JOURNAL OF MATHEMATICAL ANALYSIS AND APPLICATIONS 223, 88 95 1998 ARTICLE NO. AY985959 Etire Fuctios That Share Oe Value with Oe or Two of Their Derivatives Gary G. Guderse* Departmet of Mathematics, Ui ersity of New Orleas, New Orleas, Louisiaa 70148 ad Lia-Zhog Yag Departmet of Mathematics, Shadog Ui ersity, Jia, Shadog, 250100, P.R. Chia Submitted by Bruce C. Berdt Received November 26, 1997 We prove two uiqueess theorems for etire fuctios of fiite order that share oe fiite value with oe or two of their derivatives. 1998 Academic Press 1. INTRODUCTION AND RESULTS I this paper a meromorphic fuctio will mea meromorphic i the whole complex plae. We say that two meromorphic fuctios f ad g share a fiite value a IM Ž igorig multiplicities. whe f a ad g a have the same eros. If f a ad g a have the same eros with the same multiplicities, the we say that f ad g share the value a CM Ž coutig multiplicities.. It is assumed that the reader is familiar with the stadard symbols ad fudametal results of Nevalia theory, as foud i 5. Rubel ad C. C. Yag proved the followig result. THEOREM A 9. Let f be a ocostat etire fuctio. If f ad f share two fiite, distict alues CM, the f f. *E-mail: gguders@math.uo.edu The secod author was supported by the NSF of Shadog Provice ad the ACF of Shadog Uiversity. 0022-247X 98 $25.00 Copyright 1998 by Academic Press All rights of reproductio i ay form reserved. 88
ENTIRE FUNCTIONS AND DERIVATIVES 89 Mues ad Steimet improved Theorem A with the ext result. THEOREM B 8. Let f be a ocostat etire fuctio. If f ad f share two fiite, distict alues IM, the f f. Aother proof of Theorem B for the case whe the two shared values are both oero is i 2. A atural questio is: QUESTION 1. What ca be said whe a ocostat etire fuctio f shares oe fiite alue CM with its first deri ati e f? We eed the followig defiitio. DEFINITION. Let f be a ocostat meromorphic fuctio. The first iterated order of f, deoted by Ž f. 1, is defied by log log TŽ r, f. 1Ž f. lim sup. log r r Bruck made the followig cojecture. Cojecture 1. Let f be a ocostat etire fuctio satisfyig Ž f. 1, where Ž f. 1 is ot a positive iteger. If f ad f share oe fiite value a CM, the f a c f a for some costat c 0. Bruck showed that the cojecture is true whe a 0, by provig the ext result. THEOREM C 1. Let f be a ocostat etire fuctio satisfyig Ž f. 1, where Ž f. 1 is ot a positi e iteger. If f ad f share the alue 0 CM, the f cf for some costat c 0. Bruck also proved the ext result, which shows that the cojecture is true for a 0 provided that f satisfies the additioal assumptio Nr,0, Ž f. Sr,f, ad i this case the order restrictio o f ca be omitted. THEOREM D 1. Let f be a ocostat etire fuctio. If f ad f share the alue 1 CM, ad if NŽ r,0, f. Sr,f,the for some costat c 0. f 1 c f 1
90 GUNDERSEN AND YANG O the other had, Bruck 1 showed that the cojecture does ot hold whe Ž f. 1 is a positive iteger or, by cosiderig solutios of the followig differetial equatios: f 1 f 1 e e ad e. f 1 f 1 We prove the followig result, which shows that Bruck s cojecture holds for etire fuctios of fiite order. THEOREM 1. Let f be a ocostat etire fuctio of fiite order. If f ad f share oe fiite alue a CM, the for some costat c 0. f a c Ž 1. f a We caot delete the words fiite order from Theorem 1, by the examples of Bruck above. We also caot replace the word etire with the word meromorphic i Theorem 1. For example, if 2 e 1 f, e 1 the f ad f share the value 1 CM, but Ž. 1 does ot hold. We metio that this example occurs whe h e i Theorem 1 of 2. Jak, Mues, ad Volkma proved the ext two results. THEOREM E 6. Let f be a ocostat meromorphic fuctio, ad let a 0 be a fiite costat. If f, f, ad f share the alue a CM, the f f. THEOREM F 6. Let f be a ocostat etire fuctio, ad let a 0 be a fiite costat. If f ad f share the alue a IM, ad if f a whee er f a, the f f. We also metio the followig result of Zhog. THEOREM G 12. Let f be a ocostat etire fuctio, let a 0 be a fiite costat, ad let be a positi e iteger. If f ad f share the alue a Ž 1. Ž. CM, ad if f f a whee er f a, the f f. Theorem E suggests the followig questio of Yi ad Yag. QUESTION 2 11, p. 458. Let f be a ocostat meromorphic fuctio, let a 0 be a fiite costat, ad let ad m be positi e itegers satisfyig m. If f, f, ad f Ž m. share the alue a CM, where ad m are ot both e e or both odd, must f f?
ENTIRE FUNCTIONS AND DERIVATIVES 91 The followig example 10 shows that the aswer to Questio 2 is, i geeral, egative. Let ad m be positive itegers satisfyig m 1, ad let b 0 be a costat which satisfies b b m 1. Set a b ad b Ž. Žm. f e a 1. The f, f, ad f share the value a CM, ad f f. O the other had, we will use Theorem 1 to prove the ext result, which gives a positive aswer to Questio 2 i the case whe f is etire of fiite order ad m 1. THEOREM 2. Let f be a ocostat etire fuctio of fiite order, let a 0 be a fiite costat, ad let be a positi e iteger. If the alue a is shared by f, f, ad f Ž 1. IM, ad shared by f ad f Ž 1. CM, the f f. For etire fuctios of fiite order, Theorem 2 geeralies Theorem E. If i Theorem 2, a 0 is replaced by a 0, the it is easy to show that f cf for some costat c 0. 2. LEMMAS LEMMA A 3. Let F be a ocostat meromorphic fuctio of fiite order, ad let 0 be a gi e costat. The there exists a set E 0, 2. that has liear measure ero, such that if 0, 2. 0 E, the there is a costat R R Ž. 0 0 0 0 such that for all satisfyig arg 0 ad R, we ha e 0 F 1. F LEMMA B 4. Let F be a etire fuctio, ad suppose that F is ubouded o some ray arg. The there exists a ifiite sequece of i poits r e where r, such that F Ž. ad as. F F Ž. Ž 1 ož 1.. LEMMA 1. Let Q be a ocostat polyomial. The e ery solutio F of the differetial equatio is a etire fuctio of ifiite order. F e Q F 1 Ž 2.
92 GUNDERSEN AND YANG Proof. It is well kow that every solutio of Eq. Ž. 2 is etire. We prove Lemma 1 by cotradictio. Assume that Lemma 1 is ot true, i.e., suppose that F is a solutio of Eq. Ž. 2 that has fiite order. From Ž. 2, F 1 Q e. Ž 3. F F Let 0 be ay give costat. The from Lemma A, there exists a set E 0, 2. that has liear measure ero, such that if 0, 2. 0 E, the there is a costat R R Ž. 0 0 0 0 such that for all satisfyig arg ad R, we have 0 0 F 1. Ž 4. F. Now suppose that is ay real umber that satisfies 0, 2 E, ad for every 0, QŽ re i. e Ž.Ž. Ž. as r. The from 5, 4, ad 3, it follows that r Ž 5. FŽ re i. 0 Ž 6. as r. Now suppose that is ay real umber that satisfies 0, 2., ad for every 0. r e QŽ rei. 0 Ž 7. as r. We ow show that F is bouded o the ray arg. Assume the cotrary, i.e., suppose that F is ot bouded o the ray arg. The from Lemma B, there exists a ifiite sequece of poits i r e where r, such that F Ž. ad F F Ž. 1 ož 1. Ž 8. as. Sice F Ž., it follows from Ž 7. ad Ž 2. that FŽ.. The from Ž.Ž. 8, 7, ad Ž. 3, we obtai that F Ž. 1, which cotradicts
ENTIRE FUNCTIONS AND DERIVATIVES 93 F Ž.. This cotradictio proves that F must be bouded o the ray arg. By cosiderig the formula we obtai that H F FŽ 0. F Ž w. dw, 0 F FŽ 0. M Ž 9. for all satisfyig arg, where M MŽ. 0 is some costat. We have show that Ž. 9 holds for ay 0, 2. with property Ž 7., ad that Ž. 6 holds for ay 0, 2. E with property Ž 5.. Sice Q is a ocostat polyomial, there exist oly fiitely may real umbers i 0, 2. that do ot satisfy either Ž 7. or Ž 5.. We also ote that the set E has liear measure ero. Therefore, sice F has fiite order, it ca be deduced from Ž. 9, Ž. 6, the Phragme Lidelof theorem 7, pp. 270 271, ad Liouville s theorem, that F must be a polyomial with deg F 1. But this is impossible because Q is ocostat i Ž 2.. This cotradictio proves Lemma 1. 3. PROOF OF THEOREM 1 First suppose that a 0. Sice f has fiite order, ad sice f ad f share the value a CM, it follows from the Hadamard factoriatio theorem that f a Q e, Ž 10. f a where Q is a polyomial. Set F f a 1. The from 10, F e Q F 1. Ž 11. If Q is ocostat, the from Ž 11. ad Lemma 1 we obtai that F has ifiite order. Sice f has fiite order, this is impossible. Hece Q is a costat. The from Ž 10. we obtai Ž 1.. Now suppose that a 0. I this case we obtai Ž. 1 from Theorem C, but we will give a proof for the coveiece of the reader. I this case, f ad f share the value 0 CM, ad so both f ad f have o eros. Sice f has fiite order, it follows that f e p, where p is a ocostat polyomial. Sice f p e p has o eros, p c for some costat c 0. Thus f cf, which is Ž. 1 whe a 0.
94 GUNDERSEN AND YANG 4. PROOF OF THEOREM 2 Sice f ad f Ž 1. share the value a CM, we obtai from Theorem 1 that f Ž 1. a cž f Ž. a., Ž 12. where c 0 is some costat. We ote that f caot be a polyomial, from the give coditios. Set g f. The from Ž 12., g is a solutio of the differetial equatio g cg až 1 c., ad so g b de c Ž 13. for some costats b ad d 0. From itegratio of 13, we obtai d f P e c, Ž 14. where P is some polyomial with deg P. The from 14, c g a f Ž 1. a cde c a, ad sice a 0, we obtai that g a has ifiitely may eros. Therefore, sice f, f, ad f Ž 1. share the value a IM, this implies that the fuctio Ž g a. c Ž f a. must also have ifiitely may eros. But from Ž 14. ad Ž 13. we have Ž g a. c Ž f a. ac b a c PŽ., which implies that ac b a cp 0. Thus P is a costat. The from Ž 14., where is a costat. Hece d f e c, Ž 15. c f de c ad f Ž 1. dce c. Ž 16. Ž. Ž 1. Sice f, f, ad f share the value a IM, it follows from Ž 16. ad Ž 15. that c 1 ad 0. Thus from 15, f de ad f f. This proves Theorem 2. ACKNOWLEDGMENT The authors thak Hog-Xu Yi for valuable discussios cocerig this paper.
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