

h

)

= N(r, 0;h)+N(r,∞;h)+ S(r, h) = S(r)

and so

T (r, b) = T

(r,

1

b

)+O(1)

= T

(r, 1 + αh′

α′h

)+ O(1)

� T(r,α

α′)

+ T

(r,h′

h

)+O(1)

= T

(r,α′

α

)+ T

(r,h′

h

)+O(1) = S(r).

Since F ′/F and α′/α have no multiple pole and T (r, b′)� 2T (r, b)+ S(r, b), it followsfrom the above and (5) that

N(r, 0; f − b | � 2)

� 2N

(r, 0; α

′

α(b − 1)− b′

)+ S(r)

� 2T

(r,α′

α(b − 1)− b′

)+ S(r)

� 2T

(r,α′

α

)+ 2T (r, b − 1)+ 2T (r, b′)+ S(r)

� 2T

(r,α′

α

)+ 6T (r, b)+ S(r) = S(r).(6)


LetN0(r, 0; g′ | � 2) denote the counting function of those multiple zeros of g′ (countedwith multiplicities) which are not the zeros of g(g − 1). Then from (4) and (6) we get

N0(r, 0; g′ | � 2)�N(r, 0; f − b | � 2)+ S(r) = S(r).(7)

Now from (7) we obtain by Lemma 3

N(r, 0; g′ | � 2)�N0(r, 0; g′ | � 2)+N(r, 0; g | � 2)

+ N(r, 1; g | � 2) = S(r).(8)

Differentiating g = hf we get g′ = h′f + hf ′ = h[f ′ + h′hf ] and so

h′

hf = 1

hg′ − f ′ = f

gg′ − f ′.(9)

Let z0 be a simple zero of g′ which is not a zero of f and g. Since E1)(0; g′) ⊂E∞)(0; f ′), it follows that z0 is a zero of f ′ and so from (9) we see that z0 is a zero of h′/h.Therefore if we denote byN1)(r, 0; g′ | g �= 0) the counting function of those simple zerosof g′ which are not the zeros of g and f , we get

N1)(r, 0; g′ | g �= 0)�N(r, 0; h

′

h

)� T

(r,h′

h

)+O(1) = S(r).(10)

Again if we denote by N1)(r, 0; g′ | g = 0) the counting function of those simple zerosof g′ which are zeros of g and f then by Lemma 3 we get

N1)(r, 0; g′ | g = 0)�N(r, 0; g | � 2) = S(r).(11)

Now by (8), (10) and (11) we obtain

N(r, 0; g′) = N(r, 0; g′ | � 2)+N1)(r, 0; g′ | g �= 0)

+ N1)(r, 0; g′ | g = 0) = S(r).(12)

Similarly we can prove

N(r, 0; f ′) = S(r).(13)

Let

ψ1 =(f ′′

f ′ − 2f ′

f − 1

)−

(g′′

g′ − 2g′

g − 1

).

Since f , g share (1,m), if z0 is a simple 1-point of f then z0 is also a simple 1-pointof g. Hence in some neighbourhood of z0 we can write f (z) = 1 + (z − z0)A andg(z) = 1 + (z − z0)B, where A = A(z) and B = B(z) are analytic at z0 and A(z0) �= 0,B(z0) �= 0.


Therefore in some neighbourhood of z0 we get

ψ1(z) = 2A′ + (z− z0)A′′

A+ (z− z0)A′ − 2B ′ + (z− z0)B′′

B + (z− z0)B ′ − 2A′

A+ 2B ′

B

= 2AA′B2 − 2AA′B2 − 2A2BB ′ + 2A2BB ′ + (z− z0)C

AB{A+ (z− z0)A′}{B + (z− z0)B ′}= (z− z0)C

AB{A+ (z− z0)A′}{B + (z− z0)B ′} ,

where C = C(z) is analytic at z0. Hence a simple 1-point of f and g is a zero of ψ1.So in view of (12), (13) and Lemma 3 we get

N(r, 1; f | � 1)

�N(r, 0;ψ1)� T (r, ψ1)+O(1)

�N(r, 0; f ′)+N(r, 0; g′)+ N(r, 1; f | � 2)+N(r, 1; g | � 2)

+ N(r,∞; f | � 2)+N(r,∞; g | � 2)+ S(r)

= S(r)

and so

N(r, 1; g) = N(r, 1; f ) = N(r, 1; f | � 1)+N(r, 1; f | � 2) = S(r).(14)

Again let

ψ2 =(f ′′

f ′ − 2f ′

f

)−

(g′′

g′ − 2g′

g

).

Since f , g share (0, k), it follows that a simple zero of f and g is a zero of ψ2. So in viewof (12), (13) and Lemma 3 we get

N(r, 0; f | � 1)

�N(r, 0;ψ2)� T (r, ψ2)+O(1)

�N(r, 0; f ′)+N(r, 0; g′)+ N(r, 0; f | � 2)+N(r, 0; g | � 2)

+ N(r,∞; f | � 2)+N(r,∞; g | � 2)+ S(r)

= S(r)

and so

N(r, 0; g) = N(r, 0; f ) = N(r, 0; f | � 1)+N(r, 0; f | � 2) = S(r).(15)

Finally let

ψ3 = f ′′

f ′ − g′′

g′ .


Since f , g share (∞, 1), it follows that a simple pole of f and g is a zero of ψ3. So by(12), (13) and Lemma 3 we get

N(r,∞; f | � 1) � N(r, 0;ψ3)� T (r, ψ3)+O(1)

� N(r, 0; f ′)+N(r, 0; g′)+N(r,∞; f | � 2)

+ N(r,∞; g | � 2)+ S(r)

= S(r)

and so

N(r,∞; g) = N(r,∞; f ) = N(r,∞; f | � 1)

+ N(r,∞; f | � 2) = S(r).(16)

Now by (14), (15), (16) and the second fundamental theorem we get

T (r, f )�N(r, 0; f )+N(r, 1; f )+N(r,∞; f )+ S(r, f ) = S(r, f ),

a contradiction. Therefore f is a bilinear transformation of g. Now the result follows fromLemma 6. This proves the theorem.

A c k n o w l e d g e m e n t. The authors are thankful to the referee for valuable commentsand suggestions.

References

[1] A. H. H. Al-khaladi, Unicity theorems for meromorphic functions. Kodai Math. J. 23, 27–34 (2000).[2] W. K. Hayman, Meromorphic Functions. Oxford 1964.[3] I. Lahiri, Weighted sharing and uniqueness of meromorphic functions. Nagoya Math. J. 161, 193–206

(2001).[4] I. Lahiri, Weighted value sharing and uniqueness of meromorphic functions. Complex Variables Theory

Appl. 46(3), 241–253 (2001).[5] I. Lahiri, Weighted sharing of three values and uniqueness of meromorphic functions. Kodai Math. J. 24,

421–435 (2001).[6] I. Lahiri, On a result of Ozawa concerning uniqueness of meromorphic functions II. J. Math. Anal. Appl.

283, no. 1, 66–76 (2003).[7] I. Lahiri, Uniqueness of meromorphic functions and sharing of three values with some weight. New Zealand

J. Math. 32, no. 2, 161–171 (2003).[8] K. Tohge, Meromorphic functions covering certain finite sets at the same points. Kodai Math. J. 11,

249–279 (1988).

Received: 8 July 2004

I. Lahiri A. SarkarDepartment of Mathematics Fulia StationparaUniversity of Kalyani P.O. Fulia ColonyWest Bengal 741235 District-NadiaIndia West Bengal [email protected] [email protected]

On a uniqueness theorem of Tohge

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