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Uniqueness of meromorphic functions sharing two finite sets

  • Jun-Fan Chen EMAIL logo
Published/Copyright: October 9, 2017
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Open Mathematics
From the journal Open Mathematics Volume 15 Issue 1

Abstract

We prove uniqueness theorems of meromorphic functions, which show how two meromorphic functions are uniquely determined by their two finite shared sets. This answers a question posed by Gross. Moreover, some examples are provided to demonstrate that all the conditions are necessary.

Keywords: Meromorphic function; Shared set; Order; Uniqueness
MSC 2010: 30D35; 30D30

1 Introduction and main results

Throughout this paper, for a meromorphic function, the word “meromorphic” means meromorphic in the whole complex plane ℂ. Let 𝓜(ℂ) (resp. 𝓔(ℂ)) be the field of meromorphic (resp. holomorphic) functions in ℂ. The order λ(f) and the lower order μ(f) of f ∈ 𝓜(ℂ) are defined in turn as follows:

λ(f)=lim supr→∞log⁡T(r,f)log⁡r,μ(f)=lim infr→∞log⁡T(r,f)log⁡r.

If λ(f) < +∞, then we denote by S(r, f) any quantity satisfying S(r, f) = O(log r), r → ∞. If λ(f) = +∞, then we denote by S(r, f) any quantity satisfying S(r, f) = O(log(rT(r, f))), r → ∞, r ∉ E, where E is a set of finite linear measure not necessarily the same at every occurrence.

Denote the preimage of a subset S ⊆ ℂ∪{∞} under h ∈ 𝓜(ℂ) by

E(S,h)=⋃a∈S{z∈C|h(z)−a=0},

where each zero of h(z) − a of multiplicity l appears l times in E(S, h). The notation E(S, h) expresses the set containing the same points as E(S, h) but without counting multiplicities. Let f, g ∈ 𝓜(ℂ). If E(S, f) = E(S, g), then f and g share the set S CM (counting multiplicity). If E(S, f) = E(S, g), then f and g share the set S IM (ignoring multiplicity). For fundamental concepts and results from Nevanlinna theory and further details related to 𝓜(ℂ), see [1, 2].

In the sequel, we mainly consider a subset 𝓜1(ℂ) of 𝓜(ℂ) defined by

M1(C)={f∈M(C)|f has only finitely many poles in C}.

In 1976, Gross (see [3]) posed the following interesting question.

Question 1.1

Can one find two finite sets Si(i = 1, 2) of ℂ ∪{∞} such that any two elements f and g of a family 𝓖 ⊆ 𝓔(ℂ) satisfying E(Si, f) = E(si, g) for i = 1, 2 must be identically equal?

More generally, Question 1.1 suggested the following question.

Question 1.2

For a family 𝓖 ⊆ 𝓜(ℂ), determine subsets S1, S2, ⋯, Sq of ℂ ∪ {∞} in which the cardinality of every Si (i = 1, 2, ⋯, q) is as small as possible and minimise the number q such that any two elements f and g of 𝓖 are algebraically dependent if E(Si, f) = E(Si, g) for every i (i = 1, 2, ⋯, q), that is, if f and g share every Si(i = 1, 2, ⋯, q) CM (counting multiplicity).

In [4], Yi proved that there exist two finite sets S1 (with 1 element) and S2 (with 5 elements) of ℂ such that any two elements f and g in 𝓔(ℂ) sharing S1 and S2 CM must be identically equal, which completely answered Question 1.1. In [5] and [6], Fang and Xu and independently Yi proved that there exist two finite sets S1 (with 1 element) and S2 (with 3 elements) of ℂ such that any two elements f and g in 𝓔(ℂ) sharing S1 and S2 CM must be identically equal, which also answered Question 1.1.

For the case 𝓖 = 𝓜(ℂ), choosing Si = {ai}(i = 1, 2, ⋯, q) for distinct elements ai of ℂ ∪ {∞}, when q ≥ 4, Question 1.2 was completely settled by famous four-value theorem due to Nevanlinna (see e.g. [7] or [1, 2 However, Question 1.2 is still interesting for the cases q ≤ 3. In [8], Li and Yang proved that there exist two finite sets S1 (with 15 elements) of ℂ and S2 = {∞} such that any two elements f and g in 𝓜(ℂ) sharing S1 and S2 CM must be identically equal. In [9] and [10], Yi and independently Li and Yang proved that there exist two finite sets S1 (with 11 elements) of ℂ and S2 = {∞} such that any two elements f and g in 𝓜(ℂ) sharing S1 and S2 CM must be identically equal. In [11], Fang and Guo proved that there exist two finite sets S1 (with 9 elements) of ℂ and S2 = {∞} such that any two elements f and g in 𝓜(ℂ) sharing S1 and S2 CM must be identically equal. In [12], Yi proved that there exist two finite sets S1 (with 8 elements) of ℂ and S2 = {∞} such that any two elements f and g in 𝓜(ℂ) sharing S1 and S2 CM must be identically equal. In [4], Yi proved that there exist two finite sets S1 (with 2 element) and S2 (with9 elements) of ℂ such that any two elements f and g in 𝓜(ℂ) sharing S1 and S2 CM must be identically equal. In [13], Yi and Li recently proved that there exist two finite sets S1 (with 2 element) and S2 (with 5 elements) of ℂ such that any two elements f and g in 𝓜(ℂ) sharing S1 and S2 CM must be identically equal.

For the family 𝓖 = 𝓜1(ℂ), we solve Question 1.2 by proving the following theorems.

Theorem 1.3

Let k be a positive integer and let S1 = {α1, α2, ⋯, αk}, S2 = {β1, β2}, where α1, α2, ⋯, αk, β1, β2 are k + 2 distinct finite complex numbers satisfying

(β1−α1)2(β1−α2)2⋯(β1−αk)2≠(β2−α1)2(β2−α2)2⋯(β2−αk)2.

If two nonconstant meromorphic functions f(z) and g(z) in 𝓜1(ℂ) share S1 CM, S2IM, and if the order of f(z) is neither an integer nor infinite, then f(z) ≡ g(z).

In order to state the next result, we need the following definition related to unique range set.

Definition 1.4

For a family 𝓖 ⊆ 𝓜(ℂ), the subsets S1, S2, ⋯, Sq of ℂ ∪{∞} such that for any two elements f and g of 𝓖 the conditions E(Si, f) = E(Si, g) for every i(i = 1, 2, ⋯, q) imply f(z) ≡ g(z) are called unique range sets (URS, in brief) of meromorphic functions in 𝓜(ℂ).

For the case 𝓖 = 𝓔(ℂ) (resp. 𝓖 = 𝓜(ℂ)), q = 1 in Definition 1.4, the best lower and upper bounds of the cardinality of the set S1 known so far are 4 and 7 (resp. 5 and 11), respectively.

Choosing the family 𝓖 = 𝓜1(ℂ), q = 2 in Definition 1.4, from Theorems 1.3 we have the following result.

Theorem 1.5

Let k be a positive integer and let S1 = {α1, α2, ⋯, αk}, S2 = {β1, β2}, where α1, α2, ⋯, αk, β1, β2 are k + 2 distinct finite complex numbers satisfying

(β1−α1)2(β1−α2)2⋯(β1−αk)2≠(β2−α1)2(β2−α2)2⋯(β2−αk)2.

If the order of f(z) is neither an integer nor infinite, then the sets S1 and S2 are the URS of meromorphic functions in 𝓜1(ℂ).

Remark 1.6

The following example shows that the condition “ (β1−α1)2(β1–α2)2 ⋯ (β1–αk)2≠(β2–α1)2(β2–α2)2⋯(β2–αk)2” in Theorems 1.3-1.5 cannot be dropped. Fix a positive integer k. Let f(z) = ∑n=1∞znn3n, g(z) = −f(z), S1 = {−1, 1, −2, 2, ⋯, −k, k}, and S2 = {−(k + 1), k + 1}. Then by Lemma 2.1 in Section 2 we deduce

λ(f)=1lim infn→∞log⁡n3nnlog⁡n=lim supn→∞nlog⁡nlog⁡n3n=13.

It is easy to verify that f(z), g(z)∈ 𝓜1(ℂ), f(z) and g(z) share S1, S2 CM. But f(z) ≢ g(z).

Remark 1.7

The assumption “nonconstant meromorphic functions f(z) and g(z) in 𝓜1(ℂ) ” in Theorems 1.3-1.5 cannot be relaxed to “nonconstant meromorphic functions f(z) and g(z) in 𝓜(ℂ) as shown by the following example. Fix a positive integer k. Let f(z) = ∑n=1∞znn3n,g(z)=1f(z),S1={2,12,3,13,⋯,k,1k},S2={k+1,1k+1}. Then by Remark 1.6 we know λ(f) = 13 and so by Lemma 2.2 in Section 2 we see that g(z) has infinitely many poles in ℂ. Moreover, f(z) and g(z) share S1, S2 CM. But f(z)≢ g(z).

Remark 1.8

The assumption that the order of f(z) is neither an integer nor infinite in Theorems 1.3-1.5 is necessary. The example is as follows. Fix a positive integer k. Let f(z) = ez (resp. f(z) = eez), g(z) = g(z)=1f(z),S1={2,12,3,13,⋯,k,1k},S2={k+1,1k+1}. Then by Lemma 2.3 in Section 2 we see that λ(f) = 1 (resp. λ(f) = ∞). Moreover, all other conditions of Theorems 1.3-1.5 are satisfied. But f(z)≢ g(z).

2 Some lemmas

In this section we present some important lemmas which will be needed in the sequel.

Lemma 2.1

(see [14], p. 288). Let f(z) = ∑n=0∞ cnzn ∈ 𝓔(ℂ) be nonconstant and of finite order. Then

λ(f)=1lim infn→∞−log⁡|cn|nlog⁡n.

Lemma 2.2

(see [14], p. 293). Let f(z)∈ 𝓔(ℂ). If the order of f(z) is neither an integer nor infinite, then f(z) assumes every finite value infinitely often.

Lemma 2.3

(see [2], Theorem 1.44). Let h(z)∈ 𝓔(ℂ), and let f(z) = eh(z). Then

  1. if h(z) is a polynomial of degree deg h, then λ(f) = μ(f) = deg h;

  2. if h(z) is a transcendental entire function, then λ(f) = μ(f) = ∞.

Lemma 2.4

(see [15] or [2], Theorem 1.19). Let T1(r) and T2(r) be two nonnegative, nondecreasing real functions defined in r > r0 > 0. If T1(r) = O(T2(r))(r → ∞, r ∉ E), where E is a set with finite linear measure, then

lim supr→∞log+⁡T1(r)log⁡r≤lim supr→∞log+⁡T2(r)log⁡r

and

lim infr→∞log+⁡T1(r)log⁡r≤lim infr→∞log+⁡T2(r)log⁡r,

which imply that the order and the lower order of T1(r) are not greater than the order and the lower order of T2(r) respectively

Lemma 2.5

(see [2], Theorem 1.42). Let f(z)∈ 𝓜(ℂ). If 0 and ∞ are two Picard exceptional values of f(z), then f(z) = eh(z), where h(z) ∈ 𝓔(ℂ).

Lemma 2.6

(see [2], Theorem 1.14). Let f(z), g(z)∈ 𝓜(ℂ). Then

λ(f⋅g)≤max{λ(f),λ(g)},λ(f+g)≤max{λ(f),λ(g)}.

Lemma 2.7

(see [2], Theorem 2.20). Let a1, a2, and a3 be three distinct complex numbers in ℂ∪{∞}. If two nonconstant meromorphic functions f(z) and g(z) in 𝓜(ℂ) share a1, a2, and a3 CM, and if the order of f(z) and g(z) is neither an integer nor infinite, then f(z)≡ g(z).

3 Proofs of the theorems

3.1 Proof of Theorem 1.3

First we consider the following function

V(z)=H(z)(f(z)−α1)(f(z)−α2)⋯(f(z)−αk)(g(z)−α1)(g(z)−α2)⋯(g(z)−αk),

where H(z) is a rational function such that V(z) has neither a pole nor a zero in ℂ. It is easy to see that such an H(z) does exist since f(z), g(z)∈ 𝓜1(ℂ), and a possible pole or zero of (f(z)−α1)(f(z)−α2)⋯(f(z)−αk)(g(z)−α1)(g(z)−α2)⋯(g(z)−αk) may only come from a pole of f(z) or g(z), in view of the condition that f(z) and g(z) share S1 = {α1, α2, ⋯, αk} CM. Then by Lemma 2.5 there exists an entire function ϕ(z)∈ 𝓔(ℂ) such that

V(z)=H(z)(f(z)−α1)(f(z)−α2)⋯(f(z)−αk)(g(z)−α1)(g(z)−α2)⋯(g(z)−αk)=eϕ(z). (1)

Noting that f(z) and g(z) have only finitely many poles, we have

N(r,f)=O(log⁡r),N(r,g)=O(log⁡r). (2)

Since f(z) and g(z) share S2 = {β1, β2} IM, it follows from (2), the first and second fundamental theorems that

T(r,f)≤N¯(r,1f−β1)+N¯(r,1f−β2)+N¯(r,f)+S(r,f)≤N¯(r,1g−β1)+N¯(r,1g−β2)+O(log⁡r)+S(r,f)≤T(r,1g−β1)+T(r,1g−β2)+O(log⁡r)+S(r,f)≤2T(r,g)+O(1)+O(log⁡r)+S(r,f), (3)

r → ∞, r ∉ E. Then by (3) and Lemma 2.4 we obtain

λ(f)≤λ(g). (4)

Similarly,

λ(g)≤λ(f). (5)

Combining (4) with (5) yields

λ(g)=λ(f). (6)

From the first fundamental theorem we have

T(r,1g−αi)=T(r,g)+O(1)

for i = 1, 2, ⋯, k, which implies

λ(1g−αi)=λ(g) (7)

for i = 1, 2, ⋯, k. Moreover,

λ(f−αj)=λ(f) (8)

for i = 1, 2, ⋯, k. Clearly, λ(H) = 0 since H(z) is a rational function. Thus it follows by (1), (6), (7), (8), and Lemma 2.6 that

λ(eϕ)≤λ(f). (9)

In view of the assumption that f(z) and g(z) share S2 = {β1, β2} IM, we deduce from (1) that a zero of (f(z) − β1)(f(z) −β2) is a zero of H−1(z)eϕ(z)−1 or H−1(z)eϕ(z) − (β1−α1)(β1−α2)⋯(β1−αk)(β2−α1)(β2−α2)⋯(β2−αk) or H−1(z)eϕ(z) − (β2−α1)(β2−α2)⋯(β2−αk)(β1−α1)(β1−α2)⋯(β1−αk). We claim that one of the following three cases holds:

(i)(f(z)−α1)(f(z)−α2)⋯(f(z)−αk)(g(z)−α1)(g(z)−α2)⋯(g(z)−αk)≡1;(ii)(f(z)−α1)(f(z)−α2)⋯(f(z)−αk)(g(z)−α1)(g(z)−α2)⋯(g(z)−αk)≡(β1−α1)(β1−α2)⋯(β1−αk)(β2−α1)(β2−α2)⋯(β2−αk);(iii)(f(z)−α1)(f(z)−α2)⋯(f(z)−αk)(g(z)−α1)(g(z)−α2)⋯(g(z)−αk)≡(β2−α1)(β2−α2)⋯(β2−αk)(β1−α1)(β1−α2)⋯(β1−αk).

Otherwise all of the following three cases would hold:

(i′)(f(z)−α1)(f(z)−α2)⋯(f(z)−αk)(g(z)−α1)(g(z)−α2)⋯(g(z)−αk)≢1;(ii′)(f(z)−α1)(f(z)−α2)⋯(f(z)−αk)(g(z)−α1)(g(z)−α2)⋯(g(z)−αk)≢(β1−α1)(β1−α2)⋯(β1−αk)(β2−α1)(β2−α2)⋯(β2−αk);(iii′)(f(z)−α1)(f(z)−α2)⋯(f(z)−αk)(g(z)−α1)(g(z)−α2)⋯(g(z)−αk)≢(β2−α1)(β2−α2)⋯(β2−αk)(β1−α1)(β1−α2)⋯(β1−αk).

Then, in view of the fact that H(z) is rational, it follows by (i’) −(iii’), (1) (2), the first and second fundamental theorems that

T(r,f)≤N¯(r,1f−β1)+N¯(r,1f−β2)+N¯(r,f)+S(r,f)≤N¯(r,1H−1eϕ−1)+N¯(r,1H−1eϕ−(β1−α1)(β1−α2)⋯(β1−αk)(β2−α1)(β2−α2)⋯(β2−αk))+N¯(r,1H−1eϕ−(β2−α1)(β2−α2)⋯(β2−αk)(β1−α1)(β1−α2)⋯(β1−αk))+O(log⁡r)+S(r,f)≤T(r,1H−1eϕ−1)+T(r,1H−1eϕ−(β1−α1)(β1−α2)⋯(β1−αk)(β2−α1)(β2−α2)⋯(β2−αk))+T(r,1H−1eϕ−(β2−α1)(β2−α2)⋯(β2−αk)(β1−α1)(β1−α2)⋯(β1−αk))+O(log⁡r)+S(r,f)≤3T(r,H−1eϕ)+O(1)+O(log⁡r)+S(r,f)≤3T(r,eϕ)+O(1)+O(log⁡r)+S(r,f),

r → ∞, r ∉ E, which together with Lemma 2.4 gives

λ(f)≤λ(eϕ). (10)

Thus from (9) and (10) we have

λ(f)=λ(eϕ).

This contradicts Lemma 2.3 since the order of f(z) is neither an integer nor infinite. The claim is proved. Next we discuss the following three cases.

Case 1. Suppose that (i) occurs. Then by (i) and the assumption

(β1−α1)2(β1−α2)2⋯(β1−αk)2≠(β2−α1)2(β2−α2)2⋯(β2−αk)2

we deduce that f(z) = β1 if and only if g(z) = β1 since f(z) and g(z) share S2 = {β1, β2} IM; further, we know that f(z) = β2 if and only if g(z) = β2. This implies that f(z) and g(z) share β1, β2 IM. Again by (i) we conclude that f(z) and g(z) share β1, β2, and ∞ CM. Note that the order of f(z) is neither an integer nor infinite. Thus from (6) and Lemma 2.7 we get f(z)≡ g(z).

Case 2. Suppose that (ii) occurs. Then by (ii) and the assumption

(β1−α1)2(β1−α2)2⋯(β1−αk)2≠(β2−α1)2(β2−α2)2⋯(β2−αk)2

we deduce that f(z) = β1 if and only if g(z) = β2 since f(z) and g(z) share S2 = {β1, β2} IM; further, we know that f(z) = β2 if and only if g(z) = β1. Since the order of f(z) is neither an integer nor infinite, it follows from Lemma 2.2 that there exists z0 ∈ ℂ such that f(z0) = β2. Thus g(z0) = β1 and so by (ii) we obtain

(β1−α1)2(β1−α2)2⋯(β1−αk)2=(β2−α1)2(β2−α2)2⋯(β2−αk)2,

which contradicts the assumption.

Case 3. Suppose that (iii) occurs. Then using the same manner as in Case 2, we also get a contradiction. This completes the proof of Theorem 1.3.

3.2 Proof of Theorem 1.5

Note that if f and g share the set S CM (counting multiplicity) then f and g certainly share the set S IM (ignoring multiplicity). Then f and g satisfy the conditions in Theorem 1.3. Therefore the conclusion of Theorem 1.5 follows from Theorem 1.3. This completes the proof of Theorem 1.5.

Acknowledgement

The author would like to thank the referees for their thorough comments and helpful suggestions.

Project supported by the National Natural Science Foundation of China (Grant No. 11301076), and the Natural Science Foundation of Fujian Province, China (Grant No. 2014J01004).

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Received: 2017-5-15
Accepted: 2017-8-1
Published Online: 2017-10-9

© 2017 Chen

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 License.

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https://doi.org/10.1515/math-2017-0102
Keywords for this article
Meromorphic function; Shared set; Order; Uniqueness
Creative Commons
BY-NC-ND 4.0

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