Journal of Mathematical Analysis and Applications. Two normality criteria and the converse of the Bloch principle

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1 J. Mh. Anl. Appl ) Conens liss vilble ScienceDirec Journl of Mhemicl Anlysis nd Applicions Two normliy crieri nd he converse of he Bloch principle K.S. Chrk, J. Rieppo b,,1 Deprmen of Mhemics, Universiy of Jmmu, Jmmu , Indi b Deprmen of Physics nd Mhemics, Universiy of Joensuu, P.O. Box 111, FIN Joensuu, Finlnd ricle info bsrc Aricle hisory: Received 15 July 008 Avilble online 8 November 008 Submied by J. Xio We prove wo normliy crieri for fmily of meromorphic funcions sisfying cerin differenil condiion nd provide counerexmple o he converse of he Bloch principle. 008 Elsevier Inc. All righs reserved. Keywords: Meromorphic funcion Norml fmily Bloch principle Differenil polynomil Vlue disribuion 1. Inroducion nd resuls More hn hree decdes go, Lwrence Zlcmn [13] proved heurisic lemm chrcerizing norml fmilies of nlyic nd meromorphic funcions on plne domins. Over he yers, he lemm hs coninuously given n impc o wide vriey of opics in funcion heory nd reled res. Wih renewed ineres in norml fmilies of nlyic nd meromorphic funcions in plne domins, minly becuse of heir role in complex dynmics, i hs become quie ineresing o lk bou norml fmilies in heir own righ. Anoher vluble heurisic ool in he sudy of norml fmilies is he Bloch principle sed s follows: Le us look he semens ) If meromorphic funcion sisfies condiion P in he complex plne, hen i mus be consn funcion. b) If fmily of meromorphic funcions sisfies he condiion P in n rbirry complex domin, hen he fmily is norml. In common, wh is undersood o be he Bloch principle is when ) implies b) nd he converse of he Bloch principle is hen he semen when b) implies ). In his ricle we focus on normliy crieri h re conneced o he vlue disribuion of differenil polynomils. For exmple, Theorem A. See [4].) Le n 5 be n ineger,, b C nd 0. If, for meromorphic funcion f, f f n b for ll z C, hen f mus be consn. * Corresponding uhor. E-mil ddresses: kschrk7@rediffmil.com K.S. Chrk), jrkko.rieppo@jns.fi J. Rieppo). 1 The uhor hs been prilly suppored by he Acdemy of Finlnd grn X/$ see fron mer 008 Elsevier Inc. All righs reserved. doi: /j.jm

2 44 K.S. Chrk, J. Rieppo / J. Mh. Anl. Appl ) Theorem B. See [8,9].) Le n 3 be n ineger,, b C, 0 nd F be fmily of meromorphic funcions in domin D. If f f n b for ll f F,henF is norml fmily. E. Mues [7] hs given exmples showing h he clim of Theorem A is no vlid when n = 3ndn = 4. Th mens h he converse of he Bloch principle is no rue in hese cses. Finding pproprie generl condiions for he propery P under which ) nd b) re equivlen or generlizing he Bloch principle in some oher wy is n ineresing problem, see e.g. []. From h poin of view, i is imporn o look for vriey of exmples concerning he Bloch principle. In he lierure here re no oo mny counerexmples bou he converse of he Bloch principle he recen one is by Bo Qin Li [6]), nd hence we give he one in his ricle. Recenly I. Lhiri proved he following crierion for he normliy by using Zlcmn s lemm. Theorem C. See [5].) Le F be fmily of meromorphic funcions in complex domin D. Le, b C such h 0. Define { E f = z D: f z) } f z) = b. If here exiss posiive consn M such h f z) M for ll f F whenever z E f,henf is norml fmily. Noe h in Theorem C, when he se E f is empy for every f in fmily, hen he fmily is norml. Lhiri gve counerexmple o he converse of he Bloch principle by using his fc. In his pper, we prove wo normliy crieri of Lhiri s ype nd using one of hem we provide noher counerexmple o he generl converse of he Bloch principle. Theorem 1. Le F be fmily of meromorphic funcions in complex domin D. Le, b C such h 0. Lem 1,m,n 1,n be nonnegive inegers such h m 1 n m n 1 > 0,m 1 m 1,n 1 n, ndpu { E f = z D: f z) ) n 1 f z) ) } m 1 f z)) n f = z)) m b. If here exiss posiive consn M such h f z) M for ll f F whenever z E f,henf is norml fmily. To consider he cse m 1 n = m n 1, we need vlue disribuion resul for corresponding differenil expression. Theorem. Le, b C such h 0 nd le f be nonconsn meromorphic funcion. If n 1,n,m 1,m re posiive inegers such h m 1 n = m n 1, hen funcion Ψz) defined by Ψz) := f z) ) n 1 f z) ) m 1 f z)) n f z)) m b 1) hs finie zero. Theorem 3. Le F be fmily of meromorphic funcions in complex domin D. Le, b C such h 0. Lem 1,m,n 1,n be posiive inegers such h m 1 n = m n 1,ndpu { E f = z D: f z) ) n 1 f z) ) } m 1 f z)) n f = z)) m b. If here exiss posiive consn M such h f z) M for ll f F whenever z E f,henf is norml fmily.. Proofs of he resuls.1. Auxiliry resuls For he proof of Theorem 1 we need he following lemms: Lemm 4. [1] Tke nonnegive inegers n, n 1,..., n k wih n 1, n 1 n n k 1 nd define d = n n 1 n n k. Le f be rnscendenl meromorphic funcion wih he deficiency δ0, f )> 3. Then for ny nonzero vlue c, he funcion 3d1 f n f ) n 1...f k) ) nk c hs infiniely mny zeros. I is noed in [1] h for n he deficien condiion in Lemm 4 cn be omied. In our pplicion of his resul we consider n, n 1 1 nd n,...,n k re ll equl o 0. The version of he Zlcmn lemm given here is due o Png [10].

3 K.S. Chrk, J. Rieppo / J. Mh. Anl. Appl ) Lemm 5 Zlcmn Png lemm). Le F be fmily of meromorphic funcions in domin D. Le α R :1 < α < 1.ThenF is no norml in neighbourhood of z 0 Diffhereexis poins z k D : z k z 0 s k ; posiive rel numbers ρ k : ρ k 0 s k ; nd funcions f k F such h ρ k α fk z k ρ k ζ) gζ ) sphericlly uniformly on compc subses of C, where g is nonconsn meromorphic funcion. Lemm 6. Le f be nonconsn rionl funcion nd m,n nurl numbers. Then, he funcion f n f ) m kes every finie nonzero vlue. Proof. Denoe f z) = Az)/Bz), where A nd B re complex polynomils. Consider quniy defined by d f ) := deg A deg B. Firs we show h d f n f ) m ) 0. To do h, suppose firs d f ) = 0. Then, i is esy o see h dega B AB )< deg A deg B 1 nd hus d f n f ) m) = nd f ) md f ) = m dega B AB ) deg B ) < mdeg A deg B 1 degb) = md f ) m =m. Assume nex d f ) 0. Then, dega B AB ) = deg A deg B 1 nd since d f ) is n ineger we obin d f n f ) m) = nd f ) md f ) = nd f ) mdeg A deg B 1) = m n)d f ) m 0. Now, if g := f n f ) m voids nonzero complex vlue, hen1/g ) is polynomil, nd hus g ) =. Buhis conrdics he fc dg) 0, h ws shown bove... Proof of Theorem 1 Proof. Suppose on he conrry h F is no norml z 0 D. Then by Lemm 5, for α :1 < α < 1, here exiss sequence of complex numbers {z } in D such h z z 0, sequence of posiive numbers {ρ } such h ρ 0, nd sequence of funcions { f } in F such h g w) := ρ α f z ρ w) gw) ) sphericlly uniformly on compc subses of C s, where g is nonconsn meromorphic funcion in C. Tob- brevie he noions below, we shll denoe f z ρ w) jus by f nd k j := n j m j,for j = 1,.NowbyLemm4in combinion wih Lemm 6, here exiss w 0 C such h gw0 ) ) n 1 g w 0 ) ) m 1 gw 0 )) n g = 0. w 0 ) m 3) Since gw 0 ) 0,, g w) gw) uniformly in closed disk Δw 0 ; δ) for some δ>0. Now in Δw 0 ; δ), we hve g w) ) n 1 g w) ) m 1 g w)) n g w) ) αk m ρ m b αk = ρ 1 m 1 f ) n 1 f αk ρ αk m f ) n f ρ m ) m b αk = ρ m ρ αk 1k )m m 1 f ) n 1 f f ) n f ) m b Tking α = m 1m nd using he ssumpion m k 1 k 1 n n 1 m > 0, we see h g n 1 g ) m 1 g n g ) m is he uniform limi of m 1 n n 1 m ) k ρ 1 k f ) n 1 f f ) n f ) m b in Δw 0 ; δ). In view of 3) we pply Hurwiz s heorem o ge sequence {w } such h {w } converges o w 0 nd for ll lrge nd ζ := z ρ w, f ζ ) ) n 1 f ζ ) ) m 1 f ζ )) n f ζ )) m = b. 6) ). 4) 5)

4 46 K.S. Chrk, J. Rieppo / J. Mh. Anl. Appl ) Hence, for ll lrge, ζ := z ρ w E f h g w ) m 1 m Mρ k 1 k. nd hus here exiss n M > 0 such h f z ρ w ) M. This furher implies 7) Since g is holomorphic w 0, gw) C for some posiive consn C, ndforllw Δw 0 ;η) for some η > 0. Agin, since g g, forllɛ > 0 here exiss some 0 such h for ll w Δw 0 ;η), wehve g w) gw) < ɛ for ll 0, nd now by using 7), we find h C gw ) g w ) g w ) gw ) Mρ m 1 m k 1 k for ll 0. Th is ɛ C Mρ m 1 m k 1 k ɛ for ll 0, which is no in reson since ρ 0s..3. Proof of Theorem Proof. The lgebric complex equion x x n b = 0 /n 1 hs lwys nonzero soluion, sy x 0 C. By [3, Theorem ], [1, Corollry 3], Lemm 4 nd Lemm 6, he meromorphic funcion f n 1 f ) m 1 cnno void i nd hus here exiss z 0 C such h f z 0 )) n 1 f z 0 )) m 1 = x 0. By ssumpion, we my wrie m = n /n 1 )m 1 nd n = n /n 1 )n 1.Consequenly, Ψz 0 ) = [ f z 0 ) ) n 1 f z 0 ) ) m 1 ] nd we re done..4. The proof of Theorem 3 [ f z 0 )) n 1 f z 0 )) m 1 ] n /n 1 b = 0 Proof. Suppose on he conrry h F is no norml z 0 D. Then by Lemm 5, for α :1 < α < 1, here exiss sequence of complex numbers {z } in D such h z z 0, sequence of posiive numbers {ρ } such h ρ 0 s, nd sequence of funcions { f } in F such h g w) := ρ α f z ρ w) gw) 8) sphericlly uniformly on compc subses of C s, where g is nonconsn meromorphic funcion in C. By Theorem, here exiss w 0 C such h gw0 ) ) n 1 g w 0 ) ) m 1 gw 0 )) n g w 0 )) m b = 0. 9) Since gw 0 ) 0,, g w) gw) uniformly in closed disk Δw 0 ; δ) for some δ>0. Tking α = m 1m,wehve k 1 k g w) ) n 1 g w) ) m 1 g w)) n g w)) m b = f ) n 1 f f ) n f ) m b in Δw 0 ; δ). Thus g n 1 g ) m 1 g n g ) m b is he uniform limi of f ) n 1 f f ) n f ) m b in Δw 0 ; δ). Since we hve Eq. 9), we my pply Hurwiz s heorem o ge sequence {w } such h {w } converges o w 0 nd for ll lrge f z ρ w ) ) n 1 f z ρ w ) ) m 1 f z ρ w )) n f z ρ w )) m = b. From now on he proof will coninue word by word similrly s he proof of Theorem 1 fer Eq. 6).

5 K.S. Chrk, J. Rieppo / J. Mh. Anl. Appl ) Discussion Considering he converse of he Bloch principle nd in he view of Theorems 1, nd 3, one is emped o sk: Quesion 7. For wh vlues of m 1,n 1,m,n here exiss nonconsn meromorphic funcion f such h Ψz) := f n 1 f ) m 1 f n f ) m b hs no zeros in C for some, b C, 0? In priculr, one my pose: Quesion 8. Does here exis nonconsn meromorphic funcion f such h Ψz) := f n f ) m f b hs no zeros in C for some n 1, m 1 nd for some 0, b in C? As noher counerexmple o he converse of he Bloch principle, we give Exmple 9. Le f z) = n z, hen f z) = 1 n z 0forllz C. Now we see h Ψz) = f z) 1 f z)) 1 = n z 1) 0, n z bu Theorem 1 is rue especilly when E f is n empy se for every f in he fmily. To consider he quesions bove, le Ry 0, y 1,...,y m ), m N {0}, be rionl funcion in vribles y 0, y 1,...,y m wih consn coefficiens. Le D be domin in he complex plne conining he origin. If ll meromorphic funcions f in D sisfying R f, f,..., f m)) 0 10) form norml fmily, hen, by Mry s crierion, ny funcion g meromorphic in he whole plne such h R g, g,...,g m)) 0 11) mus sisfy ρg), where ρg) is he order of growh of he meromorphic funcion g. Indeed, ssume conrry o he sserion h ρ := ρg)>. Then here exiss sequence z k ) such h z k nd g # z k ) z k ρ/1 s k. By Mry s crierion, he fmily {g k }, where g k z) = gz k z), is no norml he origin. Bu since ech g k sisfies R g k, g ) k,...,gm) k 0, 1) we hve conrdicion wih he ssumpions. We noe h we my replce he ideniies 10), 11) nd 1) by R f, f,..., f m) ) 0, Rg, g,...,g m) ) 0 nd Rg k, g k,...,gm ) 0forllz C nd he conclusion remins he sme. k In he ligh of he preceding discussion, f in Quesion 7 s well s in Quesion 8) sisfies ρ f ). If Ψz) in Quesion 8 hs no zeros, hen we mus hve n = orn = 1 ccording o Sz 3 in [11]. Acknowledgmens The uhors would like o hnk he nonymous reviewer for suggesing necessry correcions o our mnuscrip.

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