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. ± ∈ ∈
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
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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
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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)=∅.
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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 −
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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) | |. e(z+n)2 ∼ e(z+n)2
| | But for all z K B(0, 1 ), we have ∈ ⊂ 10 n Re((z + n)2 +(z+n)) n2 + ≥ 2 and
z+n 1/2 nlog(n+1) z + n −
So we can write (2.4) as
√2π π2 ρ(f(z + n)) 2(n +1)+γ+ n <16 ≤ n 6 for all z K and every positive integer n. Hence by Marty’s theorem 0 N(f). This completes∈ the proof of the theorem. ∈
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3. Main results Now we shall prove our main result. We know that, for every meromorphic func- tion f,thesetN(f) is a periodic open set with period 1. To prove the existence part we shall construct an entire function with the help of the Weierstrass factorization theorem. Theorem 3.1. Let G be an open set in the complex plane C. Then there exists an entire function f such that N(f)=Gif and only if G is periodic with period 1.
Theorem 3.1 is also true for the set NB(f). We can prove it by a slight modifi- cation of the proof of Theorem 3.1. Theorem 3.2. Let G be an open set in the complex plane C. Then there exists an entire function f such that NB(f)=Gif and only if G is periodic with period 1. Proof of Theorem 3.1. If G = C, then we take a constant function for f.We assume G is a proper subset of C.SinceGis a periodic open set with period 1, it is enough to check for the set z G:0 Re z<1 to show N(f)=G. Because of its length we divide the proof{ into∈ three≤ steps and} put each step as a section.
3.1. Construction of an entire function. Let G be a periodic open set in the complex plane C with period 1. For each positive integer i,welet W = w : j i i { ij ≥ } be a countable dense subset of R = z Gc :0 Re z<1,i 1 Im z
Table 3.1. Sequence Z = z { ij }
C1 C2 C3 ... 3 3 z11 = w11 +2 z12 = w11 +(2 +1)! z13 = w11 +(3 +1)! ... 3 3 z22 = w12 +(2 +2)! z23 = w12 +(3 +2)! 3 3 z32 = w22 +(2 +3)! z33+=w22 +(3 +3)! 3 z43 = w13 +(3 +4)! 3 z53 = w23 +(3 +5)! 3 z63 = w33 +(3 +6)!
integer i if the set Wi contains only finitely many elements, for instance, suppose that W = w : i j k ,andifz is a translation of w where l>k,then i { ij ≤ ≤ } st il we skip the term zst.Soifwelet Cj be the number of the elements in the jth column of the Table 3.1, then we havek k j(j+1) (3.2) C . k jk≤ 2
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And for each z C ,wehave ij ∈ j j(j+1) (3.3) Re z < 1+ j3 + ! ij 2 and j(j+1) (3.4) (j3 +1)! z ∞ z (3.5) g(z)= 1 . − zij j=1 zij Cj Y Y∈ From (3.2) and (3.4) we have ∞ 1 ∞ j(j +1) 1 < 3 zij 2 (j +1)! j=1 z C j=1 X ijX∈ j | | X Therefore the function g is entire and its only zeros are at the points zij Z. Finally, we define a function f on the complex plane C by ∈ (3.6) f (z)=ezg(z). Then f is an entire function whose zero set is equal to that of g. We shall show that f is a suitable function. 3.2. For all z G,wehavez N(f).For z 0 G (we assume 0 Re z0 < 1), 1 ∈ ∈ ∈ c ≤ 0 <ε< 10 ,letF=B(z0,ε)beacompactsubsetofG.SinceG is a closed set, there is a positive number ε0 such that c (3.7) d(F, G )=ε0. Let n be a sufficiently large positive integer and suppose that k is the largest integer such that k(k+1) (3.8) 1+ k3 + ! License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1454 J. H. KIM AND LEE A. RUBEL by (3.3). Thus from the inequalities (3.4) and (3.8), we obtain z + n 1 1 = z (z + n) − z z | ij − | ij | ij | j(j+1) n 2 (j3 + )! > − − 2 3 j(j+1) j +1+(j + 2 )! (k3 + k(k+1) )! 1 (j3 + j(j+1) )! > 2 − − 2 3 j(j+1) j +1+(j + 2 )! > 1. Hence we can conclude k 1 − z + n (3.9) 1 > 1. − zij j=1 zij Cj Y Y∈ Case 2. j = k If z = w +(k3 + k(k+1) )! C , then for all z F we have ik kk 2 ∈ k ∈ z (z + n) ε0 | ik − |≥ by (3.7). Thus by (3.4), we get z + n ε (3.10) 1 0 . k(k+1) − zik ≥ 3 k +1+(k + 2 )! ( +1) Now we suppose z = w +( k3 + k k )!. Then from Table 3.1, we get ik 6 kk 2 k(k+1) Re z < 1+ k3 + 1 ! ik 2 − and k(k+1) z From (3.10) and (3.11), we conclude License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use INTEGER TRANSLATION OF MEROMORPHIC FUNCTIONS 1455 z + n ε (3.12) 1 > 0 k(k+1) − zik k +1+(k3 + )! zik Ck 2 Y∈ for all z F . Case 3.∈ j = k +1 We shall show that 3 3 (k+1)(k+2) 2 z + n 2[((k +1) +2)! ((k +1) +1)!] 2 − (3.13) 1 >ε0 − (k+1)(k+2) − zij (k+1)(k+2) 2 xij Ck+1 3 ∈Y k +2+ (k+1) + 2 ! h i for all z F . ∈ Let zαj , zβj Cj be the first and second nearest points to the compact set F +n; then from (3.7)∈ and Table 3.1, we have ε , if z = z or z , z (z + n) 0 ij αj βj | ij − |≥ (j3 +2)! (j3 +1)!, if z = z and z . ( − ij 6 αj βj So by (3.2) and (3.4), we have z + n z C zij (z + n) ij ∈ j | − | 1 j(j+1) − zij ≥ j(j+1) 2 zij Cj Q 3 Y∈ j +1+(j + 2 )! (3.14) 3 3 j(j+1) 2 2 ((j +2)! (j +1)!) 2 − >ε0 − j(j+1) . 3 j(j+1) 2 j +1+ j + 2 ! Put j = k + 1 into (3.14); then it becomes (3.13). Case 4. j>k+1 We define a sequence α by { k} ∞ z + n (3.15) ak =min 1 zF − zij nN∈(k) j=k+2 zij Cj ∈ Y Y∈ where k(k+1) (k+1)(k+2) N(k)= n:1+ k3+ ! Since F G, by the periodicity of G,wehaveF+n Gfor all positive integers n. Hence g⊂(z + n) =0forallz F.Thusa >0 for⊂ all positive integers k. | |6 ∈ k License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1456 J. H. KIM AND LEE A. RUBEL We shall show that a is eventually an increasing sequence. For convenience, { k} we write minz,n1 and minz0,n2 instead of min z F and min z0 F (and likewise n1 ∈N(k) n N∈(k+1) ∈ 2∈ for max). To prove that ak is increasing, we examine the ratio: z0+n min ∞ (1 2 a z0,n2 j=k+3 z0 Cj z k+1 ij ∈ − ij0 z+n ak ≥ max Q ∞ Q (1 1 z0,n1 j=k+2 z C z ij ∈ j − ij0 z0 (z0+n ) Q Q min ij − 2 z z0,n2 z0 Ck+3 z min ij ij ∈ ij0 ≥ z,n1 zij (z + n1) × zij (z+n1) z C maxz,n Q − ij k+2 − 1 zij Ck+3 zij ∈Y ∈ z0 (z0+n ) min ij − 2 Q z0,n2 z C x ij0 ∈ k+4 ij0 zij (z+n1) × Q − ×··· maxz,n1 z C z ij ∈ k+4 ij min z (z + n ) Q z z0,n2 z C ij0 0 2 ij ij0 k+3 | − | (3.16) min ∈ ≥ z,n1 z (z + n ) × max z (z + n ) ij 1 z,n1 Q zij Ck+3 ij 1 zij Ck+2 ∈Y − ∈ | − | minz0,n2 z C zij0 (z0 + n2) Q ij0 ∈ k+4 | − | × max z (z + n ) ×··· z,n1 Q zij Ck+4 ij 1 ∈ | − | Q zij =min min zij0 (z0 + n2) z,n1 zij (z + n1) × z0,n2 | − | zij Ck+2 − zij0 Ck+3 ∈Y ∈Y min z (z + n ) z0,n2 z C ij0 0 2 ij0 ∈ k+4 | − | × max z (z + n ) z,n1 Q zij C ij 1 ∈ k+3 | − | min z (z + n2) z0,n2 Qz0 Ck+5 ij0 0 ij ∈ | − | . × max z (z + n ) ×··· z,n1 Q zij Ck+4 ij 1 ∈ | − | Q We can easily check that the first and second terms of the last equation of (3.16) are greater than 1. From (3.4) and Table 3.1, for all z F , n1 N(k)andzij Ck+l 1 (l 4), we have ∈ ∈ ∈ − ≥ 3 (k + l 1)(k + l) z (z + n1) z k+l+ (k+l 1) + − ! | ij − |≤| ij |≤ − 2 and for z0 F , n2 N(k +1)andz0 C + (l 4), we have ∈ ∈ ij ∈ k l ≥ z0 (z + n2) Re z0 (z + n2) | ij − |≥ | ij − | (k+2)(k+3) ((k + l)3 +1)! 2 (k +2)3 + ! ≥ − − 2 >((k + l)3 +1)! (k+3)3!. − So we have an inequality, License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use INTEGER TRANSLATION OF MEROMORPHIC FUNCTIONS 1457 z0 (z0 + n2) z (z + n1) | ij − |−| ij − | > ((k + l)3 +1)! (k+3)3! − (k + l 1)(k + l) k+l+ (k+l 1)3 + − ! − − 2 (3.17) =(k+3)3! (k+l)3 +1) ((k +3)3 +1) 1 ··· − (k + l 1)(k + l) k+l+(k+l 1)3 + − ((k +3)3 +1) . − − 2 ··· It is easy to check that (k + l 1)(k + l) (k + l)3 +1>k+l+(k+l 1)3 + − − 2 for l 4. Hence the right-hand side of (3.17) is greater than 0 and we have ≥ z0 (z0 + n2) ij − > 1 z (z + n ) ij 1 − for all z,z0 F, n1 N(k), n 2 N(k +1), z ij Ck+l 1 and zij0 Ck+l.Since ∈ ∈ ∈ ∈ − ∈ Ck+l Ck+l 1 ,wehave k k≥k − k minz0,n2 z C zij0 (z0 + n2) ij0 ∈ k+l | − | > 1 max z (z + n ) z,n1 Qzij Ck+l 1 ij 1 ∈ − | − | for all l 4. Q ≥ ak+1 Now we can conclude that > 1. Therefore ak is a positive and eventually ak increasing sequence. { } From (3.9), (3.12), (3.13) and (3.15), for all z f and n N(k), we can write ∈ ∈ f(z + n) = ez+ng(z + n) | | | | k 1 − z + n z + n = ez+n 1 1 | |× − zij × − zij j=1 zij Cj zij Ck Y Y∈ Y∈ z + n ∞ z + n 1 1 × − zij × − zij zij Ck+1 j=k+2 zij Cj ∈Y Y Y∈ n ε0 (3.18) >ake 3 k(k+1)) k+1+(k + 2 )! 3 3 (k+1)(k+2) 2 [((k +1) +2)! ((k +1) +1)!] 2 − ε2 − × 0 (k+1)(k+2) 3 (k+1)(k+2) 2 k +2+ (k+1) + 2 ! h 3 3 i(k+1)(k+2) 2 3 n [((k +1) +2)! ((k +1) +1)!] 2 − >ε0ake − . (k+1)(k+2) +1 3 (k+1)(k+2) 2 k +2+ (k+1) + 2 ! Here we will show that h i (3.19) (k +1)(k+2) (k+1)(k+2) log en +1 log k +2+ (k+1)3 + ! >0. − 2 2 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1458 J. H. KIM AND LEE A. RUBEL 3 k(k+1) Since n N(k), 1+(k + 2 )! (k +1)(k+2) (k+1)(k+2) log en +1 log k +2+ (k+1)3 + ! − 2 2 k(k+1) 1 k3 + ! (k+2)2log(k +2)3! ≥ 2 −2 3 k(k+1) 1 2 3 (k+2)3 3(k+2)3 k + ! (k+2) log 2π(k +2) e− (k +2) ∼ 2 −2 k(k +1) 1 p = k3 + !+ (k+2)5 2 2 1 (k+2)2(3(k +2)3log(k + 2) + log 2π(k +2)3) −2 k(k+1) 3 p > k3 + ! (k+2)5log(k +2) 2 −2 >0 for large enough k. This implies 3 3 3 (k+1)(k+2) 2 (3.20) f(z + n) >ε a [((k +1) +2)! ((k +1) +1)!] 2 − | | 0 k − and the right-hand side of (3.20) tends to infinity as k . From (3.8), k as n ,so f(z+n) uniformly on the compact→∞ set F as n . Therefore→∞ →∞ | |→∞ →∞ z0 N(f). ∈ 3.3. If z Gc, then z N(f). Since W is a countable subset of Gc,foreach ∈ 6∈ 0 wst W and δ>0, we can choose a number δ0,0<δ0 <δ,sothatwst + δ0 = wkl (and∈= w + 1) for all w W. From the construction of the sequence Z, t 6 is the 6 kl kl ∈ smallest integer such that Ct contains the translation of wst. And the translations of wst have the form 3 (3.21) zγn = wst +(n +γ)! t(t 1) where n t and γ = −2 + s.Wefixwst and show f(zγn +δ0) as n . For each≥ positive integer n,welet | |→∞ →∞ n 1 − zγn + δ0 (3.22.1) In = 1 , − zij j=1 zij Cj Y Y∈ zγn + δ0 (3.22.2) Jn = 1 , − zij zij Cn Y∈ and ∞ zγn + δ0 (3.22.3) Kn = 1 . − zij j=n+1 zij Cj Y Y∈ License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use INTEGER TRANSLATION OF MEROMORPHIC FUNCTIONS 1459 We fix n for a moment and write ∞ zγn + δ0 g(zγn + δ0)= 1 (3.23) − zij j=1 zij Cj Y Y∈ = I J K . n × n × n We examine the moduli of In,Jn,andKn. (i) Estimation of In : Assume δ0 < 1; then by (3.4), for each zij Cj where 1 j n3! n(n 1) ≥ n +((n 1)3 + − )! − 2 > 2. So we obtain z + δ z + δ 1 γn 0 1 γn 0 − z ≥ − z ij ij > 1 and n 1 − zγn + δ0 (3.24) In = 1 > 1. | | − zij j=1 zij Cj Y Y∈ (ii) Estimation of J :Since | n| min z z : z C ,z = z {| γn − in| in ∈ n in 6 γn} > z1 z2 Re(z2 z1 ) | n − n|≥ n − n > (n3 +2)! ((n3 +1)!+1) − >(n3 +1)! and zγn + δ0 Jn = 1 | | − zin zin Cn Y∈ z + δ z + δ = 1 γn 0 1 γn 0 − zγn × − zin zin Cn zinY=∈zγn 6 δ0 z (z + δ0) in − γn , ≥ zγn × zin | | zin Cn zinY=∈zγn 6 with (3.2) and (3.4), we obtain the following inequality 3 n(n+1) 1 ((n +1)!) 2 − (3.25) J >δ0 . | n| n(n+1) 3 n(n+1) 2 n +1+(n + 2 )! License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1460 J. H. KIM AND LEE A. RUBEL (iii) K is an eventually increasing sequence: We consider the ratio (for {| n|} convenience, we let zγnˆ = zγ(n+1)): 1 zγnˆ +δ0 k ∞ z0 Cj+1 z n+1 = ij ∈ − ij0 K zγn+δ0 n j=n+1 Q Cj 1 Y zij ∈ − zij + Q 1 zγnˆ δ0 z z0 Cn+2 z = ij ij ∈ − ij0 zij (zγn + δ0) × zγn+δ0 z C Q 1 ij n+1 − zij Cn+2 zij ∈Y ∈ − + 1 zγnˆ δ0 Q z Cn+3 z ij0 ∈ − ij0 × Q zγn+δ0 ×··· z C 1 z ij ∈ n+3 − ij (z (z ˆ + δ0)) (3.26) Q z z0 Cn+2 ij0 γn = ij ij ∈ − zij (zγn + δ0) × (zij (zγn + δ0)) z C Qzij Cn+2 ij ∈Yn+1 − ∈ − z C (zij0 (zγnˆ +Qδ0)) ij0 ∈ n+3 − × (z (z + δ )) ×··· Qzij Cn+3 ij γn 0 ∈ − zij = Q (zij0 (zγnˆ + δ0)) zij (zγn + δ0) × − zij Cn+1 z C ∈Y − ij0 ∈Yn+2 z C (zij0 (zγnˆ + δ0)) z C (zij0 (zγnˆ + δ0)) ij0 ∈ n+3 − ij0 ∈ n+4 − × (z =(z + δ )) × (z (z + δ )) Qzij Cn+2 ij γn 0 Qzij Cn+3 ij γn 0 ∈ ∈ − . ×···Q Q The moduli of the first and second parts of the last equation of (3.26) are greater than 1. And for z0 C + +1 and z C + , l 2, we have ij ∈ n l ij ∈ n l ≥ z0 (zγnˆ + δ0) z0 zγnˆ +δ0 | ij − | > | ij |−| | z (z + δ0) z | ij − γn | | ij | 3 3 (n+1)(n+2) ((n + l +1) +1)! (n+1)+1+ (n+1) + 2 ! > − h 3 (n+l)(n+l+1) i n + l +1+ (n+l) + 2 ! ((n + l +1)3 +1)! (n+2) 3! > − ((n + l)3 +(n+l)2)! > 1 for sufficiently large n.Since C++1 C + , k n l k≥k n lk z C zij0 (zγnˆ + δ0) ij0 ∈ n+l+1 | − | > 1 z (z + δ ) Q zij Cn+l ij ij 0 ∈ | − | Q for all l 2. Hence we can conclude that Kn+1 > 1 for sufficiently large n. ≥ | Kn | License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use INTEGER TRANSLATION OF MEROMORPHIC FUNCTIONS 1461 Now we can estimate the value f(zγn +δ0) . From (3.23), (3.24) and (3.25), we obtain | | zγn+δ0 f(z + δ0) = e g(z + δ0) | γn | | γn | = ezγn+δ0 I J K | |·| n|·| n|·| n| (3.27) 3 n(n+1) 1 (n3+γ)! 1 ((n +1)!) 2 − e − δ0 K . ≥ n(n+1) | n| 3 n(n+1) 2 n +1+(n + 2 )! Here we let e(n3+γ)! 1 α = − . n n(n+1) 3 n(n+1) n +1+ n + 2 ! Then for large enough n, by Stirling’s formula we get n(n +1) log α (n3 + γ)! log(n3 + n2)! n ≥ − 2 3 n(n +1) 3 2 (n3+n2) 3 2 n3+n2 (n + γ)! log 2π(n + n )e− (n + n ) ∼ − 2 n(n +1) p =(n3+γ)! + (n3 +n2) 2 n(n+1) [(n3 + n2) log(n3 + n2) + log 2π(n3 + n2)] − 2 n3(n +1)2 p > (n3 + γ)! log(n3 + n2) − 2 > 0. So αn > 1 and the inequality (3.27) becomes 3 n(n+1) 1 f(z + δ0) >δ0 K ((n +1)!) 2 − . | γn | ·| n|· Since Kn is an increasing sequence, f(zγn + δ0) as n . But{| for each|} positive integer n, | |→∞ →∞ 3 f(zγn)=f(wst +(n +γ)!) =0 by the definition of the function f.Sowst does not belong to N(f). We choose c wst arbitrarily from the set W and W is a dense subset of G0. Hence no point in c G0 can be in N(f). References 1. G. D. Birkhoff, D´emostration d’un th´eor´eme ´el´ementaire sur les fonctions enti`eres,C.R.Acad. Sci. Paris 189 (1929), 473–475. 2. C. Blair and L. A. Rubel, A universal entire function, Amer. Math. Monthly 90 (1983), 331– 332. MR 85a:30046 3. J. Clunie and W. K. Hayman, The spherical derivative of integral and meromorphic functions, Comment. Math. Helv. 40 (1966), 117–148. MR 33:282 4. W. K. Hayman, Meromorphic functions, Clarendon Press, Oxford, 1964. MR 29:1337 5. D. H. Luecking and L. A. Rubel, Complex analysis, Springer-Verlag, New York, 1984. MR 86d:30002 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1462 J. H. KIM AND LEE A. RUBEL 6. F. Marty, Recherches sur la r´epartition des valeurs d’une fonction m´eromorphe, Ann. Fac. Sci. Univ. Toulouse (3) 23 (1931), 183–261. 7. Paul Montel, Le¸cons sur les familles normals de fonctions analytiques et leurs applications, Paris, 1927. Korea Military Academy, Seoul 139-799, Korea E-mail address: [email protected] Department of Mathematics, University of Illinois, Urbana, Illinois 61801 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use