一类正规定则改进英文

一类正规定则旳改善_英文_SomeImprovementtoOneNormalCriteriaZhangJie,XuYan()SchoolofMathematicsandComputerScience,NanjingNormalUniversity,Nanjing210097,ChinaAbstract:Thispaperobtainsomenormalityoffamiliesofmeromorphicfunctionsallowingthefunctionstohavezeroesbuttogiveadditionalconditions,whichgeneralizeandimprovesomeresultsofLinWeichuanandXuYan’swork.We122()())()()())()()(()(obtain:iffzf″z-af′z?0a?1,1?andfzf″z-af′z=0impliesf′z=0thenfhasn?n(αβ)(αβ)(α)()()forms:fz=expz+orfz=z+?0.AndFbeafamilyofmeromorphicfunctionsindomainD,ifeachf()k()()(())()()?Fhasonlyzeroesofmultiplicityatleastk?3andsatisfies:fz=azaz?0,implies|fz|?Aandfz()k()=0implies0<|fz|?K.ThenFisnormalinD.HereA,Karepositiveconstants.Keywords:transcendentalmeromorphicfunction,normalfamily,reside()CLCnumberO174.52DocumentcodeAArticleID044205一类正规定则旳改善张杰,徐焱()南京师范大学数学与计算机科学学院,江苏南京210097[摘要]采用改函数不取零值为可取零值加限制旳措施改善了林伟川,徐焱等人旳成果.得到:122()(())()()(())()()()若fzf″z-af′z?0a?1,1?及fzf″z-af′z=0蕴含f′z=0,则f有n?n(αβ)αβ)(α)()()(形式fz=expz+或fz=z+?0.F是区域D上旳亚纯族,若每个f?F()k)()()(())(()()旳零点重数至少是kk?3并满足:fz=azaz?0蕴含|fz|?A和fz=0蕴含()k()0<|fz|?K.则F在区域D上正规.其中A,K为正常数.[关键词]超越亚纯函数,正规族,留数0IntroductionIn1995,Bergweiler[1]obtainedthefollowingresult:1()()TheoremALetfbeatranscendentalmeromorphicfunctionoffiniteorder,anda?1,1+.iffzf″zn2(())()(αβ)αβ-af′z?0,thenfz=expz+,where,?C.Recently,LinWeiChuan[2]andYiHongXunexcludedtheadditionalorderrestrictionasfollows:1TheoremBLetfbeameromorphicfunctioninthecomplexplaneandleta?1,1?,wheren?N,n2()()(())andiffzf″z-af′z?0,thenfhasoneofthefollowingforms:)(αβ)()ifz=expz+(αβ))()iifz=z+Receiveddate:203230.()Foundationitem:SupportedbytheNationalNaturalScienceFoundationofChina10671093.Biography:ZhangJie,bornin1981,master,majoredincomplexanalysis.E2mail:zhangjielaoguai@163.comCorrespondingauthor:XuYan,bornin1964,doctor,associateprofessor,majoredincomplexanalysis.E2mail:Yanxu_njnu@1)iiif=n(αβ)z+αβwheren?N,?0,?C.MoreovertheyobtainanormalfamilyanalogueofTheoremB:1ΔTheoremCLetFbeafamilyofmeromorphicfunctionsintheunitdiskanda?1,1?,wheren?nf′2Δ)()())Δ((N,ifforeveryf?F,andfzf″z-af′z?0in,thenf:?Fisnormalin.f?n(αβ)))wenotethattheresultiiandiiiinTheoremBmaybecombinedbyz+ifweweakenthecondi22()(())()tionfzf″z-af′z?0andsowehaveTheorem2.ToobtainTheorem2,weneedthefollowingtheoremwhichgeneralizesTheoremCaccordingly:1ΔTheorem1LetFbeafamilyofmeromorphicfunctionsintheunitdiskanda?1,1?,wheren?n22()()())()())()(()(N,ifforeveryf?F,andfzf″z-af′z?0andfzf″z-af′z=0impliesf′z=0,thenf′Δ:f?Fisnormalin.fAndbasedonthisTheorem,theTheorem2canbeobtainedasfollows:1Theorem2Letfbeameromorphicfunctioninthecomplexplaneandleta?1,1?,wheren?N,n22()()(())()()(())()fzf″z-af′z?0andfzf″z-af′z=0impliesf′z=0,thenfhasoneofthefollowingforms:)(αβ)()1fz=expz+?n()(αβ)αβ)2fz=z+,wheren?N,?0,?C.In1979,Gu[3]provedaconjectureofHaymanasfollows:TheoremDLetFbeafamilyofmeromorphicfunctionsindomainD,kbeapositiveinteger,ifforever2()kyf?F,f?0,andf?1,thenFisnormalinD.Recently,Xu[4]improvedandgeneralizeditandobtained:TheoremELetkbepositiveintegersuchthatk?3andKbepositivenumberFbeafamilyofmeromor2p()hicfunctionsindomainDandazbenon2vanishinganalyticfunctioninDsupposethatforeveryf?F,andfhasonlyzeroesofmultiplicityatleastkandsatisfiesfollowingcondition:()()kk)()()()())faz?az.bfz=0implies0<|fz|?K.thenFisnormalinD.()()kk()()()()Theconditionfz?azalsomaybegeneralizedbyallowingfz=azatsomedotsbutrestrictthevaluesoffatthesedots.Andwehave:Theorem3Letkbepositiveintegersuchthatk?3andA,Kbepositivenumber,Fbeafamilyofmero2()morphicfunctionsindomainDandazbenon-vanishinganalyticfunctioninDsupposethatforeveryf?F,fhasonlyzeroesofmultiplicityatleastkandsatisfiesfollowingcondition:()k())()()az=azimplies|fz|?A.f()k)()()bfz=0implies0<|fz|?K.thenFisnormalinD.1SomeLemmasεLemma1[5]LetA,Bandbepositivenumbers.LetF={f}beafamilyofmeromorphicfunctionsindomainDwhichsatisifythefollowingcondition:)()1f′z?1)2()()iffz=0,theno<|f′z|?Bm-1Δ()Δε)ifisadiskinDandiffhasm?2zerosz,z?,z?,then|?f′z-1|?.312mjj=1—15—ThenFisnormalinD()-1pznn()fz=az+az+?+a+,wherea,a?,aareconstantswitha?0.LetLemma2[6]nn-1001nn)(qz()k()()()()pz,qzaretwoco2primepolynomialswithdegpz<degqz,letkbeapositiveinteger.iff?1,then1k1)()(fz=z+?+a+,wherea?0,bareconstants,misapositiveinteger.0mk!()az+bLemma3[2]Letfbeatranscentalmeromorphicfunctionwithfiniteorder.allofthosezerosareofmulti2()k()()()plicityatleastk,andletAbepositiverealnumber.if|fz|?Awheneverfz=0,thenforeachl,1?()l()l?k,fzassumesanyfinitenon2zerovalueinfinitelyoften.1Let{a}beanintegersequenceanda?1,1?,wheren?N,thenexistsapositiveLemma4[2]mnε()εnumbersuchthatforeacha,|aa-1-1|?.mmLemma5[7]LetkbeapositiveintegerandletFbeafamilyoffunctionsmeromorphicinadomainD,suchthateachfunctionf?Fhasonlyzerosofmultiplicityatleastk,andsupposethatthereexistsA?1such()k()()αthat|fz|?Awheneverfz=0.IfFisnotnormalatz?D,then,foreach0??k,thereexistase20ρquenceofpointsz?D,z?z,asequenceofpositivenumbers?0,andasequenceoffunctionsf?Fsuchnn0nn(ρδ)fz+nnn(δ)(δ)?glocallyuniformlywithrespecttothesphericalmetric,wheregisanonconstantthatg=nαρn##(δ)()meromorphicfunctiononC,allofwhosezeroshavemultiplicityatleastk,suchthatg?g0=kA+1.Moreover,ghasorderatmost2.LetfbemeromorphicinCandoffiniteorder,letk?3beapositiveintegerandKbeposi2Lemma6[4]()k()()tivenumber,supposethatfhasonlyzerosofmultiplicityatleastk,|fzKwheneverfz|<=0,and()k()fz?1Thenk)α(β)αβα()1fz=z-,,?C,k!?1k+1()z-c11())=,wherec,caretwodistinctcomplexnumbers.2fz1k!z-c2ProofofTheorems()fz1()ProofofTheorem1Definehz:=?,f?F()1-af′zΔThenweonlyneedtoprovethefamilyH:={h}isnormalin.f?FFromthedefinitionofh,wehave:2()()()af′z-fzf″z()h′z+1.=2(())1-af′zδ)δ)((Atfirst,ifh=0,thenf=0or?.Weconsidertwocases:11δ(δ)Ifisazerooffwithmultiplicityn,thenh′=case1.1.1-an11δ(δ)Ifisapoleoffwithmultiplicitym,thenh′=case1.2.a-1m1(δ)δ)(Hence,0<|h′|?||whenh=0.a-1(δ)Secondlyweclaimh′?1.2(δ)δ)(δ)(af′-ff″(δ)(δ)δ)δ(Ifϖsuchthath′=1,thenh′=0fromcase1.2,wehavef??,-1=2((δ))1-af′2(δ)(δ)(δ)(δ)(δ)(δ)soaf′-ff″=0,Weitimpliesf′=0byconditionandthusff″=0.considertwocases:—16—111(δ)(δ)=1,thusa=1-,thisisacontra2case2.1Iff=0,thenfromcase1.1,wehavef′=1-anndiction.(δ)(δ)(δ)(δ)Iff?0,thenf″=0andh=?thush′=??1,thisisacontradiction.case2.2m-1ΔΔΔ()Thirdly,supposethat<isadiskandhhaszeroesz,z?,z?.Asabovewehave:|h′z112m1j?j=1()ε-1|=|1-aa-1|,hereaisanintegernumber.Bylemma4,thereexistsapositivenumbersuchmmm-1()Δεthatforeachh?H,|h′z-1|?.andHisnormalinbylemma1.j?j=1ThiscompletetheproofofTheorem1.#()ProofofTheorem2AmeromorphicfunctionfiscalledYosidafunctionifitssphericalderivativefz=2()))((|f′z|/1+|fz|isuniformlyboundedonC.()1fz()=Asabovewedefinehz:.()1-af′zWeclaimthathisaYosidafunction.#)()(()Ifnot,thereexistsasequencezsuchthathz??,writehz=hz+zthen{h}isnotnormalatnnnnnz=0bymartycriterion.Howevertheorem1implies{h}isnormalatz,whichisacontradiction.0n0λ()ThushisaYosidafunction,andh?2.AstotheproofofTheorem1,wehavethathisnotatranscentalfunctionbylemma3.Thushisarationalb()βa?1,()()bylemma2,wherefunctionandh′z?1.Thenhhasformhz=az+borhz=z++00l()z+cβ,b,careconstants.1111b()β()(),thenSupposethathz=z++=+O,asz??,andsoRes,?=-1.l2()()hzzz+chz()letz,z?,zbethezeroesofhz,()herem=l+1.Wenotethathzhasonlysimplezeroesandsowe12mhave:m-111()()()|h′z|=Res,z=-Res,?=1.j??hh-1j=1()z?h0()1fzε()Butonotherhandhz=,fromtheproofofTheorem1thereexistsapositivenumbersuch()1-af′zm-1()εThisisacontradiction.zthat|?h′-1|?.jj=1()Thushz=az+b.Weconsidertwocasesasfollows:0()2fz()()(())()(αβ)Ifa=0thenisnon2zeroconstantsincefzf″z-af′z?0.Hencefz=expz+,0()f′zαwith?0.()fzγ()γ=z+c,wenotethatz=-cistheonlyzeroorpoleoff,thenmustbe?1/n,Ifa?0,then0()f′z()1f′z?n(αβ)α?n=thusandthenf=z+with?0.()z+cfzThiscompletetheproofofTheorem2.αProofofTheorem3IfFisnotnormalatz?D,thenbylemma5take=kandthereexistasequence0ρofpointsz?D,z?z,asequenceofpositivenumbers?0,andasequenceoffunctionsf?Fsuchthat:nn0nnk(δ)(ρδ)ρδ)(g=fz+/?glocallyuniformlywithrespecttothesphericalmetric,wheregisanonconstantnnnnn##(δ)()()meromorphicfunctiononC,allofwhosezeroshavemultiplicityatleastk,suchthatg?g0=kK+1+1.Moreover,ghasorderatmost2.δ(δ)δδδLetbeazeroofgandthenbyHurwitz’Theoremthereexistsasequence,?.suchthat1nn1—17—()kk(δ)(ρδ)ρ(ρδ)(ρδ)g=fz+/=0forsufficientlylargen.Thusfz+=0,hence|fz+|?Kbynnnnnnnnnnnnnnn()()()k()kkk)(δ)(ρδ)(δ)(δ)conditionb.Sinceg=fz+?g,wededucethat|g|?K.nnnnnn11()Obviouslyaz?0,?.0Wedistinguishtwocases:()kδ)δ)((δ)(Ifthereexistssuchthatg=az,itisobviousthatg??.case3.10000()k()(δ)gFirstweclaim-az?0.0()k)((δ)-az?0,andsinceghasonlyzeroeswithmultiplicityatleastk,wehaveifg0)(az0kδ)((α)g=z-0k!()k)α)((and|az|=|g|?Kfromabovediscussion.00##()α()()αAsimplecalculationshowsthat:g0?k/2if||?1andg0?|az|if||<1bothcontradicts000()#k()()()g0=kK+1+1.Sog-az?0.(δ)0δNear,wehave0()kk(δ)(ρδ))(δ)(-azg-az+?g0nnnδδδbyHurwitz’Theoremagainthereexistsasequence,?,suchthatforsufficientlylargen:nn0()k(ρδ)(δ)g-az+=0.nnnnn()k(ρδ)(ρδ)Hencefz+-az+=0.nnnnnn(ρδ))Formconditionawehavethat|fz+|?A.nnnnk(δ)(ρδ)ρδ)(Henceg=fz+/??,andg=?,whichisacontradiction.nnnnnnn0()k(δ)()()case3.2Ifg?az,wemayassumeaz=1,thenbylemma6ghasform:00k+1()z-ck11α(β)αz-,k!?1or,c?c.1z-ck!wecanexcludetheformersimilarlyasincase3.1.Wejustconsiderthelattercase.Sinceghasonl