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Integer Translation of Meromorphic Functions 1449 TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 349, Number 4, April 1997, Pages 1447{1462 S 0002-9947(97)01504-3 INTEGER TRANSLATION OF MEROMORPHIC FUNCTIONS JEONGH.KIMANDLEEA.RUBEL Abstract. Let G be a given open set in the complex plane. We prove that there is an entire function such that its integer translations forms a normal family in a neighborhood of z exactly for z in G if and only if G is periodic with period 1, i.e., z 1 G for all z G. ± 2 2 1. Introduction and preliminaries Let be a space of functions defined on an open set G. For each positive integer S n,weletTn be an operator from into itself. Our concern is the normality of the family T (f): n =0,1,2,... forS a function f . In this paper, we consider the { n } ∈S case Tn(f(z)) = f(z + n), translation by n. To discuss the normality of families of meromorphic functions, we need the fol- lowing notion. Definition 1.1. Let f be a meromorphic function in the complex plane C.We define the spherical derivative of f at z by f 0(z) ρ(f (z)) = | | . 1+ f(z)2 | | Theorem (Marty [6]). A family of functions f(z), meromorphic in a domain D,isnormalinDif and only if theF set ρ(f(z)): f is uniformly bounded on every compact subset of D. { ∈F} Now we state our definition and an example. Definition 1.2. For a meromorphic function f, we define the set N(f)tobethe set of all z0 C such that the family f(z + n): n =0,1,2,... is normal in a ∈ { } neighborhood of z0. Example 1.3. Let f(z)=zsin πz;thenN(f)=CZwhere Z is the set of all integers. \ Proof. For all integers n, f(z + n)=(z+n)sinπ(z+n)=(z+n)sinπz; thus we have =0, if z Z, f(z + n) ∈ , if z Z and n . (→∞ 6∈ →∞ Received by the editors October 17, 1994 and, in revised form, March 31, 1995. 1991 Mathematics Subject Classification. Primary 30D45. The research of the second author was partially supported by a grant from the National Science Foundation. c 1997 American Mathematical Society 1447 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1448 J. H. KIM AND LEE A. RUBEL Hence no subsequence of f(z +n): n =0,1,2,... converges on any neighborhood of any integer. But the sequence{ f(z + n): n =0,}1,2,... converges uniformly to infinity on any compact subset which{ does not contain integers.} These facts prove that N(f)=C Z. \ 2. Some properties of N(f) From the definition, the set N(f)isopeninthecomplexplaneCand is periodic with period 1. The set N(f) may not be connected. For a certain meromorphic function f,the set N(f) has infinitely many components. Example 2.1. Let f(z)=tanez; then the set N(f) is not connected. Proof. We let I = z C:Imz=kπ, k =0, 1, 2,... , c { ∈ ± ± } and we shall show that N(f)=C I.Forz=x+kπi I ,wehave \ c ∈ c f(z+n)2 = tan ex+n+kπi 2 =tan2ex+n | | | | and x+n+kπi 2 x+n+kπi x+n 2 x+n f 0(z + n) = e sec e = e sec e . | | | | So for every real number x with ex+n = kπ + π (k =0, 1, 2,...), 6 2 ± ± ex+n sec2 ex+n ρ(f(z + n)) = 1+tan2ex+n ex+n = sin2 ex+n +cos2ex+n =ex+n →∞ as n . But for an integer l every neighborhood of x + lπi contains a point →∞ x0+n π x0 + lπi such that e = kπ + 2 . Hence by Marty’s Theorem, x + lπi N(f)for all real x. 6 6∈ Now for z0 Ic,wechooseapositivenumberεso that B(z0,ε) Ic =?.LetF 6∈ ∩ be a compact subset in B(z0,ε); then for z = x + iy F ,wehave ∈ z+n (tan e )0 ρ(f(z+n)) = | | 1+ tan ez+n 2 | | ez+n sec2 ez+n (2.1) = | | 1+ tan ez+n 2 | | ez+n = | | . sin ez+n 2 + cos ez+n 2 | | | | For z = x + iy,itiseasytoshow 2 2 1 2y 2y sin z + cos z = (e + e− ) | | | | 2 and since ez+n = ex+n cos y + iex+n sin y, we can write z+n 2 z+n 2 1 2ex+n sin y 2ex+n sin y sin e + cos e = (e + e− ). | | | | 2 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use INTEGER TRANSLATION OF MEROMORPHIC FUNCTIONS 1449 Thus from (2.1), 2ex+n ρ(f(z + n)) = 2ex+n sin y 2ex+n sin y (2.2) e + e− 2ex+n . 2 sin y ex+n ≤ e | | We let δ =min sin y : z = x + iy F ;thenδ>0. We define a function g on the real line R by {| | ∈ } 2et g(t)= ; e2δet then we get t 2δet t g0(t)=2e− (1 2δe ). − 1 So the positive continuous function g takes the maximum value δe at t = log 2δ. Hence from (2.2), we have − 1 ρ(f(z + n)) g(x + n) ≤ ≤ δe for all z F and n =0,1,2,.... Therefore z0 belongs to the set N(f). This completes∈ the proof. Let be the space of all entire functions equipped with the topology of uniform convergenceA on compact subset of the complex plane C. Theorem 2.2. If is a family of entire functions which is dense in , then the family is not normalF on any open set in C. A F Proof. For a given open set Ω and z0 Ω, we choose a positive number ε so that ∈ B(z0,ε) Ω is a compact subset. By the density of , there exists a sequence g (z) ⊂ , such that F { n }⊂F 1 g (z) n(z z0) < | n − − | n for all z B(z0,ε). So we have ∈ 1 g (z0) < 0 | n | n → as n . →∞ On the other hand, for z B(z0,ε) z0 , ∈ \{ } 1 g (z) >nz z0 | n | | − |−n →∞ as n . Hence the family is not normal on Ω. An→∞ entire function f whoseF translates are dense in is called a universal entire function. In 1929, the existence of a universal entire functionA was proved by G. D. Birkhoff [1]. For a proof of the following theorem, see [5, pp. 60–61]. Theorem (Birkhoff [1]). There is an entire function whose integer translations are dense in the set of all entire functions. A Birkhoff’s Theorem and Theorem 2.2 have the following consequence. Corollary 2.3. There is an entire function f such that N(f)=?. License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1450 J. H. KIM AND LEE A. RUBEL In Definition 1.2, we considered positive integer translations only. At this time we allow the negative integer translations also. Definition 2.4. Let f be a meromorphic function in the complex plane C.We define as set N (f)bythesetofallz0 Csuch that the family f(z + n): n = B ∈ { 0, 1, 2,... is normal at z0. ± ± } By the same reasons as in N(f), the set NB(f) is a periodic open set with period 1. And clearly NB(f) is a subset of N(f). But NB(f) is different from N(f)fora certain meromorphic function f. Theorem 2.5. There is an entire function f such that N (f) = N(f). B 6 Proof. We shall show that the entire function ez2 f(z)= Γ(z) satisfies the property by proving 0 N(f) NB(f). (i) 0 N (f): Let 0 <ε< 1 and∈ n be\ a positive integer. Since 6∈ B 10 zΓ(z)=Γ(z+1), we have Γ(ε n +1) Γ(ε n)= − − ε n − Γ(ε) = (ε n)(ε n +1) (ε 1) − − ··· − and (ε n)2 e − (ε n)(ε n +1) (ε 1) (2.3) f(ε n) = − − ··· − . | − | Γ(ε) The right-hand side of (2.3) tends to infinity as n .Butf ( n) = 0 for all →∞ − positive integers. Hence 0 does not belong to NB(f). (ii) 0 N(f): Since ∈ Γ0(z) ∞ z 1 = γ + − Γ(z) − m(z + m 1) m=1 X − where 1 1 1 γ = lim 1+ + + + log n n 2 3 ··· n − →∞ =0.5772157 , ··· we get z2 e Γ0(z) f 0(z)= 2z Γ(z) − Γ(z) 2 ez ∞ z 1 = 2z + γ − . Γ(z) − m(z + m 1) m=1 ! X − License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use INTEGER TRANSLATION OF MEROMORPHIC FUNCTIONS 1451 Let K be a compact subset of B(0, 1 ). Then for z K and positive integer n, 10 ∈ 2 e(z+n) ∞ z + n 1 f 0(z + n) = 2(z + n)+γ − | | Γ(z + n) − m(z + n + m 1) m=1 ! X − 2 e(z+n) ∞ 1 2(n +1)+γ+n ≤ Γ(z + n) m2 m=1 ! X 2 e(z+n) π2 = 2(n +1)+γ+ n . Γ(z + n) 6 So we can write 2 e(z+n) π2 Γ(z+n) 2(n +1)+γ+ 6 n ρ(f(z + n)) 2 2 ≤ 1+ e(z+n) (2.4) Γ(z+n) Γ(z + n) π2 2(n +1)+ γ+ n . ≤ e(z+n)2 6 By Stirling’s formula, z+n 1/2 (z+n) Γ(z + n) √2π(z + n) − e− (2.5) | |.
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