AAA Abstract and Applied Analysis 1687-0409 1085-3375 Hindawi Publishing Corporation 368321 10.1155/2013/368321 368321 Research Article Normal Families of Meromorphic Functions Concerning Higher Derivative and Shared Values Chen Wei 1 Yuan Wenjun 2 http://orcid.org/0000-0001-7847-1644 Tian Honggen 1 Fiorenza Alberto 1 School of Mathematics Sciences Xinjiang Normal University Urumqi Xinjiang 830054 China xjnu.edu.cn 2 School of Mathematics and Information Sciences Guangzhou University Guangzhou Guangdong 510006 China gzhu.edu.cn 2013 30 10 2013 2013 03 07 2013 12 09 2013 12 09 2013 2013 Copyright © 2013 Wei Chen et al. 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.

We study the normal families related to a Hayman conjecture of higher derivative and concerning shared values and get two normal criteria. Our results improve the related theorems which were obtained independently, respectively by Fang and Yuan (2001), Yuan et al. ((2011) and (2012)), Wang et al. (2011), and Qiu et al. (2012). Meanwhile, some examples are given to show the sharpness of our results.

1. Introduction and Main Results

Let f(z) and g(z) be two nonconstant meromorphic functions in a domain D, and let a be a finite complex value. We say that f and g share a CM (or IM) in D provided that f-a and g-a have the same zeros counting (or ignoring) multiplicity in D. When a=, the zeros of f-a mean the poles of f (see ). It is assumed that the reader is familiar with the standard notations and the basic results of Nevanlinna’s value-distribution theory ( or ).

It is very interesting to find normality criteria from the point of view of shared values. In this area, Schwick  first proved an interesting result that a family of meromorphic functions in a domain is normal in which every function shares three distinct finite complex numbers with their first derivative. And later, more results about normality criteria concerning shared values have emerged; for instance, see . In recent years, this subject has attracted the attention of many researchers worldwide.

We now first introduce a normality criterion related to a Hayman normal conjecture .

Theorem 1.

Let be a family of holomorphic (meromorphic) functions defined in a domain, n, a0, and b. If f(z)+afn(z)-b does not vanish in D for each function f(z) and n2(n3), then is normal in D.

The results for the holomorphic case are due to Drasin  for n3, Pang  for n=3, Chen and Fang  for n=2, Ye  for n=2, and Chen and Gu  for the generalized result with a and b replaced by meromorphic functions. The results for the meromorphic case are due to Li , Li , and Langley  for n5, Pang  for n=4, Chen and Fang  for n=3, and Zalcman  for n=3, obtained independently.

When n=2 and is meromorphic, Theorem 1 is not valid in general. Fang and Yuan  gave an example to show this and got a special result below.

Example 2.

The family of meromorphic functions ={fj(z)=jz/(jz-1)2:j=1,2,,} is not normal in D={z:|z|<1}. This is deduced by fj#(0)=j, as j and Marty’s criterion , although for any fj(z), fj+fj2=j(jz-1)-40.

Here f#(ξ) denotes the spherical derivative (1)f#(ξ)=|f(ξ)|1+|f(ξ)|2.

Theorem 3.

Let be a family of meromorphic functions in a domain D, and a0, b. If f(z)+af2(z)-b does not vanish in D and the poles of f(z) are of multiplicity 3 for each f(z), then is normal in D.

In 2008, by the ideas of shared values, Zhang  proved the following.

Theorem 4.

Let be a family of meromorphic (holomorphic) functions in D, let n be a positive integer and let a, b be two finite complex numbers such that a0. If n4(n2) and for every pair of functions f and g in , f-afn and g-agn share the value b IM, then is normal in D.

Example 5 (see [<xref ref-type="bibr" rid="B31">8</xref>]).

The family of meromorphic functions ={fj(z)=1/j(z-1/j):j=1,2,,} is not normal in D={z:|z|<1}. Obviously fj-fj3=-z/j(z-1/j)3. So for each pair of m, j, fj-fj3 and fm-fm3 share the value 0 in D, but is not normal at the point z=0, since fj#(0)=2(j)3/(1+j), as j.

Remark 6.

Example 5 shows that Theorem 4 is not valid when n=3, and the condition n=4 is best possible for meromorphic case.

In 2011, Yuan et al.  and Wang et al.  proved the following theorems, independently, respectively.

Theorem 7 (see [<xref ref-type="bibr" rid="B26">20</xref>, <xref ref-type="bibr" rid="B20">21</xref>]).

Let be a family of meromorphic functions in D and a, b two finite complex numbers such that a0. Suppose that each f has no simple pole. If f-af3 and g-ag3 share the value b IM for every pair of functions f and g in , then is normal in D.

Theorem 8 (see [<xref ref-type="bibr" rid="B26">20</xref>]).

Let be a family of meromorphic functions in D and a and b two finite complex numbers such that a0. Suppose that each f admits zeros of multiple and the poles of multiplicity at least 3. If f-af2 and g-ag2 share the value b IM for every pair of functions f and g in , then is normal in D.

Lately, Yuan et al.  and Qiu et al.  studied this result, independently, respectively, in which the derivative f was replaced by kth derivative f(k), and they got the following results.

Theorem 9 (see [<xref ref-type="bibr" rid="B25">22</xref>]).

Let be a family of meromorphic functions in D, and let k(2) and n(k+2) be two positive integers. Let a(0) and b be two finite complex numbers. If

f(k)-afn and g(k)-agn share b IM in D for every pair of functions f and g in ,

f has no simple pole and no zero of multiplicity less than k in D for every function f, then is normal in D.

Theorem 10 (see [<xref ref-type="bibr" rid="B25">22</xref>, <xref ref-type="bibr" rid="B17">23</xref>]).

Let be a family of meromorphic functions in D, k(2) and n(k+3) be two positive integers. Let a(0) and b be two finite complex numbers. If

f(k)-afn and g(k)-agn share b IM in D for every pair of functions f and g in ,

f has no zero of multiplicity less than k in D for every function f, then is normal in D.

It is natural to ask whether the condition nk+2 or nk+3 in the previous theorems can be reduced. In this paper, we study this problem and get the following results.

Theorem 11 (main theorem).

Let D be a domain in and let be a family of meromorphic functions in D. Let k,n,d+,n3,d(k+1)/(n-2) and let a, b be two finite complex numbers with a0. Suppose that every f has all its zeros of multiplicity at least k and all its poles of multiplicity at least d. If f(k)-afn and g(k)-agn share the value b IM for every pair of functions (f,g) of , then is a normal family in D.

Remark 12.

When nk+2 or nk+3, we have d=1 or 2, respectively. It follows that Theorem 11 generalizes Theorems 4, 7, 9, and 10.

Theorem 13 (main theorem).

Let D be a domain in and let be a family of meromorphic functions in D. Let k+ and a, b be two finite complex numbers with a0. Suppose that every f has all its zeros of multiplicity at least k+1 and all its poles of multiplicity at least k+2. If f(k)-af2 and g(k)-ag2 share the value b IM for every pair of functions (f,g) of , then is a normal family in D.

Remark 14.

When k=1, it follows that Theorem 13 generalizes Theorem 8.

Example 15 (see [<xref ref-type="bibr" rid="B17">23</xref>]).

Let n,k2 be two positive integers and a a nonzero complex constant. The family of meromorphic functions is ={fj(z)=jzk-1:j=1,2,,}, D={z:|z|<1}. Obviously, for each pair of m, j, fj(k)-afjn and fm(k)-afmn share the value 0 in D, but is not normal.

Example 16.

Let k be a positive integer and a,b two nonzero complex constants such that bj=a/(-1)k((2k)!/k!)jk. The family of meromorphic functions is ={fj(z)=1/(jz-bj)k+1:j=1,2,}, D={z:|z|<1}. Obviously, for each pair of m,j,fj(k)-afj2 and fm(k)-afm2 share the value 0 in D, but is normal since f#(0) as j.

Remark 17.

Example 15 shows that the condition that f(z) admits zeros of multiplicity at least k is best in Theorem 11. For the case k=1, n=3, Example 5 shows that the condition that f(z) admits poles of multiplicity at least d is sharp in Theorem 11. For the case n=2, Example 16 shows that the condition that f(z) admits poles of multiplicity at least k+2 is sharp in Theorem 13. For the case k=1, n=2, Example 2 shows that the condition that f(z) admits zeros of multiplicity at least k+1 in Theorem 13 is sharp.

2. Preliminary Lemmas

In order to prove our results, we need the following lemmas. The first is the extended version Zalcman’s  concerning normal families.

Lemma 1 (see [<xref ref-type="bibr" rid="B30">25</xref>]).

Let be a family of meromorphic functions on the unit disc satisfying all zeros of functions in having multiplicity ≥p and all poles of functions in having multiplicity ≥q. Let α be a real number satisfying -q<α<p. Then is not normal at 0 if and only if there exist

a number 0<r<1;

points zn  with  |zn|<r;

functions fn;

positive numbers ρn0

such that gn(ζ):=ρn-αfn(zn+ρnζ) converges spherically uniformly on each compact subset of to a nonconstant meromorphic function g(ζ), whose all zeros have multiplicity ≥p and all poles have multiplicity ≥q and order is at most 2.

Lemma 2.

Let f(z) be a meromorphic function such that f(k)(z)0 and c{0}, k,n,d+ with n3, d(k+1)/(n-2). If all zeros of f are of multiplicity at least k and all poles of f are of multiplicity at least d, then (2)T(r,f)1kN(r,1f)+N¯(r,1f(k)-cfn)+S(r,f), where S(r,f)=o(T(r,f)), as r, possibly outside a set with finite linear measure.

Proof.

Set (3)Φ(z):=f(k)(z)cfn(z).

Since f(k)(z)0, we have Φ(z)0. Thus, (4)fn(z)=f(k)(z)cΦ(z). Hence, (5)nm(r,f)=m(r,fn)m(r,f(k)Φ)+log+1|c|m(r,1Φ)+m(r,f(k))+log+1|c|m(r,1Φ)+m(r,f(k)f)+m(r,f)+log+1|c|. So that (6)(n-1)m(r,f)m(r,1Φ)+m(r,f(k)f)+log+1|c|.

On the other hand, (4) gives (7)nN(r,f)N(r,fn)=N(r,f(k)Φ)N(r,f(k))+N(r,1Φ)-N¯(r,Φ=f(k)=0), where N¯(r,Φ=f(k)=0) denotes the counting function of zeros of both Φ and f(k).

We obtain (8)nN(r,f)N(r,f)+kN¯(r,f)+N(r,1Φ)-N¯(r,Φ=f(k)=0),(n-1)N(r,f)kN¯(r,f)+N(r,1Φ)-N¯(r,Φ=f(k)=0). By (4), we have (9)N¯(r,φ)+N¯(r,1φ)N¯(r,1f)+N¯(r,f)+N¯(r,Φ=f(k)=0). From (6)~(9), we obtain (10)(n-1)T(r,f)kN¯(r,f)+T(r,1Φ)-N¯(r,Φ=f(k)=0)+S(r,f)kN¯(r,f)+T(r,Φ)-N¯(r,Φ=f(k)=0)+S(r,f)kN¯(r,f)+N¯(r,1Φ)+N¯(r,Φ)+N¯(r,1Φ-1)-N¯(r,Φ=f(k)=0)+S(r,f)(k+1)N¯(r,f)+N¯(r,1f)+N¯(r,1f(k)-cfn)+S(r,f).

Since all zeros and poles of f are multiplicities at least k and d, we get (11)N¯(r,f)1dN(r,f)1dT(r,f)n-2k+1T(r,f),N¯(r,1f)1kN(r,1f). So that (12)T(r,f)1kN(r,1f)+N¯(r,1f(k)-cfn)+S(r,f).

Lemma 3.

Let f(z) be a transcendental meromorphic function such that f(k)(z)0. Let k+ and c{0}. If all zeros of f are of multiplicity at least k+1 and all poles of f are of multiplicity at least k+2, then f(k)-cf2 has infinitely many zeros.

Proof.

Suppose that f(k)-cf2 has only finitely many zeros; then N(r,1/(f(k)-cf2))=S(r,f). Clearly, an arbitrary zero of f is a zero of f(k)-cf2 since all zeros of f are of multiplicity at least k+1; then we can deduce that f has only finite zeros, so N(r,1/f)=O(logr)=S(r,f).

Set (13)Φ(z):=f(k)(z)cf2(z). Similarly, with the proof of Lemma 2, we can get (14)T(r,f)(k+1)N¯(r,f)+N¯(r,1f)+N¯(r,1f(k)-cf2)+S(r,f).

Since all poles of f are multiplicities at least k+2, we obtain (15)N¯(r,f)1k+2N(r,f)1k+2T(r,f), so that (16)T(r,f)(k+2)N¯(r,1f)+(k+2)N¯(r,1f(k)-cf2)+S(r,f)=S(r,f). This is contradicting with the fact that f is transcendental.

Hence, Lemma 3 is proved completely.

Lemma 4.

Let f(z) be a nonconstant rational function such that f(k)(z)0. Let a{0}, and k,n,m+ with n2 and m(k+2)/(n-1). If f0 and all poles of f are of multiplicity at least m, then f(k)-afn has at least two zeros.

Proof.

Suppose, to the contrary, that f(k)-afn has at most one zero.

Since f0, we get that f is a rational but not a polynomial.

Case  1. If f(k)-afn has only zero z0 with multiplicity l, set (17)f(z)=A(z-z1)β1(z-z2)β2(z-zt)βt, where A is a nonzero constant and βi(k+2)/(n-1)(i=1,2,,t).

For the sake of simplicity, we denote (18)β1+β2++βt=q.

From (17), we have (19)f(k)=g(z)(z-z1)β1+k(z-z2)β2+k(z-zt)βt+k, where g(z) is a polynomial such that deg(g(z))k(t-1).

From (17) and (19), we get (20)f(k)-afn=g(z)(z-z1)β1+k(z-z2)β2+k(z-zt)βt+k-aAn(z-z1)nβ1(z-z2)nβ2(z-zt)nβt=[g(z)(z-z1)(n-1)β1-k(z-z2)(n-1)β2-k(z-zt)(n-1)βt-k-aAn]×((z-z1)nβ1(z-z2)nβ2(z-zt)nβt)-1.

By the assumption that f(k)-afn has exactly one zero z0 with multiply l, we have (21)f(k)-afn=C(z-z0)l(z-z1)nβ1(z-z2)nβ2(z-zt)nβt, where C is a nonzero constant. Thus, (22)C(z-z0)lg(z)(z-z1)(n-1)β1-k(z-zt)(n-1)βt-k-aAn.

Differentiating (22), we obtain (23)Cl(z-z0)l-1(z-z1)(n-1)β1-k-1(z-zt)(n-1)βt-k-1×[j=1,jitg(z)(z-z1)(z-zt)+g(z)i=1t((n-1)βi-k)j=1,jit(z-zj)].

For the sake of simplicity, we denote (24)g1(z)=Cl(z-z0)l-1,g2(z)=(z-z1)(n-1)β1-k-1(z-zt)(n-1)βt-k-1×[j=1,jitg(z)(z-z1)(z-zt)+g(z)i=1t((n-1)βi-k)j=1,jit(z-zj)]. Hence, (25)g1(z)g2(z).

Since (n-1)βi-k-11, we have g2(zi)=0. But g1(zi)0(i=1,2,,t), a contradiction.

Case  2. If f(k)-afn has no zeros, then l=0 for (21). We have (26)f(k)-afn=C(z-z1)nβ1(z-z2)nβ2(z-zt)nβt, where C is a nonzero constant. Thus, (27)Cg(z)(z-z1)(n-1)β1-k(z-zt)(n-1)βt-k-aAn,g(z)(z-z1)(n-1)β1-k(z-zt)(n-1)βt-kC+aAn.

Obviously, g(z)(z-z1)(n-1)β1-k(z-zt)(n-1)βt-k is not a constant, a contradiction.

Lemma 4 is proved.

Lemma 5.

Let f(z) be a rational function and a{0}, and k,n,m+ with n2 and m(k+2)/(n-1). If all zeros of f are of multiplicity at least k+1 and all poles of f are of multiplicity at least m, then f(k)-afn has at least two distinct zeros.

Proof.

Suppose, to the contrary, that f(k)-afn has at most one zero.

Case  I. When f is a nonconstant polynomial, noting that all zeros of f have multiplicity at least k+1, we know that f(k)-afn must have zeros. We claim that f has exactly one zero. Otherwise, we can get that f(k)-afn has at least two zeros, which contradicts our assumption.

Set (28)f(z)=B(z-z0)s, where sk+1, B is a nonzero constant. Then (29)f(k)(z)-afn(z)=B(z-z0)s-k×[s(s-1)(s-k+1)-a(z-z0)(n-1)s+k].

Since s-k1, we obtain that s(s-1)(s-k+1)-a(z-z0)s+k has at least one zero which is not z0 from (29). Therefore, f(k)-afn has at least two distinct zeros, a contradiction.

Case  II. When f is rational but not a polynomial, we consider two cases.

Case  1. Suppose that f(k)-afn has only zero z0 with multiplicity at least l. If f0, by Lemma 4, we get a contradiction. So f has zeros, and then we can deduce that z0 is the only zero of f. Otherwise, f(k)-afn has at least two distinct zeros, a contradiction.

We set (30)f(z)=A(z-z0)s(z-z1)β1(z-z2)β2(z-zt)βt, where A is a nonzero constant and sk+1, βim((k+2)/(n-1))(i=1,2,,t).

For the sake of simplicity, we denote (31)β1+β2++βt=q.

From (31), we have (32)f(k)=A(z-z0)s-kg(z)(z-z1)β1+k(z-z2)β2+k(z-zt)βt+k, where g(z) is a polynomial with deg(g)k(s+t+1).

From (30) and (32), we get (33)f(k)-afn=A(z-z0)s-kg(z)(z-z1)β1+k(z-z2)β2+k(z-zt)βt+k-An(z-z0)ns(z-z1)nβ1(z-z2)nβ2(z-zt)nβt=(A(z-z0)s-k×[g(z)(z-z1)(n-1)β1-k(z-z2)(n-1)β2-k(z-zt)(n-1)βt-k-aAn-1×(z-z0)(n-1)s+k])×((z-z1)nβ1(z-z2)nβ2(z-zt)nβt)-1.

By assumption that f(k)-afn has exactly one zero z0 with multiplicity l, we have (34)f(k)-afn=B(z-z0)l(z-z1)nβ1(z-z2)nβ2(z-zt)nβt, where B is a nonzero constant. Thus, (35)B(z-z0)lA(z-z0)s-k×[g(z)(z-z1)(n-1)β1-k(z-zt)(n-1)βt-k-aAn-1(z-z0)(n-1)s+k].Case  1.1. If l>s-k, from (35), we can deduce that z0 is a zero of (z-z1)(n-1)β1-k(z-zt)(n-1)βt-k, a contradiction.

Case  1.2. If l=s-k, from (35), it follows that (36)g(z)(z-z1)(n-1)β1-k(z-zt)(n-1)βt-k-aAn-1(z-z0)(n-1)s+kBA.

Differentiating (36), we have (37)(z-z1)(n-1)β1-k-1(z-zt)(n-1)βt-k-1×[j=1,jitg(z)(z-z1)(z-zt)+g(z)i=1t((n-1)βi-k)j=1,jit(z-zj)]a((n-1)s+k)An-1(z-z0)(n-1)s+k-1.

For the sake of simplicity, we denote (38)g1(z)=(z-z1)(n-1)β1-k-1(z-zt)(n-1)βt-k-1×[j=1,jitg(z)(z-z1)(z-zt)+g(z)i=1t((n-1)βi-k)j=1,jit(z-zj)],g2(z)=a((n-1)s+k)An-1(z-z0)(n-1)s+k-1. Thus, (39)g1(z)g2(z).

Since (n-1)βi-k-11, we get g1(zi)=0, but g2(zi)0(i=1,2,,t), a contradiction.

Case  2. If f(k)-afn has no zeros, then l=0 for (34). Similarly, as the proof of Case  1, we also have a contradiction.

The proof is completed.

3. Proofs of Theorems Proof.

In Theorem 11, suppose that is not normal in D. Then there exists at least one point z0 such that is not normal at the point z0. Without loss of generality, we assume that z0=0. By Lemma 1, there exist points zj0, positive numbers ρj0, and functions fj such that (40)gj(ξ)=ρjk/(n-1)fj(zj+ρjξ)g(ξ) locally uniformly with respect to the spherical metric, where g is a nonconstant meromorphic function in and whose poles and zeros are of multiplicity at least d and k, respectively. Moreover, the order of g is at most 2.

From (40), we know that (41)gj(k)(ξ)=ρjnk/(n-1)fj(k)(zj+ρjξ)g(k)(ξ),(42)gj(k)(ξ)-agjn(ξ)-ρjnk/(n-1)b=ρjnk/(n-1)(fj(k)(zj+ρjξ)-afjn(zj+ρjξ)-b)g(k)(ξ)-agn(ξ) also locally uniformly with respect to the spherical metric.

Case  1. If k=1, by Theorems 4 and 7, the Theorem 11 assumes.

Case  2. If k=2, n4, by Theorems 9 and 10, the Theorem 11 assumes.

Case  3. If k=2, n=3 or k3, 3nk+1 (If nk+2, by Theorems 9 and 10, we can get Theorem 11).

If g(k)(ξ)-agn(ξ)0, since all poles of g have multiplicity at least d, we have (43)nT(r,g)=T(r,gn)=T(r,g(k))+O(1)=m(r,g(k))+N(r,g(k))+O(1)m(r,g)+N(r,g)+kN¯(r,g)+S(r,g)T(r,g)+k(n-2)k+1T(r,g)+S(r,g)(n-1)T(r,g)+S(r,g). Therefore, g(ξ) is a constant, a contradiction. So g(k)(ξ)-agn(ξ)0.

By Lemma 2, we have (44)T(r,g)1kN(r,1g)+N¯(r,1g(k)-agn)+S(r,g)1kT(r,1g)+N¯(r,1g(k)-agn)+S(r,g). Then (45)(1-1k)T(r,g)N¯(r,1g(k)-agn)+S(r,g),(46)T(r,g)(1+1k-1)N¯(r,1g(k)-agn)+S(r,g). If g(k)(ξ)-agn(ξ)0, then (46) gives that g(ξ) is also a constant. Hence, g(k)(ξ)-agn(ξ) is a nonconstant meromorphic function and has at least one zero.

Next we prove that g(k)(ξ)-agn(ξ) has just a unique zero. On the contrary, let ξ0 and ξ0* be two distinct zeros of g(k)(ξ)-agn(ξ), and choose δ(>0) small enough such that D(ξ0,δ)D(ξ0*,δ)=ϕ, where D(ξ0,δ)={ξ:|ξ-ξ0|<δ} and D(ξ0*,δ)={ξ:|ξ-ξ0*|<δ}. From (42), by Hurwitz’s theorem, there exist points ξjD(ξ0,δ), ξj*D(ξ0*,δ) such that for sufficiently large j(47)fj(k)(zj+ρjξj)-afjn(zj+ρjξj)-b=0,fj(k)(zj+ρjξj*)-afjn(zj+ρjξj*)-b=0.

By the hypothesis that for each pair of functions f and g in , f(k)-afn and g(k)-agn share b in D, we know that for any positive integer m(48)fm(k)(zj+ρjξj)-afmn(zj+ρjξj)-b=0,fm(k)(zj+ρjξj*)-afmn(zj+ρjξj*)-b=0.

Fix m, take j, and note zj+ρjξj0, zj+ρjξj*0; then fm(k)(0)-afmn(0)-b=0. Since the zeros of fm(k)-afmn-b have no accumulation point, so (49)zj+ρjξj=0,zj+ρjξj*=0. Hence, ξj=-zj/ρj, ξj*=-zj/ρj. This contradicts ξjD(ξ0,δ), ξj*D(ξ0*,δ) and D(ξ0,δ)D(ξ0*,δ)=ϕ. So g(k)(ξ)-agn(ξ) has just a unique zero, which can be denoted by ξ0.

Noting that g has poles and zeros of multiplicities at least d and k, respectively, (46) deduces that g(ξ) is a rational function with degree at most 2.

If g(ξ) is a polynomial, noting that degg2 and the multiplicities of zeros are at least k, we have k=2 and n=3. Hence, there exist ξ1 and c0 such that g(ξ)=c(ξ-ξ1)2, and then g′′(ξ)-ag3(ξ)=c(2-ac2(ξ-ξ1)6) has 6 distinct zeros, a contradiction.

Suppose that g(ξ) is not a polynomial, k=2 and n=3. Then the multiplicities of poles of g(ξ) are at least (k+1)/(n-2)=3, which implies that degg3, a contradiction.

Suppose that g(ξ) is not a polynomial, k3 and 3nk+1; we distinguish two cases.

Case  i. If g(ξ) has zeros, since all zeros of g(ξ) have multiplicity at least k(3), it follows that degg3, a contradiction.

Case  ii. If g(ξ)0, then (46) should be as follows (50)T(r,g)N¯(r,1g(k)-agn)+S(r,g).

From (50), we can see that g(ξ) is a rational function with degree at most 1. Since all poles of g(ξ) have multiplicity at least d((k+1)/(n-2)(k+1)/(k-1)>1), which gives that degg2, a contradiction.

This completes the proof of Theorem 11.

Proof.

In Theorem 13, suppose that is not normal in D. Then there exists at least one point z0 such that is not normal at the point z0. Without loss of generality, we assume that z0=0. By Lemma 1, there exist points zj0, positive numbers ρj0, and functions fj such that (51)gj(ξ)=ρjkfj(zj+ρjξ)g(ξ) locally uniformly with respect to the spherical metric, where g is a nonconstant meromorphic function in and whose poles and zeros are of multiplicity at least k+2 and k+1, respectively. Moreover, the order of g is at most 2.

From (51), we know (52)gj(k)(ξ)=ρj2kfj(k)(zj+ρjξ)g(k)(ξ),(53)gj(k)(ξ)-agj2(ξ)-ρj2kb=ρj2k(fj(k)(zj+ρjξ)-afj2(zj+ρjξ)-b)g(k)(ξ)-ag2(ξ) also locally uniformly with respect to the spherical metric.

If g(k)(ξ)-ag2(ξ)0, since all poles of f have multiplicity at least k+2, we can deduce that g(ξ) is an entire function easily. Thus, (54)2T(r,g)=T(r,g2)=T(r,g(k))+O(1)=m(r,g(k))+N(r,g(k))+O(1)m(r,g)+N(r,g)+kN¯(r,g)+S(r,g)T(r,g)+S(r,g). Therefore, g(ξ) is a constant, a contradiction. So g(k)(ξ)-ag2(ξ)0. By Lemmas 3, 4, and 5, g(k)(ξ)-ag2(ξ) has at least two distinct zeros. Proceeding as in the later proof of Theorem 11, we will get a contradiction. The proof is completed.

Similarly, as the proof of Theorem 11, when f is a holomorphic function, we can get the following theorem which has been gotten in .

Theorem 18 (see [<xref ref-type="bibr" rid="B17">23</xref>]).

Let D be a domain in and let be a family of holomorphic functions in D. Let k,n+, n2 and let a, b be two finite complex numbers with a0. Suppose that every f has all its zeros of multiplicity at least k. If f(k)-afn and g(k)-agn share the value b IM for every pair of functions (f,g) of , then is a normal family in D.

Acknowledgments

This paper is supported by the Nature Science Foundation of China (11271090), Nature Science Foundation of Guangdong Province (S2012010010121), and Graduate Research and Innovation Projects of Xinjiang Province (XJGRI2013131). This work was supported by the Visiting Scholar Program of Chern Institute of Mathematics at Nankai University. The second author would like to express his hearty gratitude to Chern Institute of Mathematics which provided very comfortable research environments to him where he worked as a Visiting Scholar.

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