TWO MEROMORPHIC FUNCTIONS SHARING SOME PAIRS OF SMALL FUNCTIONS REGARDLESS OF MULTIPLICITIES

February 28, 2014 16:36 WSPC/S0129-167X 133-IJM 1450014

International Journal of MathematicsVol. 25, No. 2 (2014) 1450014 (16 pages)c© World Scientific Publishing CompanyDOI: 10.1142/S0129167X14500141

TWO MEROMORPHIC FUNCTIONS SHARING SOMEPAIRS OF SMALL FUNCTIONS REGARDLESS

OF MULTIPLICITIES

SI DUC QUANG

Department of MathematicsHanoi National University of Education

136 Xuan Thuy Street, Cau GiayHanoi, Vietnam

[email protected]

LE NGOC QUYNH

Faculty of Education, An Giang University

18 Ung Van Khiem, Dong XuyenLong Xuyen, An Giang, Vietnam

[email protected]

Received 15 December 2013Accepted 14 January 2014

Published 11 February 2014

In this paper, we prove that two meromorphic functions f and g must be linked by aquasi-Mobius transformation if they share a pair of small functions ignoring multiplicitiesand share other four pairs of small functions with multiplicities truncated by 2. We alsoshow that two meromorphic functions which share q (q ≥ 6) pairs of small functionsignoring multiplicities are linked by a quasi-Mobius transformation. Moreover, in ourresults, the zeros with multiplicities more than a certain number are not needed tobe counted in the condition sharing pairs of small functions of meromorphic functions.These results are generalization and improvements of some recent results.

Keywords: Meromorphic function; small function; Mobius transformation.

Mathematics Subject Classification 2010: Primary: 32H20, 32A22; Secondary: 30D35

1. Introduction

Let f be a nonzero holomorphic function on C. For each z0 ∈ C, expanding f asf(z) =

∑∞i=0 bi(z − z0)i around z0, then we define ν0

f (z0) := min{i : bi �= 0}. Let k

be a positive integer or +∞. We set

ν0f,≤k(z) :=

{ν0

f (z) if ν0f (z) ≤ k,

0 if ν0f (z) > k.

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http://dx.doi.org/10.1142/S0129167X14500141


S. D. Quang & L. N. Quynh

Let ϕ be a nonconstant meromorphic function on C with reduced representationϕ = (ϕ1 : ϕ2), where ϕ1, ϕ2 are holomorphic functions on C having no common zerosand ϕ = ϕ1

ϕ2. We define ν0

ϕ := ν0ϕ1

, ν0ϕ,≤k = ν0

ϕ1,≤k, ν∞ϕ := ν0

ϕ2and ν∞

ϕ,≤k = ν0ϕ2,≤k.

The proximity function of ϕ is defined by:

m(r, ϕ) :=∫ 2π

0

log+|ϕ(reiθ)|dθ (r > 1),

here log+x = {0, logx} for x ∈ (0,∞). Then the Nevanlinna characteristic functionof ϕ is defined by

T (r, ϕ) := m(r, ϕ) + N(r, ν∞ϕ ),

where by N(r, ν) we denote the counting function of the divisor ν.Let f, a be two meromorphic functions on C. The function a is said to be small

(with respect to f) if ‖ T (r, a) = o(T (r, f)). Here, the notion ‖ P means that theassertion P holds for all r ∈ [1,∞) outside a finite Borel measure set. We denoteby Rf the field of all small (with respect to f) functions on C.

For two meromorphic functions f and g on C, the map f is said to be a quasi-Mobius transformation of g if there exist small (with respect to g) functions αi (1 ≤i ≤ 4) such that f = α1g+α2

α3g+α4. If all functions αi (1 ≤ i ≤ 4) are constants then we

say that the map f is a Mobius transformation of g.In 1997, Czubiak and Gundersen [1] proved the following.

Theorem A. Let f and g be two nonconstant meromorphic functions that sharesix pairs of values (ai, bi), 1 ≤ i ≤ 6, IM, where ai �= aj , bi �= bj whenever i �= j, i.e.

min{ν0f−ai

, 1} = min{ν0g−bi

, 1} (1 ≤ i ≤ 6).

Then f is a Mobius transformation of g.

The example given by Li and Yang [3] show that two meromorphic functionssharing five pairs of values may not be linked by a Mobius transformation. There-fore, the number of pairs of small functions in Theorem A cannot be replaced by asmaller one.

In 2012, Quang [7] improved and generalized the above result to the case of twomeromorphic functions sharing pairs of small functions as follows.

Theorem B. Let f and g be two nonconstant meromorphic functions on C. Let(ai, bi)7i=1 be seven pairs of small (with respect to f) functions on C, where ai �= aj ,

bi �= bj , whenever i �= j. Let k be a positive integer or may be +∞. Assume that

min{ν0f−ai,≤k, 1} = min{ν0

g−bi,≤k, 1} (1 ≤ i ≤ 7).

If k ≥ 17, then f is a quasi-Mobius transformation of g.

In this paper, we will improve and generalize the above results to thefollowing.

Theorem 1.1. Let f and g be two nonconstant meromorphic functions on C. Let{(ai, bi)}q

i=1 (q ≥ 6) be q pairs of small (with respect to f) functions on C, where

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Two Meromorphic Functions Sharing Some Pairs of Small Functions

ai �= aj , bi �= bj whenever i �= j. Let ki (i = 1, . . . , q) be q positive integers or +∞such that

q∑i=1

1ki

<2q − 10

5+

q − 2k0

,

where k0 = max1≤i≤q ki. Assume that

min{ν0f−ai,≤ki

, 1} = min{ν0g−bi,≤ki

, 1} (1 ≤ i ≤ q).

Then f is a quasi-Mobius transformation of g.

When k1 = k2 = · · · = kq = k, Theorem 1.1 immediately implies the followingcorollary.

Corollary 1.1. Let f and g be two nonconstant meromorphic functions on C. Let(ai, bi)

qi=1 (q ≥ 6) be q pairs of small (with may be respect to f) functions on C,

where ai �= aj , bi �= bj, whenever i �= j. Let k be a positive integer or +∞. Assumethat

min{ν0f−ai,≤k, 1} = min{ν0

g−bi,≤k, 1} (1 ≤ i ≤ q).

If k > 5q−5 , then f is a quasi-Mobius transformation of g.

We distinguish here some cases where the condition k > 5q−5 in Corollary 1.1 is

satisfied:

(1) q = 6 and k ≥ 6,(2) q = 7 and k ≥ 3,(3) q = 8, 9, 10 and k ≥ 2,(4) q > 10 and k ≥ 1.

In the case of meromorphic functions sharing five pairs of small functions, wewill prove the following.

Theorem 1.2. Let f and g be two nonconstant meromorphic functions on C. Let{(ai, bi)}5

i=1 be five pairs of small (with respect to f) functions on C, where ai �= aj ,

bi �= bj whenever i �= j. Let k5 ≥ 2 be a positive integer or +∞. Let ki (1 ≤ ki ≤ 4)be positive integers or +∞, such that

5∑i=1

1ki

<3

361+

843361k0

+240

361k5,

where k0 = max1≤i≤5 ki. Assume that

min{ν0f−a5,≤k5

, 1} = min{ν0g−b5,≤k5

, 1} and

min{ν0f−ai,≤ki


, 2} (1 ≤ i ≤ 4).

Then f is a quasi-Mobius transformation of g.

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When k1 = k2 = · · · = k5 = k, Theorem 1.2 immediately implies the followingcorollary.

Corollary 1.2. Let f and g be two nonconstant meromorphic functions on C. Let{(ai, bi)}5

i=1 be five pairs of small (with respect to f) functions on C, where ai �= aj ,

bi �= bj whenever i �= j. Let k be a positive integer or may be +∞. Assume that

min{ν0f−a5,≤k, 1} = min{ν0

g−b5,≤k, 1}and min{ν0

f−ai,≤k, 2} = min{ν0g−bi,≤k, 2} (1 ≤ i ≤ 4).

If k ≥ 241, then f is a quasi-Mobius transformation of g.

2. Basic Notions and Auxiliary Results from Nevanlinna Theory

For a divisor ν on C and for two positive integers k, M (may be M = ∞), we definethe counting function of ν by

ν(M)(z) = min{M, ν(z)},n(t) =

∑|z|≤t

ν(z), n(M)(t) =∑|z|≤t

ν(M)(z).

Define

N(r, ν) =∫ r

1

n(t)t

dt (1 < r < ∞).

Similarly, we define N(r, ν(M)) and denote it by N (M)(r, ν).For brevity we will omit the character (M) if M = ∞.For two divisors ν and µ on C, we define

Sν �=µ(z) =

{1 if ν(z) �= µ(z)

0 if ν(z) = µ(z)and N(r, ν �= µ) = N(r, Sν �=µ).

Theorem 2.1 ([8, Corollary 1]). Let f be a nonconstant meromorphic functionon C. Let a1, . . . , aq (q ≥ 3) be q distinct small (with respect to f) meromorphicfunctions on C. Then, for each ε > 0, the following holds

‖ (q − 2 − ε)T (r, f) ≤q∑

i=1

N (1)(r, ν0f−ai

) + o(T (r, f)).

Let {Hi}qi=1 (q ≥ N+2) be a set of q hyperplanes in PN (C). We say that {Hi}q

i=1

are in general position if for any 1 ≤ i1 < · · · < iN+1 ≤ q we have⋂N+1

j=1 Hij = ∅.Theorem 2.2 ([5, Theorem 3.1]). Let f : C → PN (C) be a linearly nondegen-erate holomorphic mapping. Let {Hi}q

i=1 (q ≥ N + 2) be a set of q hyperplanes inPN (C) in general position. Then

‖ (q − N − 1)Tf (r) ≤q∑

i=1

N (N)(r, f∗Hi) + o(T (r, f)),

where f∗Hi denotes the pull back divisor of Hi by f .

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Lemma 2.1 ([4, Lemma 5]). Suppose that f and g are nonconstant meromorphicfunctions, F = F (f, g) is a polynomial in f and g with coefficients being smallfunctions with respect to f and g. The degree of F about f is p, and the degreeabout g is q. Then we have

T (r, F ) ≤ pT (r, f) + qT (r, g) + o(T (r, f)).

Lemma 2.2. Let f be a nonconstant meromorphic function on C. Let a be a smallfunction with respect to f, and k is a positive integer or +∞, then

N (1)(r, ν0f−a,≤k) ≥ k + 1

kN (1)(r, ν0

f−a) − 1k

T (r, f).

Proof. We have

N (1)(r, ν0f−a,≤k) = N (1)(r, ν0

f−a) − N (1)(r, ν0f−a,>k)

≥ N (1)(r, ν0f−a) − 1

k + 1N(r, ν0

f−a,>k)

= N (1)(r, ν0f−a) − 1

k + 1(N(r, ν0

f−a) − N(r, ν0f−a,≤k))

≥ N (1)(r, ν0f−a) − 1

k + 1T (r, f) +

1k + 1

N (1)(r, ν0f−a,≤k).

This implies that

N (1)(r, ν0f−a,≤k) ≥ k + 1

kN (1)(r, ν0

f−a) − 1k

T (r, f).

The lemma is proved.

Lemma 2.3 ([7, Lemma 3.1]). Let f and g be two meromorphic functions onC. Let {ai}3

i=1 and {bi}3i=1 be two sets of small (with respect to f) meromorphic

functions on C with ai �= aj and bi �= bj for all 1 ≤ i < j ≤ 3. Assume that‖ T (r, f) = O(T (r, g)) and ‖ T (r, g) = O(T (r, f)) and f is not quasi-Mobius trans-formation of g. Then

T (r, f) + T (r, g) ≥3∑

i=1

N(r, min{ν0f−ai

, ν0g−bi

}) + o(T (r, f)).

3. Proof of Main Theorems

In order to prove the main theorems, we need the following lemmas.

Lemma 3.1. Let f and g be nonconstant meromorphic functions on C. Let {ai}qi=1

and {bi}qi=1 (q ≥ 5) be two sets of small (with respect to f) functions on C, with

ai �= aj , bi �= bj for all 1 ≤ i < j ≤ q, and ki (1 ≤ i ≤ q) be positive integers or +∞

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such that∑q

i=11

ki+1 ≤ q − 2. Assume that

min{ν0f−ai,≤ki


, 1} (1 ≤ i ≤ q).

Then ‖ T (r, f) = O(T (r, g)) and ‖ T (r, g) = O(T (r, f)). In particular, Rf = Rg.

Proof. By Theorem 2.1 and Lemma 2.2, for every ε > 0, we have

‖(q − 2 − ε)T (r, f) ≤q∑

i=1

N (1)(r, ν0f−ai

) + O

(max1≤i≤q

{T (r, ai)})

≤q∑

i=1

(ki

ki + 1N (1)(r, ν0

f−ai,≤ki) +

1ki + 1

T (r, f))

+ o(T (r, f)).

Thus∥∥∥∥∥(

q − 2 − ε −q∑

i=1

1ki + 1

)T (r, f) ≤

q∑i=1

ki

ki + 1N (1)(r, ν0

f−ai,≤ki) + o(T (r, f))

=q∑

i=1

ki

ki + 1N (1)(r, ν0

g−bi,≤ki) + o(T (r, f))

≤q∑

i=1

ki

ki + 1(T (r, g) + T (r, bi)) + o(T (r, f))

=q∑

i=1

ki

ki + 1T (r, g) + o(T (r, f)),

for every ε > 0. Choosing ε < q − 2 −∑qi=1

1ki+1 , we have ‖ T (r, f) = O(T (r, g)).

Similarly, we have ‖ T (r, g) = O(T (r, f)). The lemma is proved.

Lemma 3.2. Let f and g be two meromorphic functions on C. Let {ai}5i=1 and

{bi}5i=1 be two sets of small (with respect to f) meromorphic functions on C with

ai �= aj and bi �= bj for all 1 ≤ i < j ≤ 5. Assume that ‖ T (r, f) = O(T (r, g)) and‖ T (r, g) = O(T (r, f)) and f is not quasi-Mobius transformation of g. Then one ofthe following two assertions holds:

(i) T (r, f) = 2T (r, g) + o(T (r, f)) or T (r, g) = 2T (r, f) + o(T (r, f)),(ii) 3

2 (T (r, f) + T (r, g)) ≥∑5i=1 N(r, min{ν0

f−ai,≤ki, ν0

g−bi,≤ki}) + o(T (r, f)).

Proof. Suppose that the first assertion does not hold. We are now going to provethe second assertion.

It is easy to see that there exist six small functions ci (i = 1, . . . , 6) (at leastone of them is not identically zero) such that the following function

F := F (f, g) = c1fg2 + c2fg + c3g2 + c4g + c5f + c6

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satisfies F (ai, bi) ≡ 0 for i = 1, . . . , 5. By Lemma 2.1, we have

T (r, F ) ≤ T (r, f) + 2T (r, g) + o(T (r, f)).

If F ≡ 0, then (c1g2 + c2g + c5)f ≡ −(c3g

2 + c4g + c6). Note that at least one of ci

is not zero, therefore c1g2 + c2g + c5 �≡ 0. Hence,

f = −c3g2 + c4g + c6

c1g2 + c2g + c5.

Since f is not a quasi-Mobius transformation of g, the right-hand side of the aboveequation is irreducible. Therefore, T (r, f) = 2T (r, g) + o(T (r, f)). This contradictsthe supposition. Therefore F �≡ 0.

We denote by S the reduced divisor defined by

Supp (S) =⋃

1≤i<j≤5

(Supp (ν0(f−ai)

) ∩ Supp (ν0(f−aj)

)).

Then we have

N(r, S) ≤∑

1≤i<j≤5

N (1)(r, ν0ai−aj

) ≤∑

1≤i<j≤5

T (r, ai − aj) = o(T (r, f)).

Fix an index i0 (1 ≤ i0 ≤ 5). For a zero z of the function (f − ai0) withmultiplicity at most ki0 (then z is also zero of g− bi0 with multiplicity at most ki0)such that z �∈ Supp (S), we have

f (k)(z) = a(k)i0

(z) and g(k)(z) = b(k)i0

(z)

for all 0 ≤ k ≤ min{ν0f−ai0 ,≤ki0

(z), ν0g−bi0 ,≤ki0

(z)}−1, where by h(k) we denote thederivative of the order k of a meromorphic function h. Therefore, we have

F (f, g)(k)(z) = F (ai0 , bi0)(k)(z) = 0,

for all 0 ≤ k ≤ min{ν0f−ai0 ,≤ki0

(z), ν0g−bi0 ,≤ki0

(z)}− 1. Hence, z is a zero of F (f, g)

with multiplicity at least min{ν0f−ai0 ,≤ki0

(z), ν0g−bi0 ,≤ki0

(z)}, i.e.

ν0F (f,g)(z) ≥ min{ν0

f−ai0 ,≤ki0(z), ν0

g−bi0 ,≤ki0(z)}.

This implies that

ν0F (f,g)(z) ≥

5∑i=1

min{ν0f−ai,≤ki

(z), ν0g−bi,≤ki

(z)}.

Then, for every z ∈ C, we have

ν0F (f,g)(z) ≥

5∑i=1

min{ν0f−ai,≤ki

(z), ν0g−bi,≤ki

(z)} − 4k0S(z).

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By integrating both sides of the above inequalities and by Lemma 2.1, we obtain

T (r, f) + 2T (r, g) ≥ T (r, F ) ≥ N(r, ν0F (f, g))

≥5∑

i=1

N(r, min{ν0f−ai,≤ki

, ν0g−bi,≤ki

}) + o(T (r, f)).

Similarly, we have

T (r, g) + 2T (r, f) ≥5∑

i=1


, ν0g−bi,≤ki

}) + o(T (r, f)).

By summing-up both sides of the above two inequalities, we obtain

32(T (r, f) + T (r, g)) ≥

5∑i=1


, ν0g−bi,≤ki

}) + o(T (r, f)).

The lemma is proved.

Proof of Theorem 1.1. Suppose that f is not a quasi-Mobius transformation ofg. It is easy to see that

q∑i=1

1ki + 1

≤ q

2< q − 2.

Then, by Lemma 3.1, we have

‖ T (r, f) = O(T (r, g)) and ‖ T (r, g) = O(T (r, f)).

We first suppose that

T (r, f) = 2T (r, g) + o(T (r, f)).

Then by Lemma 2.3, for each subset {i1, i2, i3} ⊂ {1, . . . , q}, we have

32T (r, f) ≥

3∑j=1

N(r, min{ν0f−aij

,≤kij, ν0

g−bij,≤kij

}) + o(T (r, f))

≥3∑

j=1

N (1)(r, ν0f−aij

,≤kij) + o(T (r, f)).

By summing-up both sides of the above inequality over all subsets {i1, i2, i3} ⊂{1, . . . , q} and applying Lemma 2.2, we have∥∥∥∥ 3

2T (r, f) ≥ 3

q

q∑i=1

N (1)(r, ν0f−ai,≤ki

) + o(T (r, f))

≥ 3q

q∑i=1

(ki + 1

kiN (1)(r, ν0

f−ai) − 1

kiT (r, f)

)+ o(T (r, f))

≥ 3q

((q − 2 − ε)

(1 +

1k0

)−

q∑i=1

1ki

)T (r, f) + o(T (r, f)),

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for every positive number ε. Letting r → +∞, we get

32≥ 3

q

((q − 2 − ε)

(1 +

1k0

)−

q∑i=1

1ki

).

Letting ε → 0, we getq∑

i=1

1ki

≥ q − 42

+q − 2k0

.

From the assumption of the theorem, we have

2q − 105

+q − 2k0

>

q∑i=1

1ki

≥ q − 42

+q − 2k0

.

Thus 2q−105 > q−4

2 ⇔ 4q − 20 > 5q − 20. This is a contradiction.Therefore, we have T (r, f) �= 2T (r, g) + o(T (r, f)). Similarly, we also have

T (r, g) �= 2T (r, f) + o(T (r, f)).Consider five pairs of meromorphic functions {(aij , bij )}5

j=1 among q pairs{(ai, bi)}q

i=1. Then at least one of two assertions of Lemma 3.2 holds. By the proofabove, we see that the assertion (i) does not hold. Therefore the assertion (ii) holds.Then, we have

32(T (r, f) + T (r, g)) ≥

5∑j=1

N(r, min{ν0f−aij

,≤kij, ν0

g−bij,≤kij

}) + o(T (r, f)).

Summing-up both sides of this inequality over all subsets {i1, . . . , i5} ⊂ {1, . . . , q},we get∥∥∥∥ 3

2(T (r, f) + T (r, g)) ≥ 5

q

q∑i=1


, ν0g−bi,≤ki

}) + o(T (r, f))

≥ 52q

q∑i=1

(N (1)(r, ν0f−ai,≤ki

) + N (1)(r, ν0g−bi,≤ki

)) + o(T (r, f))

≥ 52q

q∑i=1

(ki + 1

kiN (1)(r, ν0

f−ai) − 1

kiT (r, f)

+ki + 1

kiN (1)(r, ν0

g−bi) − 1

kiT (r, g)

)+ o(T (r, f))

≥ 52q

((q − 2 − ε)

(1 +

1k0

)−

q∑i=1

1ki

)(T (r, f)

+ T (r, g)) + o(T (r, f)),

for every positive number ε. Letting r → +∞, we get

32≥ 5

2q

((q − 2 − ε)

(1 +

1k0

)−

q∑i=1

1ki

).

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Letting ε → 0, we getq∑

i=1

1ki

≥ 2q − 105

+q − 2k0

.

This contradicts the assumption of the theorem.Hence f must be a quasi-Mobius transformation of g. The theorem is proved.

Proof of Theorem 1.2. Suppose that f is not a quasi-Mobius transformation ofg. It is easy to see that

q∑i=1

1ki + 1

≤ q

2< q − 2.

Then, by Lemma 3.1, we have

‖ T (r, f) = O(T (r, g)) and ‖ T (r, g) = O(T (r, f)).

Similarly as in the proof of Theorem 1.1, we first suppose that

T (r, f) = 2T (r, g) + o(T (r, f)).

Then by Lemma 2.3, for each subset {i1, i2, i3} ⊂ {1, . . . , 5}, we have

32T (r, f) ≥

3∑j=1

N(r, min{ν0f−aij

,≤kij, ν0

g−bij,≤kij

}) + o(T (r, f))

≥3∑

j=1

N (1)(r, ν0f−aij

,≤kij) + o(T (r, f)).

By summing-up both sides of the above inequality over all subsets {i1, i2, i3} ⊂{1, . . . , 5} and applying Lemma 2.2, we have∥∥∥∥ 3

2T (r, f) ≥ 3

5

5∑i=1


) + o(T (r, f))

≥ 35

5∑i=1

(ki + 1

kiN (1)(r, ν0

f−ai) − 1

kiT (r, f)

)+ o(T (r, f))

≥ 35

((3 − ε)

(1 +

1k0

)−

5∑i=1

1ki

)T (r, f) + o(T (r, f)),

for every positive number ε. Letting r → +∞ and then letting ε → 0, we get5∑

i=1

1ki

≥ 12

+3k0

.

From the assumption of the theorem, we have

3361

+843

361k0+

240361k5

>

5∑i=1

1ki

≥ 12

+3k0

.

Thus 3361 + 240

361k5> 1

2 , hence k5 < 9671 < 2. This is a contradiction.

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Therefore, we have T (r, f) �= 2T (r, g) + o(T (r, f)). Similarly, we also haveT (r, g) �= 2T (r, f) + o(T (r, f)).

Hence, the assertion (i) does not hold. This implies that the assertion (ii) holds.Then we have

‖ 3(T (r, f) + T (r, g)) ≥ 25∑

i=1


, ν0g−bi,≤ki

}) + o(T (r, f))

≥ N (1)(r, ν0f−a5,≤k5

) +4∑

i=1


)

+ N (1)(r, ν0g−b5,≤k5

) +4∑

i=1

N (2)(r, ν0g−bi,≤ki

) + o(T (r, f)).

(3.1)

For each 1 ≤ i ≤ 4, we denote by Si the reduced divisor with the support definedby

Supp (Si) = {z; ν0f−ai

(z) > ki} ∪ {z; ν0g−bi

(z) > ki} ∪ {z; 2 ≤ ν0f−ai

(z) ≤ ki}.We find it easy to see that

N(r, Si) ≤ N (1)(r, ν0f−ai,>1) + N (1)(r, ν0

g−bi,>ki)

= N (2)(r, ν0f−ai,≤ki

) − N (1)(r, ν0f−ai,≤ki

)

+ N (1)(r, ν0f−ai,>ki

) + N (1)(r, ν0g−bi,>ki

),

and N(r, Si) ≤ N (2)(r, ν0g−bi,≤ki

) − N (1)(r, ν0g−bi,≤ki

)

+ N (1)(r, ν0g−bi,>ki

) + N (1)(r, ν0f−ai,>ki

).

Then, the inequality (3.1) implies that

‖ 3(T (r, f) + T (r, g)) ≥5∑

i=1


) − 24∑

i=1

N (1)(r, ν0f−ai,>ki

)

+ 24∑

i=1

N(r, Si) +5∑

i=1


)

− 24∑

i=1

N (1)(r, ν0g−bi,>ki

) + o(T (r, f)). (3.2)

On the other hand, we have


) ≤ 1ki + 1

N(r, ν0f−ai,>ki

) =1

ki + 1(N(r, ν0

f−ai) − N(r, ν0

f−ai,≤ki))

≤ 1ki + 1

(T (r, f) − N (1)(r, ν0f−ai,≤ki

))

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≤ 1ki + 1

(T (r, f) − ki + 1

kiN (1)(r, ν0

f−ai) +

1ki

T (r, f))

=1ki

T (r, f)− 1ki

N (1)(r, ν0f−ai

).

Then, we have∥∥∥∥∥5∑

i=1


) − 24∑

i=1


)

≥5∑

i=1

(ki + 1

kiN (1)(r, ν0

f−ai) − 1

kiT (r, f)

)

− 24∑

i=1

(1ki

T (r, f)− 1ki

N (1)(r, ν0f−ai

))

≥(

1 +1k0

)(3 − ε)T (r, f)−

5∑i=1

1ki

T (r, f)

− 24∑

i=1

1ki

T (r, f) +2k0

(2 − ε)T (r, f) + o(T (r, f))

=

(3 − ε +

7 − 3ε

k0−

5∑i=1

3ki

+2k5

)T (r, f) + o(T (r, f)). (3.3)

for every ε > 0. Similarly, we also have∥∥∥∥∥5∑

i=1


) − 24∑

i=1

N (1)(r, ν0g−bi,>ki

)

≥(

3 − ε +7 − 3ε

k0−

5∑i=1

3ki

+2k5

)T (r, g) + o(T (r, f)), (3.4)

for every ε > 0. Therefore, combining (3.2)–(3.4) we obtain

‖ 3(T (r, f) + T (r, g)) ≥(

3 − ε +7 − 3ε

k0−

5∑i=1

3ki

+2k5

)(T (r, f) + T (r, g))

+ 24∑

i=1

N(r, Si) + o(T (r, f)).

Thus

24∑

i=1

N(r, Si) ≤(

5∑i=1

3ki

− 7k0

− 2k5

+ ε +3ε

k0

)(T (r, f) + T (r, g)) + o(T (r, f)),

(3.5)

for every ε > 0.

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We set:

• c1 = a3−a2a2−a1

, c2 = a3−a1a2−a1

, c′1 = b3−b2b2−b1

, c′2 = b3−b1b2−b1

,

• F1 = c1(f − a1), F2 = c2(f − a2), G1 = c′1(g − b1), G2 = c′2(g − b2),• h1 = F1

G1, h2 = F2

G2, h3 = F1−F2

G1−G2= f−a3

g−b3,

• α = c1(a4−a1)c2(a4−a2) , β = c′1(b4−b1)

c′2(b4−b2),

• h4 = F1−αF2G1−βG2

= c1(a1−a2)(f−a4)/(a4−a2)c′1(b1−b2)(g−b4)/(b4−b2)

= (a3−a2)(b4−b2)(b3−b2)(a4−a2) · f−a4

g−b4.

It is easy to see that c1 �= c2, c′1 �= c′2, α �= 1, β �= 1 and all ci, c′i (1 ≤ i ≤ 2) are

small with respect to f and

‖ N (1)(r, ν0hi

) + N (1)(r, ν∞hi

) = N (1)(r, ν0f−ai

�= ν0g−bi

) + o(T (r, f))

≤ N(r, Si) + o(T (r, f)), (3.6)

for each 1 ≤ i ≤ 4.From the definition of functions Fi, Gi (1 ≤ i ≤ 2), we have the following

equations system:

F1 − h1G1 = 0,

F2 − h2G2 = 0,

F1 − F2 − h3G1 + h3G2 = 0,

F1 − αF2 − h4G1 + h4βG2 = 0.

This implies that

det

1 0 −h1 00 1 0 −h2

1 −1 −h3 h3

1 −α −h4 h4β

= 0.

Then

(1 − α)h1h2 − h1h3 + βh1h4 + αh2h3 − h2h4 + (1 − β)h3h4 = 0. (3.7)

Denote by I the set of all subsets I = {i, j} of the set {1, 2, 3, 4}. For I ∈ I, wedefine the function hI as follows:

h{1,2} = (1 − α)h1h2, h{1,3} = −h1h3, h{1,4} = βh1h4,

h{2,3} = αh2h3, h{2,4} = −h2h4, h{3,4} = (1 − β)h3h4.

Then we have ∑I∈I

hI = 0.

Take I0 ∈ I. Then

hI0 = −∑I �=I0

hI .

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Denote by t the minimum number satisfying the following: There exist t elementsI1, . . . , It ∈ I and t nonzero constants bv ∈ C (1 ≤ v ≤ t) such that hI0 =∑t

v=1 bvhIv .

By the minimality of t, the family {hI1 , . . . , hIt} is linearly independent over C.Case 1. t = 1. Then hI0

hI1∈ C\{0}.

Case 2. t ≥ 2.

Consider the holomorphic mapping h : C → Pt−1(C) with a reduced representa-tion h = (dhI1 : . . . : dhIt), where d is a meromorphic function on C.

Take z �∈ (Zero(α)∪Zero(1−α)∪Pole(α)∪Zero(β)∪Zero(1− β)∪Pole(β)). Ifz is a zero of dhIv , then z must be either zero or pole of some hi. This yields that∥∥∥∥∥ N

(1)dhIv

(r) ≤4∑

i=1

(N (1)(r, ν0hi

) + N (1)(r, ν∞hi

)) ≤4∑

i=1

N(r, Si) + o(T (r, f)).

By the Second Main theorem, we have

‖ Th(r) ≤t∑

v=1

N(t−1)dhIv

(r) + N(t−1)dhI0

(r) + o(Tf (r))

≤ (t − 1)t∑

v=1

N(1)dhIv

(r) + (t − 1)N (1)dhI0

(r) + o(Tf (r))

≤ (t − 1)(t + 1)4∑

i=1

N(r, Si) + o(T (r, f)) ≤ 244∑

i=1

N(r, Si) + o(T (r, f)).

(3.8)

Here we note that since t ≤ 5, therefore (t − 1)(t + 1) ≤ 24.We define following rational functions:

H1(X, Y ) =c1(X − a1)c′1(Y − b1)

, H2(X, Y ) =c2(X − a2)c′2(Y − b2)

,

H3(X, Y ) =X − a3

Y − b3,

H4(X, Y ) =(a3 − a2)(b4 − b2)(b3 − b2)(a4 − a2)

· X − a4

Y − b4.

For each I ⊂ {1, . . . , 5}, put Ic = {1, . . . , 5}\I. For 0 ≤ u, v ≤ t, u �= v, andi ∈ ((Iv ∪ Iu)\(Iu ∩ Iv))c, we see that∥∥∥∥ T

(r,

hIu

hIv

)= T

(r,

∏j∈Iu

hj∏j∈Iv

hj

)+ o(T (r, f))

≥ N(r, ν0Q

j∈Iu\Ivhj

Qj∈Iv\Iu

hj−

Qj∈Iu\Iv

Hj (ai,bi)Q

j∈Iv\IuHj (ai,bi)

)+ o(T (r, f))

≥ N (1)(r, ν0f−ai,≤ki

) + o(T (r, f)).

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Similarly, we have ‖ T (r, hIu

hIv) ≥ N (1)(r, ν0

g−bi,≤ki) + o(T (r, f)). Therefore∥∥∥∥ T

(r,

hIu

hIv

)≥ 1

2(N (1)(r, ν0

f−ai,≤ki) + N (1)(r, ν0

g−bi,≤ki)) + o(T (r, f)). (3.9)

Since (I0 ∪ I1\(I0 ∩ I1))c ∪ (I1 ∪ I2\(I1 ∩ I2))c ∪ (I2 ∪ I0\(I2 ∩ I0))c = {1, . . . , 5},for each i ∈ {1, . . . , 5} there exist u, v ∈ {0, 1, 2} such that i ∈ (Iu ∪ Iv\(Iu ∩ Iv))c,and then we have

T (r, h) ≥ T

(r,

hIu

hIv

)≥ 1

2(N (1)(r, ν0

f−ai,≤ki) + N (1)(r, ν0

g−bi,≤ki)) + o(T (r, f)).

Summing-up both sides of the above inequality over all i = 1, . . . , 5, we obtain

‖ 5T (r, h) ≥ 12

5∑i=1

(N (1)(r, ν0f−ai,≤ki

) + N (1)(r, ν0g−bi,≤ki

)) + o(T (r, f))

≥ 12

5∑i=1

(ki + 1

kiN (1)(r, ν0

f−ai) − 1

kiT (r, f)

)

+12

5∑i=1

(ki + 1

kiN (1)(r, ν0

g−bi) − 1

kiT (r, g)

)+ o(T (r, f))

≥ 12

((1 +

1k0

)(3 − ε) −

5∑i=1

1ki

)(T (r, f) + T (r, g)) + o(T (r, f)),

(3.10)

for every ε > 0.

From (3.8) and (3.10), it follows that∥∥∥∥∥ 5.244∑

i=1

N(r, Si) ≥ 12

((1 +

1k0

)(3 − ε) −

5∑i=1

1ki

)

× (T (r, f) + T (r, g)) + o(T (r, f)).

Combining (3.5) and the above inequality, we obtain

5.12

(5∑

i=1

3ki

− 7k0

− 2k5

+ ε +3ε

k0

)(T (r, f) + T (r, g))

≥ 12

((1 +

1k0

)(3 − ε) −

5∑i=1

1ki

)(T (r, f) + T (r, g)) + o(T (r, f)), (3.11)

for every ε > 0. Letting r → +∞ and then letting ε → 0, we get

5.12

(5∑

i=1

3ki

− 7k0

− 2k5

)≥ 1

2

(3 +

3k0

−5∑

i=1

1ki

).

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Thus5∑

i=1

1ki

≥ 3361

+843

361k0+

240361k5

.

This is a contradiction.Then from Case 1 and Case 2, it follows that for each I ∈ I, there is J ∈ I\{I}

such that hI

hJ∈ C\{0}. We consider the following two cases:

Case a. There exist I = {i, j}, J = {i, l}, j �= l, hI

hJ= constant. Then hj = ahl

with a is a nonzero meromorphic function in Rf . Therefore, f is a quasi-Mobiustransformation of g. This contradicts the supposition that f is not a quasi-Mobiustransformation of g

Case b. There exist nonzero constants b, c such that h{1,2} = bh{3,4} andh{1,3} = ch{2,4}, i.e.

(1 − α)h1h2 = b(1 − β)h3h4 and h1h3 = ch2h4.

Then (h1h4

)2 = bc(1−β)1−α ∈ Rf . This implies that h1

h4∈ Rf . Hence f is a quasi-Mobius

transformation of g. This is a contradiction.From the above two cases, we get the contradiction to the supposition. Hence f

is a quasi-Mobius transformation of g.

Acknowledgments

The research of the authors was supported in part by a NAFOSTED grant ofVietnam.

References

[1] T. Czubiak and G. Gundersen, Meromorphic functions that share pairs of values,Complex Var. Elliptic Eqs. 34 (1997) 35–46.

[2] G. Gundersen, Meromorphic functions that share three or four values, J. London Math.Soc. 20 (1979) 457–466.

[3] P. Li and C. C. Yang, On two meromorphic functions that share pairs of small func-tions, Complex Var. Elliptic Eqs. 32 (1997) 177–190.

[4] P. Li and C. C. Yang, Meromorphic functions that share some pair of small functions,Kodai Math. J. 32 (2009) 130–145.

[5] J. Noguchi, A note on entire pseudo holomorphic curves and the proof of Cartan–Nochka’s theorem, Kodai Math. J. 28 (2005) 336–346.

[6] J. Noguchi and T. Ochiai, Introduction to Geometric Function Theory in Several Com-plex Variables, Translations of Mathematical Monographs, Vol. 80 (American Mathe-matical Society, Providence, RI, 1990).

[7] S. D. Quang, Two meromorphic functions share some pairs of small functions, ComplexAnal. Oper. Theory 7 (2013) 1357–1370.

[8] K. Yamanoi, The second main theorem for small functions and related problems, ActaMath. 192 (2004) 225–294.

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TWO MEROMORPHIC FUNCTIONS SHARING SOME PAIRS OF SMALL FUNCTIONS REGARDLESS OF MULTIPLICITIES