AAA Abstract and Applied Analysis 1687-0409 1085-3375 Hindawi Publishing Corporation 896596 10.1155/2012/896596 896596 Research Article The Uniqueness of Analytic Functions on Annuli Sharing Some Values Xu Hong-Yan 1 Xuan Zu-Xing 2 Wong Patricia J. Y. 1 Department of Informatics and Engineering Jingdezhen Ceramic Institute Jingdezhen 333403 China jci.edu.cn 2 Beijing Key Laboratory of Information Service Engineering Department of General Education Beijing Union University No. 97 Bei Si Huan Dong Road Chaoyang District, Beijing 100101 China buu.edu.cn 2012 13 8 2012 2012 28 03 2012 11 07 2012 2012 Copyright © 2012 Hong-Yan Xu and Zu-Xing Xuan. 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.

The purpose of this paper is to deal with the shared values and uniqueness of analytic functions on annulus. Two theorems about analytic functions on annulus sharing four distinct values are obtained, and these theorems are improvement of the results given by Cao and Yi.

1. Introduction

In this paper, we will study the uniqueness problem of analytic functions in the field of complex analysis and adopt the standard notations of the Nevanlinna theory of meromorphic functions as explained (see ).

We use to denote the open complex plane, ¯ to denote the extended complex plane, and 𝕏 to denote the subset of . For a¯, we say that f(z)-a and g(z)-a have the same zeros with the same multiplicities (ignoring multiplicities) in 𝕏 (or ) if two meromorphic functions f and g share the value a  CM  (IM) in 𝕏 (or ). In addition, we also use f=ag=a in 𝕏 (or ) to express that f and g share the value a  CM in 𝕏 (or ), f=ag=a in 𝕏 (or ) to express that f and g share the value a  IM in 𝕏 (or ), and f=ag=a in 𝕏 (or ) to express that f=a implies g=a in 𝕏 (or ).

In 1929, Nevanlinna (see ) proved the following well-known theorem.

Theorem 1.1 (see [<xref ref-type="bibr" rid="B12">4</xref>]).

If f and g are two nonconstant meromorphic functions that share five distinct values a1,a2,a3,a4, and a5  IM in , then f(z)g(z).

After his theorem, the uniqueness theory of meromorphic functions sharing values in the whole complex plane attracted many investigations (see ). In 2003, Zheng  studied the uniqueness problem under the condition that five values are shared in some angular domain in . There were many results in the field of the uniqueness with shared values in the complex plane and angular domain, see (). The whole complex plane and angular domain all can be regarded as simply connected region. Thus, it is interesting to consider the uniqueness theory of meromorphic functions in the multiply connected region. Here, we will mainly study the uniqueness of meromorphic functions in doubly connected domains of complex plane . By the doubly connected mapping theorem  each doubly connected domain is conformally equivalent to the annulus {z:r<|z|<R}, 0r<R+. We consider only two cases: r=0, R=+ simultaneously and 0<r<R<+. In the latter case the homothety zz/rR reduces the given domain to the annulus {z:1/R0<|z|<R0}, where R0=R/r. Thus, every annulus is invariant with respect to the inversion z1/z in two cases.

In 2005, Khrystiyanyn and Kondratyuk [14, 15] proposed the Nevanlinna theory for meromorphic functions on annuli (see also ). We will show the basic notions of the Nevanlinna theory on annuli in the next section. In 2009, Cao et al. [17, 18] investigated the uniqueness of meromorphic functions on annuli sharing some values and some sets and obtained an analog of Nevanlinna’s famous five-value theorem as follows.

Theorem 1.2 (see [<xref ref-type="bibr" rid="B6">18</xref>, Theorem  3.2]).

Let f1 and f2 be two transcendental or admissible meromorphic functions on the annulus 𝔸={z:1/R0<|z|<R0}, where 1<R0+. Let aj(j=1,2,3,4,5) be five distinct complex numbers in ¯. If f1,f2 share aj  IM for j=1,2,3,4,5, then f1(z)f2(z).

Remark 1.3.

For the case R0=+, the assertion was proved by Kondratyuk and Laine .

From Theorem 1.2, we can get the following results easily.

Theorem 1.4.

Under the assumptions of Theorem 1.2, if f1,f2 are two transcendental or admissible analytic functions on annulus 𝔸 and f1,f2 share aj  IM for j=1,2,3,4, then f1(z)f2(z).

In fact, we will prove some general theorems on the uniqueness of analytic functions on the annuli sharing four values in this paper (see Section 3), and these theorems improve Theorem 1.4.

2. Basic Notions in the Nevanlinna Theory on Annuli

Let f be a meromorphic function on the annulus 𝔸={z:1/R0<|z|<R0}, where 1<R<R0+. We recall the classical notations of the Nevanlinna theory as follows: (2.1)N(R,f)=0Rn(t,f)-n(0,f)tdt+n(0,f)logR,m(R,f)=12π02πlog+|f(Reiθ)|dθ,T(R,f)=N(R,f)+m(R,f), where log+x=max{logx,0} and n(t,f) is the counting function of poles of the function f in {z:|z|t}. We here show the notations of the Nevanlinna theory on annuli. Let (2.2)N1(R,f)=1/R1n1(t,f)tdt,N2(R,f)=1Rn2(t,f)tdt,m0(R,f)=m(R,f)+m(1R,f),N0(R,f)=N1(R,f)+N2(R,f), where n1(t,f) and n2(t,f) are the counting functions of poles of the function f in {z:t<|z|1} and {z:1<|z|t}, respectively. The Nevanlinna characteristic of f on the annulus 𝔸 is defined by (2.3)T0(R,f)=m0(R,f)-2m(1,f)+N0(R,f) and has the following properties.

Proposition 2.1 (see [<xref ref-type="bibr" rid="B8">14</xref>]).

Let f be a nonconstant meromorphic function on the annulus 𝔸={z:1/R0<|z|<R0}, where 1<R<R0+. Then,

T0(R,f)=T0(R,1/f),

max{T0(R,f1·f2),T0(R,f1/f2),T0(R,f1+f2)}T0(R,f1)+T0(R,f2)+O(1).

By Proposition 2.1, the first fundamental theorem on the annulus 𝔸 is immediately obtained.

Theorem 2.2 (see [<xref ref-type="bibr" rid="B8">14</xref>] (the first fundamental theorem)).

Let f be a nonconstant meromorphic function on the annulus 𝔸={z:1/R0<|z|<R0}, where 1<R<R0+. Then (2.4)T0(R,1f-a)=T0(R,f)+O(1) for every fixed a.

Khrystiyanyn and Kondratyuk also obtained the lemma on the logarithmic derivative on the annulus 𝔸.

Theorem 2.3 (see [<xref ref-type="bibr" rid="B9">15</xref>] (lemma on the logarithmic derivative)).

Let f be a nonconstant meromorphic function on the annulus 𝔸={z:1/R0<|z|<R0}, where R0+, and let λ>0. Then,

in the case R0=+, (2.5)m0(R,ff)=O(log(RT0(R,f))) for R(1,+) except for the set ΔR such that ΔRRλ-1dR<+;

if R0<+, then (2.6)m0(R,ff)=O(log(T0(R,f)R0-R)) for R(1,R0) except for the set ΔR such that ΔRdR/(R0-R)λ-1<+.

We denote the deficiency of a¯={} with respect to a meromorphic function f on the annulus 𝔸 by (2.7)δ0(a,f)=δ0(0,f-a)=liminfrR0m0(r,1/(f-a))T0(r,f)=1-limsuprR0N0(r,1/(f-a))T0(r,f) and denote the reduced deficiency by (2.8)Θ0(a,f)=Θ0(0,f-a)=1-limsuprR0N¯0(r,1/(f-a))T0(r,f), where (2.9)N¯0(r,1f-a)=N¯1(R,1f-a)+N¯2(R,1f-a)=1/R1n¯1(t,1/(f-a))tdt+1Rn¯2(t,1/(f-a))tdt in which each zero of the function f-a is counted only once. In addition, we use n¯1k)(t,1/(f-a)) (or n¯1(k(t,1/(f-a))) to denote the counting function of poles of the function 1/(f-a) with multiplicities k (or >k) in {z:t<|z|1}, each point counted only once. Similarly, we can give the notations N¯1k)(t,f), N¯1(k(t,f), N¯2k)(t,f), N¯2(k(t,f), N¯0k)(t,f), and N¯0(k(t,f).

Khrystiyanyn and Kondratyuk  first obtained the second fundamental theorem on the the annulus 𝔸. Later, Cao et al.  introduced other forms of the second fundamental theorem on annuli as follows.

Theorem 2.4 (see [<xref ref-type="bibr" rid="B6">18</xref>, Theorem  2.3] (the second fundamental theorem)).

Let f be a nonconstant meromorphic function on the annulus 𝔸={z:1/R0<|z|<R0}, where 1<R0+. Let a1,a2,,aq be q distinct complex numbers in the extended complex plane ¯. Let λ0. Then,

(q-2)T0(R,f)<j=1qN0(R,1/(f-aj))-N0(1)(R,f)+S(R,f),

(q-2)T0(R,f)<j=1qN¯0(R,1/(f-aj))+S(R,f),

where (2.10)N0(1)(R,f)=N0(R,1f)+2N0(R,f)-N0(R,f), and (i) in the case R0=+, (2.11)S(R,f)=O(log(RT0(R,f))) for R(1,+) except for the set ΔR such that ΔRRλ-1dR<+; (ii) if R0<+, then (2.12)S(R,f)=O(log(T0(R,f)R0-R)) for R(1,R0) except for the set ΔR such that ΔRdR/(R0-R)λ-1<+.

Definition 2.5.

Let f(z) be a nonconstant meromorphic function on the annulus 𝔸={z:1/R0<|z|<R0}, where 1<R0+. The function f is called a transcendental or admissible meromorphic function on the annulus 𝔸 provided that (2.13)limsupRT0(R,f)logR=,1<R<R0=+, or (2.14)limsupRR0T0(R,f)-log(R0-R)=,1<R<R0<+, respectively.

Thus, for a transcendental or admissible meromorphic function on the annulus 𝔸, S(R,f)=o(T0(R,f)) holds for all 1<R<R0 except for the set ΔR or the set ΔR mentioned in Theorem 2.3, respectively.

3. The Main Theorems and Some Lemmas

Now we show our main results, which improve Theorem 1.4.

Theorem 3.1.

Let f,g be two analytic functions on the annulus 𝔸={z:1/R0<|z|<R0}, where 1<R0+, and let aj(j=1,2,3,4) be four distinct values. If f and g share the two distinct values a1,a2  CM in 𝔸 and f=a3g=a3 in 𝔸 and f=a4g=a4 in 𝔸, and f is transcendental or admissible on 𝔸, then f(z)g(z).

Theorem 3.2.

Under the assumptions of Theorem 3.1, with CM replaced by IM, we have either f(z)g(z) or (3.1)fa3g-a1a2g-a4, and a1+a2=a3+a4, a3, and a4 are exceptional values of f and g in 𝔸, respectively.

Remark 3.3.

It is easily seen that Theorems 3.1 and 3.2 are improvement of Theorem 1.4.

To prove the above theorems, we need some lemmas as follows.

Lemma 3.4.

Let f,g be two distinct analytic functions on the annulus 𝔸={z:1/R0<|z|<R0}, where 1<R0+, and let aj(j=1,2,3,4) be four distinct complex numbers. If f=ajg=aj in 𝔸 for j=1,2,3,4 and if f is transcendental or admissible on 𝔸, then g is also transcendental or admissible.

Proof.

By the assumption of Lemma 3.4 and applying Theorem 2.4(ii), we can get(3.2)3T0(R,f)j=14N¯0(R,1f-aj)+S(R,f)j=14N¯0(R,1g-aj)+S(R,f)4T0(R,g)+S(R,f).

Therefore (3.3)T0(R,f)4T0(R,g)+o(T0(R,f)) holds for all 1<R<R0 except for the set ΔR or the set ΔR mentioned in Theorem 2.3, respectively. Then, from Definition 2.5, we get that g is transcendental or admissible on 𝔸.

Lemma 3.5.

Suppose that f is a transcendental or admissible meromorphic function on the annulus 𝔸={z:1/R0<|z|<R0}, where 1<R0+. Let P(f)=a0fp+a1fp-1++ap(a00) be a polynomial of f with degree p, where the coefficients aj(j=0,1,,p) are constants, and let bj(j=1,2,,q) be q(qp+1) distinct finite complex numbers. Then, (3.4)m0(R,P(f)f(f-b1)(f-b2)(f-bq))=S(R,f).

Proof.

From Theorem 2.3 and the definition of m0(R,f), transcendental and admissible function, we can get this lemma by using the same argument as in Lemma  4.3 in .

Lemma 3.6.

Let f,g be two distinct analytic functions on the annulus 𝔸={z:1/R0<|z|<R0}, where 1<R0+. Suppose that f and g share a1,a2  IM in 𝔸, and f=a3g=a3 in 𝔸 and f=a4g=a4 in 𝔸, and aj(j=1,2,3,4) are four distinct finite complex numbers. If f is a transcendental or admissible function on 𝔸, then g is also transcendental or admissible, and

T0(R,g)=2T0(R,f)+S(R),

T0(R,f-g)=3T0(R,f)+S(R);

T0(R,f)=N¯0(R,1/(f-a3))+N¯0(R,1/(f-a4))+S(R),

T0(R,f)=N¯0(R,1/(f-aj))+S(R),j=1,2,

T0(R,g)=N¯0(R,1/(g-aj))+S(R),j=3,4,

T0(R,f)=T0(R,f)+S(R),T0(R,g)=T0(R,g)+S(R),

where S(R):=S(R,f)=S(R,g).

Proof.

By the assumption of this lemma and by Theorem 2.4(ii), we have T0(R,f)3T0(R,g)+S(R,f) and T0(R,g)3T0(R,f)+S(R,g). Thus, we can get S(R,f)=S(R,g).

Let (3.5)η:=fg(f-g)(f-a3)(f-a4)(g-a1)(g-a2).

From the conditions of this lemma, we can get that η is analytic on 𝔸 and η0 unless fg. By Lemma 3.5, we have m0(R,η)=S(R,f)+S(R,g)=S(R). Thus, we can get S(R,η)=S(R).

Since f,g are two nonconstant analytic functions on annulus 𝔸 and share a1,a2  IM in 𝔸 and f=a3g=a3 and f=a4g=a4 in 𝔸, again by Theorem 2.4, we have (3.6)3T0(R,f)j=14N¯0(R,1f-aj)+S(R,f),(3.7)N¯0(r,1f-g)+S(R,f)=T0(R,f-g)+S(R,f),(3.8)T0(R,f)+T0(R,g)+S(R),(3.9)T0(R,g)N¯0(R,1g-a1)+N¯0(R,1g-a2)+S(R,g),(3.10)=N¯0(R,1f-a1)+N¯0(R,1f-a2)+S(R),(3.11)2T0(R,f)+S(R).

From (3.8) and (3.11), we can get (i), and from (3.7), (3.8), and (i), we can get (ii), and from (3.6), (3.8), (3.10), (3.11), and (i), we can get (iii). Thus, we can deduce that (iv) and (v) hold easily from (3.6)–(3.11) and (i)–(iii). Now, we will prove that (vi) holds as follows.

First, we can rewrite (3.5) as (3.12)f=fgη(g-a1)(g-a2)+fg(a3f+a4f-a3a4-fg)η(f-a3)(f-a4)(g-a1)(g-a2).

From (3.12) and Lemma 3.5, we can get m0(R,f)m0(R,f)+S(R,f). Since f is analytic on 𝔸, we have (3.13)T0(R,f)T0(R,f)+2m(1,f)-2m(1,f)+S(R,f). From the fact that f is transcendental or admissible, we have (3.14)T0(R,f)T0(R,f)+S(R,f)+O(1)=T0(R,f)+S(R,f). On the other hand, since m0(R,f)=m0(R,f(f/f))m0(R,f)+m0(R,f/f)+O(1), from Theorem 2.3, we have m0(R,f)m0(R,f)+S(R,f). Thus, we can get (3.15)T0(R,f)T0(R,f)+S(R,f)+O(1)=T0(R,f)+S(R,f).

From (3.14), (3.15) and the fact that f is transcendental or admissible, we can get T0(R,f)=T0(R,f)+S(R,f). Similarly, we can get T0(R,g)=T0(R,g)+S(R,g).

Thus, we complete the proof of this lemma.

4. The Proof of Theorem <xref ref-type="statement" rid="thm3.1">3.1</xref>

Suppose fg. By the assumption of Theorem 3.1, we can get the conclusions (i)–(vi) of Lemma 3.6 and that g is transcendental or admissible on 𝔸. Set (4.1)ψ1=f(f-a3)(f-a1)(f-a2)-g(g-a3)(g-a1)(g-a2),ψ2=f(f-a4)(f-a1)(f-a2)-g(g-a4)(g-a1)(g-a2).

By Lemma 3.4, we can get (4.2)m0(R,ψi)=S(R,f)+S(R,g)=S(R),i=1,2. Moreover, we can prove N0(R,ψi)=O(1)(i=1,2). In fact, the poles of ψi in 𝔸 only can occur at the zeros of f-aj and g-aj(i,j=1,2) in 𝔸. Since f,g share a1,a2  CM in 𝔸, we can see that if z0𝔸 is a zero of f-aj with multiplicity m(1), then z0𝔸 is a zero of g-aj(j=1,2) with multiplicity m(1). Suppose that (4.3)f-aj=(z-z0)mαj(z),g-aj=(z-z0)mβj(z), where αj(z),βj(z) are analytic functions in 𝔸 and αj(z0)0,βj(z0)0(j=1,2); by a simple calculation, we have (4.4)ψi(z0)=K(αj(z0)αj(z0)-βj(z0)βj(z0))(i,j=1,2), where K is a constant. Therefore, we can get that ψi(i=1,2) are analytic in 𝔸. Thus, from (4.2), we can get T0(R,ψi)=m0(R,ψi)+O(1)=S(R)(i=1,2).

If ψi0,i=1,2, then we have (4.5)N¯0(R,1f-a3)N0(R,1ψ1)T0(R,ψ1)+S(R,f)+O(1)=S(R),N¯0(R,1f-a4)N0(R,1ψ2)T0(R,ψ2)+S(R,f)+O(1)=S(R). From (4.5) and Lemma 3.6(iv), we have T0(R,f)S(R). Thus, since f,g are transcendental or admissible functions on 𝔸, that is, f and g are of unbounded characteristic, and from the definition of S(R), we can get a contradiction.

Assume that one of ψ1 and ψ2 is identically zero, say ψ10; then we have (4.6)N¯0(2(R,1g-a4)=N¯0(2(R,1f-a4).

From (3.5), we can see that g(z1)=a4 implies that f(z1)=a4 for such z1𝔸 satisfying η(z1)0. Since T0(R,η)=S(R), we have (4.7)N¯01)(R,1g-a4)=N¯01)(R,1f-a4)+S(R).

From (4.6) and (4.7), we can get (4.8)N¯0(R,1g-a4)=N¯0(R,1f-a4)+S(R).

Similarly, when ψ20, we can get (4.9)N¯0(R,1g-a3)=N¯0(R,1f-a3)+S(R).

From (4.8), (4.9), and Lemma 3.6(i), (v), we can get (4.10)2T0(R,f)=N¯0(R,1f-a3)+S(R) or (4.11)2T0(R,f)=N¯0(R,1f-a4)+S(R). Since f,g are transcendental or admissible functions on the annulus 𝔸, we can get a contradiction again.

Thus, we complete the proof of Theorem 3.1.

5. The Proof of Theorem <xref ref-type="statement" rid="thm3.2">3.2</xref>

Suppose that fg. By Theorem 2.4(ii) and the fact that f is transcendental or admissible on 𝔸, we have (5.1)2T0(R,f)+N¯0(R,1g-a4)N¯0(R,1f-a1)+N¯0(R,1f-a2)+N¯0(R,1f-a3)+N¯0(R,1g-a4)+S(R,f)N¯0(R,1f-g)+S(R,f)T0(R,f)+T0(R,g)+S(R,f)+S(R,g). Therefore, we have (5.2)T0(R,f)+N¯0(R,1g-a4)T0(R,g)+S(R,f)+S(R,g). Similarly, we have (5.3)T0(R,g)+N¯0(R,1f-a3)T0(R,f)+S(R,g)+S(R,f).

From (5.2) and (5.3), we can see that T0(R,f)=T0(R,g)+S(R,f)+S(R,g), and (5.4)N¯0(R,1f-a3)=S(R,f)+S(R,g),N¯0(R,1g-a4)=S(R,f)+S(R,g).

Thus, from (5.2), (5.3), and the definition of S(R), we can get that g is also transcendental or admissible on 𝔸 when f is transcendental or admissible on 𝔸.

From (5.1)–(5.4), we can also get (5.5)2T0(R,f)=N¯0(R,1f-a1)+N¯0(R,1f-a2)+S(R).

From (5.5), we can see that “almost all” of zeros of f-ai(i=1,2) in 𝔸 are simple. Similarly, “almost all” of zeros of g-ai(i=1,2) in 𝔸 are simple, too. Let (5.6)φ1=(a1-a3)f(f-a2)(f-a1)(f-a3)-(a1-a4)g(g-a2)(g-a1)(g-a4),φ2=(a2-a3)f(f-a1)(f-a2)(f-a3)-(a2-a4)g(g-a1)(g-a2)(g-a4). By Lemma 3.5, we can get that m0(R,φi)=S(R)(i=1,2). Since f,g share a1,a2  IM in 𝔸 and from (5.2), we have N0(R,φi)=S(R)(i=1,2). Therefore, we can get T0(R,φi)=S(R),(i=1,2).

If φ10, then we have N¯0(R,1/(f-a2))N¯0(R,1/φ1)=S(R). Thus, from (5.5), we can get a contradiction easily. Similarly, when φ20, we can get a contradiction, too. Hence, φ1,φ2 are identically equal to 0. Then, we have (φ1-φ2)/(a1-a2)0, that is, (5.7)ff-a3-gg-a4-ff-a1+gg-a1-ff-a2+gg-a20, which implies that (5.8)f-a3g-a4(g-a1)(g-a2)(f-a1)(f-a2)c, where c is a nonzero constant. Rewrite (5.8) as (5.9)g2-(a1+a2-cγ(f)f-a3)g+a1a2+ca4γ(f)f-a30, where γ(f):=(f-a1)(f-a2). The discriminant of (5.9) is (5.10)Δ(f)=(a1+a2-cγ(f)f-a3)2-4(a1a2+ca4γ(f)f-a3)=(f)(f-a3)2, where (5.11)(z)=((a1+a2)(z-a3)-cγ(z))2-4a1a2(z-a3)2-4ca4γ(z)(z-a3) is a polynomial of degree 4 in z. If a is a zero of (z) in 𝔸, obviously aa3. Then, from (5.9), f(z)=a implies that (5.12)g(z)=12(a1+a2-cγ(a)a-a3):=b. Set (5.13)ϕ1:=fg(f-g)(f-a1)(g-a2)(f-a3)(g-a4),ϕ2:=fg(f-g)(f-a2)(g-a1)(f-a3)(g-a4),ϕ=ϕ2ϕ1=(f-a1)(g-a2)(f-a2)(g-a1).

By Lemma 3.5, we can get m0(R,ϕi)=S(R)(i=1,2). And by a simple calculation, we can get N0(R,ϕi)=S(R)(i=1,2). Then we have T0(R,ϕi)=S(R)(i=1,2), thus we have T0(R,ϕ)=S(R).

Assume that f is not a Möbius transformation of g; then ϕ is a nonconstant function. Since (5.14)(a1)=((a1+a2)(a1-a3))2-4a1a2(a1-a3)2=(a1-a3)2(a1-a2)20,(a2)=((a1+a3)(a2-a3))2-4a1a2(a2-a3)2=(a2-a3)2(a1-a2)20. From aai(i=1,2) and (5.8), we can get (5.15)N¯0(R,1f-a)N¯0(R,1ϕ-ξ)T0(R,ϕ)=S(R), where ξ=(a-a1)(b-a2)/(a-a2)(b-a1). Since f is transcendental or admissible analytic in 𝔸, by Theorem 2.4(ii) and (5.4), we can get (5.16)T0(R,f)N¯0(R,1f-a3)+N¯0(R,1f-a)+S(R)=S(R).

Since f,g are transcendental or admissible functions on 𝔸, from the above inequality, we can get a contradiction. Therefore, we can get that f is a Möbius transformation of g on 𝔸. Since f,g are transcendental or admissible functions on 𝔸, by a simple calculation, we can get easily that a1+a2=a3+a4 and (5.17)fa3g-a1a2g-a4. Furthermore, a3,a4 are Picard exceptional values of f and g in 𝔸, respectively.

Thus, we complete the proof of Theorem 3.2.

Acknowledgments

Z.-X. Xuan is supported in part by Beijing Municipal Research Foundation for The Excellent Scholars Ministry (2011D005022000009); Science and Technology Research Program of Beijng Municipal Commission of Education (KM201211417011); Funding Project for Academic Human Resources Development in Beijing Union University. This work was supported by the NSF of Jiangxi of China (Grant no. 2010GQS0119).

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