复旦学报(自然科学版)

Article I D :042727104(2005)0320435203R eceived d ate :2004209202Biography :C AI Hui 2ping (1962—),femal ,ass ociate profess or ;W ANGJun (1976—),female ,postdoctor.U niqueness of E ntire Functions and Their Differential PolynomialsCAI Hui 2ping 1,2,WAN G J un 3(1.Department o f Mathematics ,Fudan Univer sity ,Shanghai 200433,China ;2.Department o f Mathematics ,Shijiazhuang College ,Shijiazhuang 050035,China ;3.Institute o f Mathematics ,Fudan Univer sity ,Shanghai 200433,China )Abstract :T he uniqueness problem on entire functions that share only one small function with their derivatives and their linear dif 2ferential polyn omials are studied.T he result im proves a theorem due to Jank ,Mues and V olkmann in tw o ways.K eyw ords :entire functions ;linear differential polyn omial ;sharing small function ;uniquenessC LC number :O 174.52 Document code :A Let f and g be tw o nonconstant merom orphic functions in the com plex plane.It is assumed that the reader is fa 2miliar with the standard symbols and fundamental results of Nevanlinna ’s theory 1,2.We use S (r ,f )to denote any quantity that satis fies S (r ,f )=o (T (r ,f ))as r →∞,possibly outside a set of r of finite linear measure.A mero 2m orphic function αis called a small function with respect to f if T (r ,α)=S (r ,f )holds.We say that f and g share a small function α(possibly constant )I M (ignoring multiplicities )if f (z )-α(z )and g (z )-α(z )have the same zeros.F or the case that f (z )-α(z )and g (z )-α(z )have the same zeros with the same multiplicities ,then we say f and g share αC M (counting multiplicities ).Rubel and Y ang proved the follwing well 2known result.Theorem 13　Let f be a nonconstant entire function.I f f and f ′share tw o finite ,distinct values C M ,then f ≡f ′.In 1986,Jank ,et al weakened the number of sharing values in theorem 1as follows.Theorem 24　Let f be a nonconstant entire function.I f f and f ′share a finite ,nonzero value a I M ,and if f ″(z )=a whenever f (z )=a ,then f ≡f ′.From the hypothesis of theorem 2,it is easy to see that f and f ′share a C M.Naturally ,the question whether value a in theorem 2can be replaced by small function αwas posed.On the other hand ,if f ″is replaced by linear differen 2tial polynomials of f ,is it still true that f ≡f ′?Recently ,Wang Jun obtained s ome results on this problem for entire functions of small growth 5.The main purpose of this paper is to investigate the above problems without the condition of small growth.In the sequel ,we setL (f )=a 2f ″+a 1f ′+a 0f ,where a j (j =0,1,2)are constants and a 2≠0.In fact ,we prove the following theorem.Theorem 3　Let f be a nonconstant entire function ,and let α(z )be a small function of f satis fying α α′.Suppose that a 2+a 1+a 0=1.I f f and f ′share αC M ,and if L (f )-α(z )=0whenever f (z )-α(z )=0,then f ≡f ′.R em ark 1　Let f (z )=e e z+e z and α(z )=e z .Clearly ,f ,f ′and f ″satis fy the hypothesis of theorem 3.H owev 2er ,f ≡f ′.The exam ple shows that the assum ption α α′is necessary.Now we prove theorem 3by contradiction.In the sequel ,we suppose that f f ′.Under the assum ption of theorem 第44卷　第3期2005年6月复旦学报(自然科学版)Journal of Fudan University (Natural Science )V ol.44N o.3Jun.20053and by the Hadamard factorization theorem ,we havef ′(z )-α(z )f (z )-α(z )=e Q (z ),(1)where Q (z )is an entire function and e Q 1.Rewrite (1)asf ′-α′f -α+α′-αf -α=e Q ,(2)f ′=e Q f +(1-e Q )α.(3)Since f ′-α=(f -α)′-(α-α′),s o the zero of f -αwith multiplicity k ≥2must be the zero of α-α′with mul 2tiplicity k -1.This im pliesN (2r ,1f -α≤2N r ,1α-α′=S (r ,f ),N —r ,1f -α=N r ,1f -α+S (r ,f ),where N (2r ,1f -αdenotes the counting function of the zeros of f -αwith multiplicity no less than 2.Since α α′,from (2)we know m (r ,e Q )=m r ,1f -α+S (r ,f ).In the following ,we discuss tw o cases that m (r ,e Q )=S (r ,f )and m (r ,e Q )≠S (r ,f ).Case 1　m (r ,e Q )=S (r ,f ).This leads to N —r ,1f -α=T (r ,f )+S (r ,f ).Differentiating both sides of (3),we have f ″=(e 2Q +Q ′e Q )f +e Q (1-e Q -Q ′)α+(1-e Q )α′.(4)This and L (f )-α(z )=0whenever f (z )-α(z )=0im ply that the zeros of f -αmust be the zeros ofβ(z )≡a 2e Q α+a 2(1-e Q )α′+(a 1+a 0)α-α.Clearly ,T (r ,β)=S (r ,f ).I f β 0,we have T (r ,f )=N —r ,1f -α+S (r ,f )≤N —r ,1β+S (r ,f )≤T (r ,β)+S (r ,f )=S (r ,f ).It is a contradiction.Hence ,β≡0.This leads toα′α≡1-a 1-a 0-a 2e Q a 2(1-e Q ).(5)From the hypothesis a 2+a 1+a 0=1,we get 1-a 1-a 0-a 2e Q =a 2(1-e Q ).S o (5)means α′α≡1,a contradic 2tion.Case 2　m (r ,e Q )≠S (r ,f ).Setγ(z )=(α-α′)(L (f )-L (α))-(α-L (α))(f ′-α′)f -α.(6)We divide our argument into tw o subcases.Subcase 1　γ≡0.From (6),we have(α-α′)(L (f )-L (α))-(α-L (α))(f ′-α′)≡0.Substituting (3)and (4)into the above equation ,we obtain(α-α′){(a 2e Q (e Q +Q ′-1)α+α2(e Q -1)α′-α1(eQ -1)α}+(α-L (α))((1-e Q )α-α′)≡634 复旦学报(自然科学版)第44卷　{(α-α′)(a 2e 2Q +a 1e Q +a 2Q ′+a 0)-(α-L (α))e Q }f.(7)Since m (r ,e Q )≠S (r ,f ),we get (α-α′)(a 2e 2Q +a 1e Q +a 2Q ′+a 0)-(α-L (α))e Q 0.In fact ,if n ot ,2m (r ,e Q )=m (r ,a 2e 2Q )+O (1)=m r ,α-L (α)α-α′-a 1e Q -a 2Q ′-a 0≤m (r ,e Q )+S (r ,f ).It leads to m (r ,e Q )=S (r ,f ),a contradiction.Hence ,we can s olve out f from (7).It is easy to see T (r ,f )=S (r ,f ),which is im possible.Subcase 2　γ 0.Since L (f )-α=0whenever f -α=0,we know that the zeros of f -αare als o the zeros of (α-α′)・(L (f )-L (α))-(α-L (α))(f ′-α′).Hence ,the possible poles of γarise from the zeros of f -αwith multiplicity k ≥2.According to the above argument ,clearly we haveT (r ,γ)=m (r ,γ)+S (r ,f )≤S (r ,f ).Rewrite (6)as f -α=1γ((α-α′)(L (f )-L (α))-(α-L (α))(f ′-α′)).Again substituting (3)and (4)into this equation ,we obtainα+1γ{(α-α′)((a 2e Q (e Q +Q ′-1)α+a 2(e Q -1)α′+a 1(e Q -1)α)+(α-L (α))((1-e Q )α- α′)}≡1-1γ((α-α′)(a 2e 2Q +a 1e Q +a 2Q ′+a 0)-(α-L (α))e Q ) f.(8)Using similar argument of (7)to (8),we can als o get m (r ,e Q )=S (r ,f )or T (r ,f )=S (r ,f ),a contradiction.This com pletes the proof of theorem 3.R eference :1　Hayman W K.Merom orphic function M.Ox ford :Clarendon Press ,1964.2　Y i H X ,Y ang C C.Uniqueness theory of merom orphic functionsM.Beijing :Science Press ,1995.3　Rubel L A ,Y ang C C.Values shared by an entire function and its derivative A.Buckholtz J D ,Su ffridge TJ ,eds.C on 2ference of C omplex Analysis C.Berlin :S pringer 2Verlag ,1977.1012103.4　Jank G,Mues E ,V olkmann L.Merom orphe Funktionen ,die mit ihre ersten und zweiten ableitung einen endichen Wert teilenJ .Complex Variables Theory Appl ,1986,6:51271.5　Wang J ,Y i H X.Entire functions of small growth that share one value with its linear differential polynomialsJ .K odaiMath J ,2003,26:2452258.整函数与其微分多项式的唯一性蔡惠萍1,2,王 3(1.复旦大学数学系,上海　200433;2.石家庄学院数学系,石家庄　050035;3.复旦大学数学研究所,上海　200433)摘　要:研究整函数及其导数与线性微分多项式分担一个小函数的唯一性问题,所得结果从两方面改进了Jank ,Mues 和V olkmann 的定理.关键词:整函数;线性微分多项式;分担小函数;唯一性中图分类号:O 174.52 文献标识码:A 734　第3期蔡惠萍等:整函数与其微分多项式的唯一性 。