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Advances in Pure Mathematics, 2011, 1, 210-217 doi:10.4236/apm.2011.14037 Published Online July 2011 (http://www.SciRP.org/journal/apm) Copyright © 2011 SciRes. APM Normal Criteria and Shared Values by Differential Polynomials* Jihong Wang1, Qian Lu2, Qilong Liao3 1Department of Mathematics, Southwest University of Science and Technology, Mianyang, China 2Department of Mathematics, Southwest University of Science and Technology, Mianyang, China 3Department of Material Science and Engineer, Southwest University of Science and Technology, Mianyang, China E-mail: wangjihong@swust.edu.cn, luqiankuo1965@ho tmail.com, liaoql@swust.edu.cn Received April 5, 2011; revised April 28, 2011; accepted May 8, 2011 Abstract For a family F of meromorphic functions on a domain D, it is discussed whether F is normal on D if for every pair functions f z, g zF, n f af ,b and share value on D when , where a, b are two complex numbers, . Finally, the following result is obtained:Let n agg d2, 3n 0,a F be a family of meromorphic functions in D, all of whose poles have multiplicity at least 4 , all of whose zeros have multi- plicity at least 2. Suppose that there exist two functions az not idendtically equal to zero, analytic in D, such that for each pair of functions dz f and g in F , 2 f az f and 2 g azg share the function . If has only a multiple zeros and dz az fz whenever az 0 , then F is normal in D. Keywords: Normal Family, Meromorphic Function, Shared Value, Differential Polynomial 1. Introduction and the Main Result In 1959,Hayman[4] proved Theorem 1.1. Let f be meromorphic functions in C, n be a positive integer and a, b be two constant such that , . If 5n0,aandb n f af b then f is a constant. Corresponding to Theorem 1.1 there is the following theorems which confirmed a Hayman’s well-known con- jecture about normal families in [5]. Theorem 1.2. Let F be a meromorphic function family in D, be a positive integer and a, b be two constant such that . If and for each function n 0,aandb 3n f Fn , f af b , then F is normal in D. This result is due to S. Y. Li [8](), X. J. Li [9] (), X. C. Pang [10](), H. H. Chen and M. L. Fang [2](). 5n 5n4n 3n In 2001, M. L. Fang and W. J. Yuan [3] obtained Theorem 1.3. Let F be a meromorphic function family in D, a, b be two constants such that . If, for each function 0,aandb f F ,2 f af b and the poles of f z are of multiplicity 3 at least, then F is nor- mal in D. Let D be a domain in C, f z be meromorphic on D,and. aC 1: f Eaf aDZDfz a Two functions f and g are said to share the value a if f Ea Ega. For a case in Theorem 1.2, Q. C. Zhang [14] improved Theorem 1.2 by the idea of shared values and obtained the following result. 4n Theorem 1.4. Let F be a family of meromorphic func- tions in D, n be a positive integer and a, b be two con- stant such that , 4n0,aandb n . If, for each pair of functions f and g in F, f af and n g ag share the value b, then F is normal in D. In this paper, we shall discuss a condition on which F still is normal in D for the case and obtain the following result. 2n3 Theorem 1.5. Let F be a family of meromorphic func- tions in D, all of whose poles have multiplicity 2 at least, and a, b be two constant such that . 0,aandb *Supported by China Industrial Technology Development Program (B3120110001). J. H. WANG ET AL. 211 If, for each pair of functions f and g in F, 3 f af and 3 g ag share the value b in D, then F is normal in D. We denote # 2 1 f z fz f z for the spherical de- rivatives of f z.The following example imply that the restriction of poles in Theorem 1.5 is necessary. Example 1. [14] Let :1Dzz and n F f, where 1,,1,2, (1 ) n fz zDn nz n Then for each pair m, n, 3 mm f f and 3 nn f f z share the value 0 in D. But F is not normal at 0 since # n1fn. But we also have the following examples which imply that on the same as restriction of poles in Theorem 1.5 F is not normal in D if for each pair of functions f and g in F, 2 f af and 2 g ag share the value b on D. Example 2. [3] Let 2 1 n fznz zn for 1, 2,n , and :1zz Clearly, 4 210 nn fz fnzn , and n f z only a double pole and a simple zero. Since #0 n fn , as from Marty’s criterion we n have that n f z is not normal in In fact, in the- present paper we also obtain two results as follows. Theorem 1.6. Let F be a family of meromorphic func- tions in D, all of whose poles have multiplicity 4 at least, all of whose zeros have multiplicity 2 at least, and a, b be two constant such that . If, for each pair of functions f and g in F, 0,aandb 2 f af and 2 g ag share the value b in D, then F is normal in D. Theorem 1.7. Let F be a family of meromorphic func- tions in D, all of whose poles have multiplicity at least 4 , all of whose zeros have multiplicity at least 2. Suppose that there exist two functions not idendtically equal to zero, analytic in D, such that for each pair of functions f and g in F, az dz 2 f az f and 2 g az 0az g share the function in D. If az has only a multiple zeros and whenever dz fz then F is normal in D. The following example shows that the condition fzwhen in Theorem 1.7 is necessary. 0az Example 3. [7] Let :1Dzz and n F f where 4 1,,1,2, n fz zDn nz . We take 3 4az z and 0dz . Clearly, F fails to be nor- mal at 0z However, all poles of n f zare of multi- plicity 4, and for each pair m, n, 2 mm f az f and 2 nn f az f share analytic functions in dz . 2. Lemmas To prove the above theorems, we need some lemma as follows: Lemma 2.1. ([1,2]) Let f zbe a meromorphic function in C, n be a positive integer and b be a non-zero constant. If n f fb , then f is a constant. Moreover if f is a transcendental meromorphic function, then n f fz assumes every finite non-zero value finitely often. Lemma 2.2. ([1]) Let f z be a transcendental me- romorphic function with finite order in C. If f z has only multiple zeros, then it’s first derivative f assumes every finite value except possibly zero infinitely often. Lemma 2.3. ([12]) Let f z be a non-polynomials rational function in C. If f z has only zeros of multi- plicity 2 at least, then 2 cz d faz b where a, b, c, d are four constants, 0, 0ac . Lemma 2.4. ([4]) If f zbe a transcendental mero- morphic function in C, then either f z assumes every finite value infinitely often or every derivative()l f as- sumes every finite value except possibly zero infinitely often. If f z is a non-constant rational function and f za , a is a finite value, then ()l f assumes every finite value except possibly zero at least once. Lemma 2.5. ([11]) Let f z k be a transcendental meromorphic function with finite order, all of whose ze- roes are of multiplicity at least , and let 1 Pz be a polynomial, Pz is not idendtically equal to zero. Then ()k f zPz has infinitely many zeros often. Lemma 2.6. ([6]) Let f z be a non-polynomial ra- tional functions in C, all of whose zeroes are of multi- plicity at least 4. Then r f z z has a zeros at least often. Lemma 2.7. ([13]) Let F be a family of meromorphic functions on the unit disc , all of whose zeroes have multiplicity p at least, all of whose poles have multiplic- ity q at least. Let be a real number satisfying pq . Then F is not normal at a point 0 z if and only if there exist 1) points n z , ; 0n zz 2) functions n f F ; and 3) positive numbers 0 n such that Copyright © 2011 SciRes. APM 212 J. H. WANG ET AL. nn nnn fzg g spherically uniformly on each compact subset of C, where g is a non-constant meromorphic function satisfying the zeros of g are of multiplicities p at least and the poles of g are of multiplicities q at least. Moreover, the order of g is not greater than 2. 3. Proofs of Theorem 1.5.-1.7. 3.1. Proof of Theorem 1.5. Suppose that there exists one point 0 such that F is not normal at point 0. Without loss of generality we assume that . By Lemma 2.7, there exist points, ,0n, functions zD z 00z z n z zn f F and positive numbers 0 n such that 1 1n jjjjj gfzg (3.1) spherically uniformly on each compact subset of C, where g is a non-constant meromorphic function with order , all of whose poles are of multiplicities k at least. 2 From (3.1) we have 1 1 n n n jjjj jjj n nn n jjj fzaf zb g agb gag (3.2) By the same method as [14], from Lemma 2.1 it is not difficult to find that n g ag has just a unique zero 0 . Set 1g again, if then 3n 2nn gaga n thus 2n an has just a unique zero 0 . Thus 0 is a multiple pole of or else a zero of . 2 na If 0 is a multiple pole of , since 2nn a has only one zero 0 , then . By Lemma 2.1 again, 20 na is a constant which contradicts with g is not any constant. So we have that has no multiple poles and a have only a unique zero. By Lemma 2.1, and Lemma 2.4, we have is not transcendental. If is non-constant polynomial, then 2 0 l naA . Since all zeros of are of multiplicity 2, then . 3l Denoting for 1n1n , 11 nn , we have 0 l A a and 1 0 l Al . Since all zeros of are of multiplicity, then 214n 0 0, . If 00 , then 0 0 which contradicts with 0a 0 . So is a constant. Next we prove that there exists no rational functions such as . Noting that 11 nn and has no multiple pole, we may set 12 1 m mm s n n t 1, 1 12 1 12 , () s 1 n A (3.3) where A is a non-zero constant, s t,12 ,,, s mm m 2,,)js are s positive integers,j,(1 . For a convenience of stating, we denote 21 , , mn 12 s mm mm (3.4) then 21mns . From (3.3), we have 11 1 1 m n A 1 s m s n t h 1 1 , p q (3.5) where 1hmtn 12st st a 20 st a 1 11 m 1 s m s ph 1 , n t a 11 n q 0 (3.6) are three polynomials. Since has only a unique zero then there exists a non-zero constant B such that 0 12 l nn t , n B a (3.7) so 1 02 11 n nn t Bp 1 12 , l (3.8) where plnt 1tt b 21t0 b is a poly- nomial. From (3.5) we also have 1 m A 2 13 1 s m s n t p 2 1 1 n (3.9) where 3 p is a polynomial also. We denote degp for the degree of a polynomial p , from (3.5) and (3.6) we may obtain deg deg hst pm 11 deg qn 1 1, t t (3.10) Copyright © 2011 SciRes. APM J. H. WANG ET AL. 213 t From (3.8), (3.9) and (3.10) we may obtain 2 deg ,p (3.11) 3 deg2 22.pts (3.12) Since has only a unique zero a 0 and 21(1,2,,), j mjs then 0j (1 rom (3.8), (3.9) and (3.11) it follows ,2,,j. )s F that then , (3.13) Since , then w 3 deg 1pl 2 2degmsp t 21 j mn 2 21mns, so by (3.13) e have s t. If lnmt, fro (3.8), (3.9) and (3.12), we have Then, . Combining with above inequality 3 11deg 222ntlpts 21ts 2 s t, wout a contradiction. nt, then from (3.5) and (3.7) w e bring ab If e have that is . If , then 1 this is impossible. Thus, l deg degpq 11 degmsh nt 2s t . So 1mtn degh 1 deg deg 22 mtnsnth tn st h stst 1mtn re, and deg 1hst.Therefo 11 . Then m 2s t t n e have m21s , this contradicts to 2 . Again from (3.8) and (3.9), wt s t. This comple ptes theroof of Theorem 1.5. .2. Proof of Theorem 1.6. or any points , Without loss of generality, we 3 F0 zD ose thatset 00z. Supp F is not normal at 00z , then by L 2.7, we have that there exist a uence n emma subseq f F, points sequence 0 zD , and a positive num- bers n , 0 n , such that 1,z nnnnn gf g (3.14) spherically uniformly on each compact subset of C, where g is a non-constant meromorphic function with or, all of whose poles are ofmultiplicities at least 2, all hose zeros are of multiplicities at least 4. From (3.14) we have der 2 of w 2 22 nn n 1 g a gad gg (3.15) If , then 0ga 0 g ac , this contra- dito which allfcts zeros o g have multiplicity at least 4. If for any point C , 0a , then By Lemma 2.2, we have that g g is not transcendental in C, so g is non-constantal function in C. By Lemmwe also have that ration a 2.3 3 d ab c g a contradictions. Theref ore, 2 gag have a zeros. We may claim that 2 gag h as a uni- que zero 0 . Otherwis* e, suppose that 00 , are two distinguis of h zeros 2 gag sitive nu then there exists a pomber0 such that * 00 ,,NN .On the other han by Hur- n find two point sequences d, witzrem we c’s Theoa 0, nN , ** 0, nN Such that ** 00 , nn , and nn 22 0 nn n ggad * mm 2*2 0 mmm ggad then, we have 2 n n af z 0 nnnnn d n fz , *2 m m af z *0 mmmmm d m fz ev . Fr ypothesis thery pair functions om the hat for f, g in F, 2 fz af and 2 g zag share comp mber dave lex nu in D, we h z af 2 n n z 0 nnnnn d m f , *2 m n af z *0 n nnnn fz d m . Fix , the0m, let n n 2 00 mm fafd. Since 2 mm f zafzd n w 0, m has no accumulation points, so for sufficiently large have * mm zz e 0 nnn then * , nn nn nn zz Thiradicts to s cont * 00 ,,NN . Thus, 2 gag has zero0 a unique . Further- either 0 move thatre, we ha is a me poles of ultipl g or 0 is a unzero of ique g a . If 0 is a le poles of gmultip , then 0ga , C for any . By Lemma 2 Lemm mediately deduce that .2 anda 2.3, we im- g must be a constant in C, Copyright © 2011 SciRes. APM 214 J. H. WANG ET AL. which contradicts to g is a non-constant mero- morphic functions in C. erefore, g Th has only a simple poles and g a has a un0 ique . But since g has otiple poles, so we that nly a mule hav g is entire in C and g a has a unique 0 . Also by Lemma 2.2, hat we have t g is a nstant polynomials, all of whose zeros mul- tiplicity at least 4. Setting 1 non-coare of 12 2 s m , 2 s m m gA m s 12 11mm we have g 1 A h 12 12ss a 02s a ,A Where , hm 2 0, 01 ,, s aa t ,s are a 1, 2, are some complex constan s, j mj s positive integers, 4 j m, and 1 js j j . m mThus, we have gaB l 0 , ga where 3l. So we have that 0 Bl 1l g 0 0g . If 0g, then 00 g 0 ctions. . Bu 0 t 0 efore, F ga is norm , a contr a adi Ther l at 0z . .3. or a 3 Proof of ny he Theorem az .7 1.7. 0, by the FzD ore , ifplete we may give the com proof of Tm 1 same argument as Theorem 1.6, we emit the detail. In the sequel, we shall prove that F is normal at which 0az . Set r az zbz, where bz is analytic 0, 01b positive integer, 2r. at z , r is a 1 1 :, r F FF tion zzf fz F z uncFor every f F z in 1 F , from the hypothesis in Theorem 1.7, we ca that all zeros of n see F z are of order at least 4, all poles of F z are of micity at least 2. Suppose ultipl that 1 F ex is not normal at z n 0, then by Lemma 2.7, there ists a subsequence 1 F F, a point sequence ,1 nn zz r,and a positive nequence n umber s , n 1z 0 , such th nn at n n r nn 11 n zf n nn z n gF g on com act subsets of rmly p rically uni (3.16) C, wsphe fohere g is a non-constant meromorphic function on C, whose zeros are of multiplicity at least 4, and all of whose poles are multiple. Moreover, g all of has an or- der at most 2. Now we distinguish two cases: Case 1. nn z . Without loss of a generalization, w exe assume that thereists a point zsuch that ,1 n zzzr , we have 1 1 () 11 rr n nn r g 22 2 2 nnn n nn n nn nn nn r n nn n fz g zg zg gz r g z note 1 S (3.17) For the sake of convenience, we defor of the set all zeros of g ,2 S for the set of all zeros of g , and 3 S for t sof all poles of gheet . Since 22 nn gg lim ngg , 11 lim ng gn , and uni- formly on compact subsets of 1 \CS lim 0 nnn r z ly on coms of C, thus uniformpact subset lim nn n fz n , uniformlympact sub- on co sets of 123 \CS SS. Thus, it is not difficult to see that 2 2 2 1 nnn nnn nnnnn nn nnnnn zazfz azfzdz dz azfzd g bz nn nn nn z f (3.18) uniformly on compact subsets of. If \CS 123 S S 10 g bz , then g bz , for any 123 \CS SS . Thus, g bz for any C . By Lemma 2.5, we can see that g is not ndental in C, but is a rational functioo from Lemma 2.3, we deduce that g transce n. Als is constant, which contradicts to the fact that g is non-constant. On the other hand, it is easy to see that g is not iden- tically equal to bz . Hence, g b e as the z has one zeros at least in Crguments in Theorem 1.5 and Theorem 1.6, we deduce that . In fact, by the sam a g has a unique zero 0 . By Lemma 2.5, we ca that n see g is not trdental in C, so g anscen is non- consttional function in C. For a non-cnt poly-ant raonsta Copyright © 2011 SciRes. APM J. H. WANG ET AL. 215 nomials g , and noting that g has only a zero with multiplicity at least 4, we hav e 0 l,3 g bz Bl Thus, 1 0 l gBl . Hence, g has a zero 0 at most. If 0 is a zero of g , then g 0 00 0gg . But 00gbz , In the sequel, a contradiction. we denote r the degree of a po p fo deg lynomial p . If g i s n on polynomials ra- tional functions, then we set 11 m 2 12 12 12 , s t m s n nn t (3.19) Wher t. t (3.20) Then, m 2 j n, st jk qn gA e j m 4,1,2,,js; 1, 2,,j 11 4, 2 jk mms 1 1 1 1 A 1 1 1 1 s t m m s n n t h g1 1 p q (3.21) where 2 12 0 st st s t q a , deg hs hm pA a 1 t 11 2 12 1 11 s m mm sh 1 11 t n t q 12 11 2 nn Since g bz has a uno ique zer0 , so we set 1 0 1 l nt. Then 1 1 () t n n t B z (3.22) where Bnonzero constafrom (3.22), we gb is a have 1 1 02 n t Bp 1tt b 2 2 1 t l n (3.23) where 1 bis a poly- hat g 2 deg 20 t plqt nomial, pt. follow tFrom (3.21), it 11 2 m A 1 13 2 2 1 n p g 2 s t m s n t 3 222 223 02 1 st st 23 s t pmqmq cc is also a polynomial, 3 deg22 2pst. cases to derivative a cWe distinguish fiveontradiction: Subcase 1.1. mq . Then from (3.21), we have lqt . So, 222 deg, 1pti it , 00 deg1, 11hst hhst and 333 deg22 2,122 2pstitst From (3.23) and (3.24), we have. S also fro 23 1ii e o m (3.23) and (3.24), we also hav 3 deg . Thus, we have 3 22122lsti st 1 i lp 2 Since lq . t t and2qt, then we have 2 2si . On the ot, from and (3.24), we her hand (3.23) also have 2 2degms p . Since 4ms, we have 2 2 s ti . . Subcase 1.2. m This is impossible 1q . Then lqt, 22 deg pti , 2 it1 , deg 1hst and 333 deg222, 1222psti tst Similarly to Subcase (1.1), from (3.23) and (3.24), we also have that 23 1ii . Also from (3(3 3 1deglp , .23) and .24), we have then, we have 2 21ts i On the ot- larly to the argucase (1.1), from (3.23) and (3.24), we also have her hand, simi ment of Sub 22 2degmsp ti , then 2 21 s ti . This also is 3. 2mq impossible. Subcase 1. . Then we still have 222 1lp tst 3 ,deg,1,degqttii th and , 3 deg22 2pst . Therefor, 222lst, so 2s 2t . Similarly, we 2 2sti, t 2 2 have 2mshen s ti . . This is a contradiction e 1.4. 1mq Subcas . Thenlqt , deg 1hst , 32, deg22tps and 22 deg pt2 ,i i0t . 2.ti From (3.23) and (3.24), we have 2ms Thus,2 2 s tiand 2 21 .ts i This i Subcase 1.5. m s impossible. . Then , 2qlqt (3.24) where deg 1hs t , 3 deg 2pst22 d 2 gpt, ande . From (3.23) and (3.24), we have 3 1deg22 1lpst and 2 2degp t . ms 21ts So, we have that 2 and s t. This is a con- tradiction. Case 2. Suppose that there exists a complex number C and a subsequence of sequence 1 nn z , still noting it 1 znn , such that 1 znn . Wea con- have Copyright © 2011 SciRes. APM 216 J. H. WANG ET AL. verges 11 ˆ nnnnnnnnnn F Fzz gg H (3.25) spherically uniform on compact subsets of C. Clearly, all zeros of ˆ g are of multiplicity at least 4, all poles of ˆ g arultiplicity at least 2. For each 00e of m , it to see that there exists a neighborhood ), is easy(0 N of 0 , such that ˆ rr n Hg , the con bei spherically , vergence ng uniform on 0 N . For 00 , since 0 is the pole of g , then there exists 0 , such tt ha ˆ 1g is an on alytic 2:D 2 , 1Hn aytic on re anal :2D 2 for suffi ciently la - rge n. Since 1H r nnnn f then 00 is a zero of ˆ 1g has order at least r, we cace that n dedu 1r n H converges uniform to ly ˆ 1rg on 2:2D Hence, we have 1 1ˆ rr nn r nnn GH fg (3.26) spherically uniform on compact subsets of C. It follows that 00G from f whenever 0a for D all of ze , hence ros of G have orst 4, oles of der at lea all of p G have o least 2. Noting that rder at 22rr Gb Gd 22 2 nn nnn r nnn nnnn r fa f d GG (3.27) If , then 20 r GG r G 0, so r G , 1 0 1 r GC r for any C . Since 00G , then. Also 00C since G ha 0, thi s the zeroiplicity at least 4, then G s is a contradiction. Therefore, 2r GG s of mult is not identically equal to zero. If 2r GG for any C , then G has no multiple poles and. Note that 0G r G les, so G has only multiple po at G i on C e th s entire. Also by Lemma 2.5, we hav is not tran- scendental in C, and then G isynomial. Thus, a pol 0 r GC , where 00C. We ha ve 1r Gr , then from 00G and a multiplicities of every zeros of G it follows that 0G for any C , this Hence, is impossible. 2 GG r has some zeros. In fact, by the sagument as the C may dethat me arase 1, we duce 2r GG ha s a unique zero 0 . Thus, we ha0 ve that either is multi- ple poles of G or 0 is a unique zero of Gr . ilarlySim, if 0 is multiple poles of G , from that 2r GG has a unique zero 0 it atfollows th r G for any C . By Lemma 2.5, we have that G is not transcendentain by al. Ag Lemm G a 2.6, we have that is a constant, is which a contradiction. Hence, G has nle pole and o multip r G has a e zero 0 uniqu . Thus, G is entire on C and r G has a unique zero 0 . By Lemma 2.5, we have that G must be a polyno- mial. Setting 2 2, 1 1 s m m s gA ) where, 12 ,,, m (3.28 s mmm are s posntegers, 4m itive ij 1, 2,,js , 1 s j j mm G 0 l rB l is a positiv ) , .29 (3) ve where e integer, l Bl 3, we ha 1l 1l r Gr 0 , (3.30) 1 2 (3 1 0 1l r GrrBl . (3.31) For 00G , we have 00 and 0 j . From (3.29) it follows that 0j , d (3.31) 1, 2,,js . or 1, From ha (3.29), (3.30)a, fn2,js, , we ve l 0 r jj B jj rBl 1Bll 1 jj (3.32) (3.33) 0 1 1l r r 2 0 l 2 rr , we have (3.34) From (3.32) and (3.33) , j rl 0,1r j 2,,s (3.35) If lr , then 00 , this is im lr possible. Therefore, we have , and so Copyright © 2011 SciRes. APM J. H. WANG ET AL. Copyright © 2011 SciRes. APM 217 [2] H. H. Chen and M. L. Fang, “On the Value Distribution of n f f ,” Science 12 s r rl 0 in China Series A, Vol. 38, No. 7, c Functions,” Indian Journal of Ma- of Mathematics, 1995, pp. 789-798. [3] M. L. Fang and W. J. Yuan, “On the Normality for Fami- lies of Meromorphi From (3.33) and (3.34), we also have , 12 1 s r rl 0 th thematics, Vol. 43, 2001, pp. 341-350. [4] W. K. Hayman, “Picard Values of Meromorphic Func- tions and the Its Derivatives,” Annals en 00 1rr . Thus, we have 00 , a contra- dict Finally, we prove that F is normal at the origin. For any function sequence ion. Vol. 70, 1959, pp. 9-42. doi:10.2307/1969890 [5] W. K. Hayman, “Meromorphic Functions,” Clarendon, Oxford, 1964. n f z in F, since 1 F is nor- at , theositivmal n there exist a pe number 0z[6] X. J. Huang and Y. X. Gu, “Normal Families of Mero- morphic Funct ce k n F of n F suc12 and subsequenh that ions with Multiple Zeros and Poles,” Journal of Mathematical Analysis and Applications, Vol. 295, No. 2, 2004, pp. 611-619. doi:10.1016/j.jmaa.2004.03.041 [7] X. J. Huang and Y. X. Gu, “Norm morphic Functions,” Results in k n F converges uniformly to a meromorphic function hz or on 0, 2N . Noting 0 n F, we de- duce thate exists a postive number 0M such theri that al Families of Mero- Mathematics, Vol. 49, k n F zM for any 0,zN . Again noting 2006, pp. 279-288. doi:10.1007/s00025-006-0224-2 [8] S. Y. Li, “On Normal Criterion of Meromorphic Func- tions,” Journal of Fujian Normal University, Vol that ve t for 0 k n f we hahat k n fz all 0,zN , that is, k n f z is analytic in 0, . 25 hina Series A, Vol. 28, 1985, pp. ence in China Series A, Vol. 33, No. 5, 1990, orphic Functions Omitting a Function ii,” tions with Multiple Ze- erican Mathematical Society, ” Journal of Mathe- N . 1984, pp. 156-158. [9] X. J. Li, “Proof of Hayman’s Conjecture on Normal Fam- ilies,” Science in C There k n, we hav fore, for all e 112 ,2 k k nrr n z z M zF r fz 596-603. [10] X. C. Pang, “On Normal Criterion of Meromorphic Func- tions,” Sci k n f z pp. 521-527. [11] X. C. Pang, D. G. Yang and L. Zalcman, “Normal Fami- lies of Merom z By Montel’s Theorem, is normal at , 0 and thus F is normal at . The complete proof of Theorem 1.7 is given. 4. ulo the referee for a numb elpful suggestions to improve the paper. Singularities of Finite Order,” ática Iberoamericana, Vol. 11, N 373. 0z Computational Methods and Function Theory, Vol. 2, No. 1, 2002, pp. 257-265. [12] Y. F. Wang and M. L. Fang, “Picard Values and Normal Families of Meromorphic Func Acknowledgements The authors are gratef ter of ros,” Acta Mathematica Sinica, Chinese Series, Vol. 41, No. 4, 1998, pp. 743-748. [13] L. Zalcman, “Normal Families: New Perspectives,” Bul- letin (New Series) of the Am h . References 5 [1] W. Bergweiler and A. Eremenko, “On the the Inverse to a Meromorphic Function of Vol. 35, No. 3, 1998, pp. 215-230. [14] Q. C. Zhang, “Normal Families of Meromorphic Func- tions Concerning Sharing Values, matical Analysis and Applications, Vol. 338, No. 1, 2008, pp. 545-551. doi:10.1016/j.jmaa.2007.05.032 Revista Matem 1995, pp. 355- o. 2, |