In this paper, we study the problem of meromorphi functions sharing a small function with its derivative and prove one theorem. The theorem improves the results of Jin-Dong Li and Guang-Xin Huang [10]

Harina, P. Waghamore.

Rajeshwari, S.

ePrints@Bangalore University

Bulletin of Mathematical Analysis and ApplicationsISSN: 1821-1291, URL: http://www.bmathaa.orgVolume 8 Issue 1(2016), Pages 27-36.ON MEROMORPHIC FUNCTIONS THAT SHARE A SMALLFUNCTION WITH ITS DERIVATIVESHARINA P. WAGHAMORE AND RAJESHWARI S.(COMMUNICATED BY DAVID KALAJ)Abstract. In this paper, we study the problem of meromorphic functionssharing a small function with its derivative and prove one theorem. The the-orem improves the results of Jin-Dong Li and Guang-Xin Huang [10].1. IntroductionLet f be a nonconstant meromorphic function defined in the whole complex planeC. It is assumed that the reader is familiar with the notations of the Nevanlinnatheory such as T (r, f), N(r, f) and so on, that can be found, for instance in [1].Let f and g be two nonconstant meromorphic functions. Let a be a finite complexnumber. We say that f and g share the value a CM(counting multiplicities) if f−aand g− a have the same zeros with the same multiplicites and we say that f and gshare the value a IM(ignoring multiplicities) if we do not consider the multiplicities.When f and g share 1 IM, let z0 be a 1-points of f of order p, a 1-points of g oforder q, we denote by N11(r,1f−1 ) the counting function of those 1-points of f andg where p = q = 1; and N(2E (r,1f−1 ) the counting function of those 1-points of fand g where p = q ≥ 2. NL(r, 1f−1 ) is the counting function of those 1-points ofboth f and g where p > q. In the same way, we can define N11(r,1g−1 ), N(2E (r,1g−1 )and NL(r,1g−1 ). If f and g share 1 IM, it is easy to see thatN(r,1f − 1) = N11(r,1f − 1) +NL(r,1f − 1) +NL(r,1g − 1) +N(2E (r,1g − 1)= N(r,1g − 1)Let f be a nonconstant meromorphic function. Let a be a finite complex number,and k be a positive integer, we denote by Nk)(r,1f−a )(orNk)(r,1f−a )) the countingfunction for zeros of f − a with multiplicity ≤ k (ignoring multiplicities), and byN(k(r,1f−a )(orN (k(r,1f−a )) the counting function for zeros of f−a with multiplicity2000 Mathematics Subject Classification. 35A07, 35Q53.Key words and phrases. Uniqueness, Meromorphic function, Weighted sharing.c©2016 Universiteti i Prishtine¨s, Prishtine¨, Kosove¨.Submitted September 12, 2015. Published March 3, 2016.2728 HARINA P. WAGHAMORE AND RAJESHWARI S.atleast k(ignoring multiplicities). SetNk(r,1f − a ) = N(r,1f − a ) +N (2(r,1f − a ) + ...+N (k(r,1f − a )Θ(a, f) = 1− lim supr−→∞N(r, 1f−a )T (r, f), δ(a, f) = 1− lim supr−→∞N(r, 1f−a )T (r, f).We further defineδk(a, f) = 1− lim supr−→∞Nk(r,1f−a )T (r, f).Clearly0 ≤ δ(a, f) ≤ δk(a, f) ≤ δk−1(a, f)... ≤ δ2(a, f) ≤ δ1(a, f) = Θ(a, f)Definition 1.1(see[3]). Let k be a nonnegative integer or infinity. For a ∈ C wedenote by Ek(a, f) the set of all a-points of f, where an a-point of multiplicity mis counted m times if m ≤ k and k + 1 times if m > k. If Ek(a, f) = Ek(a, g), wesay that f, g share the value a with weight k.We write f, g share (a, k) to mean that f, g share the value a with weight k;clearly if f, g share (a, k), then f, g share (a, p) for all integers p with 0 ≤ p ≤ k.Also, we note that f, g share a value a IM or CM if and only if they share (a, 0) or(a,∞), respectively.A meromorphic function a is said to be a small function of f where T (r, a) =S(r, f), that is T (r, a) = o(T (r, f)) as r →∞, outside of a possible exceptional setof finite linear measure. Similarly, we can define that f and g share a small functiona IM or CM or with weight k.R.Bruck [4] first considered the uniqueess problems of an entire function sharingone value with its derivative and proved the following result.Theorem A. Let f be a non-constant entire function satisfying N(r, 1f ′ ) = S(r, f).If f and f ′ share the value 1 CM, then f′−1f−1 ≡ c for some nonzero constant c.Bruck [4] further posed the following conjecture.Conjecture 1.1. Let f be a non-constant entire function, ρ1(f) be the first iteratedorder of f . If ρ1(f) is not a positive integer or infinite, f and f′ share the value 1CM, then f′−1f−1 ≡ c for some nonzero constant c.Yang [5] proved that the conjecture is true if f is an entire function of finite order.Yu [6] considered the problem of an entire or meromorphic function sharing onesmall function with its derivative and proved the following two theorems.Theorem B. Let f be a non-constant entire function and a ≡ a(z)( 6≡ 0,∞) be ameromorphic small function. If f − a and f (k) − a share 0 CM and δ(0, f) > 34 ,then f ≡ f (k).Theorem C. Let f be a non-constant non-entire meromorphic function and a ≡a(z)( 6≡ 0,∞) be a meromorphic small function. If(i) f and a have no common poles.(ii) f − a and f (k) − a share 0 CM.(iii) 4δ(0, f) + 2(8 + k)Θ(∞, f) > 19 + 2k,then f ≡ f (k) where k is a positive integer.In the same paper, Yu [6] posed the following open questions.(i) can a CM shared be replaced by an IM share value ?(ii) Can the condition δ(0, f) > 34 of theorem B be further relaxed ?(iii) Can the condition (iii) in theorem C be further relaxed ?ON MEROMORPHIC FUNCTIONS THAT SHARE ... 29(iv) Can in general the condition (i) of theorem C be dropped ?In 2004, Liu and Gu [7] improved theorem B and obtained the following results.Theorem D. Let f be a non-constant entire function and a ≡ a(z)( 6≡ 0,∞) be ameromorphic small function. If f − a and f (k) − a share 0 CM and δ(0, f) > 12 ,then f ≡ f (k).Lahiri and Sarkar [8] gave some affirmative answers to the first three questionsimposing some restrictions on the zeros and poles of a. They obtained the followingresults.Theorem E. Let f be a non-constant meromorphic function, k be a positive inte-ger, and a ≡ a(z)( 6≡ 0,∞) be a meromorphic small function. If(i) a has no zero (pole) which is also a zero (pole) of f or f (k) with the samemultiplicity.(ii) f − a and f (k) − a share (0, 2)(iii) 2δ2+k(0, f) + (4 + k)Θ(∞, f) > 5 + k then f ≡ f (k).In 2005, Zhang [?] improved the above results and proved the following theorem.Theorem F. Let f be a non-constant meromorphic function, k(≥ 1), l(≥ 0) beintegers. Also let a ≡ a(z)( 6≡ 0,∞) be a meromorphic small function. Suppose thatf − a and f (k) − a share (0, l). Ifl ≥ 2 and(3 + k)Θ(∞, f) + 2δ2+k(0, f) > k + 4 (1.1)or l = 1 and(4 + k)Θ(∞, f) + 3δ2+k(0, f) > k + 6 (1.2)or l = 0 and(6 + 2k)Θ(∞, f) + 5δ2+k(0, f) > 2k + 10 (1.3)then f ≡ f (k).In 2015, Jin-Dong Li and Guang-Xiu Huang [?] proved the following Theorem.Theorem G. Let f be a non-constant meromorphic function, k(≥ 1), l(≥ 0) beintegers. Also let a ≡ a(z)(6≡ 0,∞) be a meromorphic small function. Suppose thatf − a and f (k) − a share (0, l). Ifl ≥ 2 and(3 + k)Θ(∞, f) + δ2(0, f) + δ2+k(0, f) > k + 4 (1.4)l = 1 and(72+ k)Θ(∞, f) + 12Θ(0, f) + δ2(0, f) + δ2+k(0, f) > k + 5 (1.5)or l = 0 and(6 + 2k)Θ(∞, f) + 2Θ(∞, f) + δ2(0, f) + δ1+k(0, f) + δ2+k(0, f) > 2k + 10 (1.6)then f ≡ f (k).In this paper we pay our attention to the uniqueness of more generalised formof a function namely fm and (fn)(k) sharing a small function for two arbitrarypositive integer n and m.Theorem 1.1. Let f be a non-constant meromorphic function, k(≥ 1), l(≥ 0)be integers. Also let a ≡ a(z)( 6≡ 0,∞) be a meromorphic small function. Suppose30 HARINA P. WAGHAMORE AND RAJESHWARI S.that fm − a and (fn)(k) − a share (0, l). Ifl ≥ 2 and(k + 4)Θ(∞, f) + (k + 5)Θ(0, f) > 2k + 9−m (1.7)l = 1 and(k +92)Θ(∞, f) + (k + 112)Θ(0, f) > 2k + 10−m (1.8)or l = 0 and(2k + 7)Θ(∞, f) + (2k + 8)Θ(0, f) > 4k + 15−m (1.9)then fm ≡ (fn)(k).Corollary 1.2. Let f be a non-constant meromorphic function, m, k(≥ 1), l(≥ 0)be integers. Also let a ≡ a(z)( 6≡ 0,∞) be a meromorphic small function. Supposethat fm − a and (fn)(k) − a share (0, l). Ifl ≥ 2 and Θ(0, f) > 45or l = 1 and Θ(0, f) > 911or l = 0 and Θ(0, f) > 78 − 18 [7Θ(∞, f)− 7Θ(0, f)]then fm ≡ (fn)(k).2. LemmasLemma 2.1 (see [10]). Let f be a non-constant meromorphic function, k, pbe two positive integers, thenNp(r,1f (k)) ≤ Np+k(r, 1f) + kN(r, f) + S(r, f)clearly N(r, 1f(k)) = N1(r,1f(k))Lemma 2.2 (see [10]). LetH = (F ′′F ′− 2F′F − 1)− (G′′G′− 2G′G− 1) (2.1)where F and G are two non constant meromorphic functions. If F and G share 1IM and H 6≡ 0, thenN11(r,1F − 1) ≤ N(r,H) + S(r, F ) + S(r,G)Lemma 2.3 (see [11]). Let f be a non-constant meromorphic function and letR(f) =Σnk=0akfkΣmj=0bjfjbe an irreducible rational function in f with constant coefficients ak and bj wherean 6= 0 and bm 6= 0. ThenT (r,R(f)) = dT (r, f) + S(r, f),where d = max{n,m}.ON MEROMORPHIC FUNCTIONS THAT SHARE ... 313. Proof of the Theorem 1.2Let F = fma and G =(fn)(k)a . Then F and G share (1, l), except the zeros andpoles of a(z). Let H be defined by (2.1)Case 1. Let H 6≡ 0.By our assumptions, H have poles only at zeros of F ′ and G′ and poles of Fand G, and those 1-points of F and G whose multiplicities are distinct from themultiplicities of corresponding 1-points of G and F respectively. Thus, we deducefrom (2.1) thatN(r,H) ≤ N (2(r, 1H) +N (2(r,1G) +N(r,H)+N0(r,1F ′) +N0(r,1G′) +NL(r,1F − 1)+NL(r,1G− 1)(3.1)here N0(r,1F ′ ) is the counting function which only counts those points such thatF ′ = 0 but F (F − 1) 6= 0.Because F and G share 1 IM, it is easy to see thatN(r,1F − 1) = N11(r,1F − 1) +NL(r,1F − 1) +NL(r,1G− 1) +N(2E (r,1G− 1)= N(r,1G− 1)(3.2)By the second fundamental theorem, we see thatT (r, F ) + T (r,G) ≤ N(r, F ) +N(r,G) +N(r, 1F)+N(r,1G) +N(r,1F − 1) +N(r,1G− 1)−N0(r, 1F ′)−N0(r, 1G′) + S(r, F ) + S(r,G)(3.3)Using Lemma 2.2 and (3.1), (3.2) and (3.3) We getT (r, F ) + T (r,G) ≤ 3N(r, F ) +N2(r, 1F) +N2(r,1G)+N11(r,1F − 1) + 2N(2E (r,1G− 1)+ 3NL(r,1F − 1) + 3NL(r,1G− 1) + S(r, F ) + S(r,G)(3.4)We discuss the following three sub cases.Sub case 1.1. l ≥ 2. Obviously.N11(r,1F − 1) + 2N(2E (r,1G− 1) + 3NL(r,1F − 1) + 3NL(r,1G− 1)≤ N(r, 1G− 1) + S(r, F )≤ T (r,G) + S(r, F ) + S(r,G)(3.5)32 HARINA P. WAGHAMORE AND RAJESHWARI S.Combining (3.4) and (3.5), we getT (r, F ) ≤ 3N(r, F ) +N2(r, 1F) +N2(r,1G) + S(r, F ) (3.6)that isT (r, fm) ≤ 3N(r, fm) +N2(r, 1fm) +N2(r,1(fn)(k)) + S(r, f)By Lemma 2.1 for p = 2, we getmT (r, f) ≤ (k + 5)N(r, 1f) + (k + 4)N(r, f) + S(r, f)So(k + 4)Θ(∞, f) + (k + 5)Θ(0, f) ≤ 2k + 9−mwhich contradicts with (1.7).Sub case 1.2. l = 1. It is easy to see thatN11(r,1F − 1) + 2N(2E (r,1G− 1) + 2NL(r,1F − 1) + 3NL(r,1G− 1)≤ N(r, 1G− 1) + S(r, F )≤ T (r,G) + S(r, F ) + S(r,G)(3.7)NL(r,1F − 1) ≤12N(r,FF ′)≤ 12N(r,F ′F) + S(r, F )≤ 12[N(r,1F) +N(r, F )] + S(r, F ).(3.8)Combining (3.4) and (3.7) and (3.8), we getT (r, F ) ≤ N2(r, 1F) +N2(r,1G) +72N¯(r, F ) +12N¯(r,1F) + S(r, F ) (3.9)that ismT (r, f) ≤ N2(r, 1fm) +N2(r,1(fn)(k)) +72N¯(r, fm) +12N¯(r,1fm) + S(r, f).By Lemma 2.1 for p = 2, we getmT (r, f) ≤ (k + 92)N(r, f) + (k +112)N(r,1f) + S(r, f)So(k +92)Θ(∞, f) + (k + 112)Θ(0, f) ≤ 2k + 10−mwhich contradicts with (1.8).Sub case 1.3. l = 0. It is easy to see thatN11(r,1F − 1) + 2N(2E (r,1G− 1) +NL(r,1F − 1) + 2NL(r,1G− 1)≤ N(r, 1G− 1) + S(r, F )≤ T (r,G) + S(r, F ) + S(r, F )(3.10)ON MEROMORPHIC FUNCTIONS THAT SHARE ... 33NL(r,1F − 1) ≤ N(r,1F − 1)−N(r,1F − 1≤ N(r, FF ′) ≤ N(r, F′F) + S(r, F )≤ N(r, 1F) +N(r, F ) + S(r, F ).(3.11)Similarly, we haveNL(r,1G− 1) ≤ N(r,1G) +N(r,G) + S(r, F )≤ N1(r, 1G) +N(r, F ) + S(r,G).(3.12)Combining (3.4) and (3.10)− (3.12), we getT (r, F ) ≤ N2(r, 1F) +N2(r,1G) + 2N(r,1F)+ 6N(r, F ) +N1(r,1G) + S(r, F )(3.13)that ismT (r, f) ≤ N2(r, 1fm) +N2(r,1(fn)(k)) + 2N(r,1fm)+ 6N(r,1fm) +N1(r,1(fn)(k)) + S(r, f).By Lemma 2.1 for p = 2 and for p = 1 respectively, we getmT (r, f) ≤ (2k + 8)N(r, 1f) + (2k + 7)N(r, f).So(2k + 7)Θ(∞, f) + (2k + 8)Θ(0, f) ≤ 4k + 15−mwhich contradicts with (1.9).Case 2. Let H ≡ 0.on integration we get from (2.1)1F − 1 ≡CG− 1 +D, (3.14)where C, D are constants and C 6= 0. we will prove that D = 0.Sub case 2.1. Suppose D 6= 0. If z0 be a pole of f with multiplicity p such thata(z0) 6= 0,∞, then it is a pole of G with multiplicity np + k respectively. Thiscontradicts (3.14). It follows that N(r, f) = S(r, f) and hence Θ(∞, f) = 1. Also itis clear that N(r, f) = N(r,G) = S(r, f). From (1.7)-(1.9) we know respectively(k + 5)Θ(0, f) > k + 5−m (3.15)(k +112)Θ(0, f) > k +112−m (3.16)and(2k + 8)Θ(0, f) > 2k + 8−m (3.17)Since D 6= 0, from (3.14) we getN(r,1F − (1 + 1D ))= N(r,G) = S(r, f)34 HARINA P. WAGHAMORE AND RAJESHWARI S.Suppose D 6= −1.Using the second fundamental theorem for F we getT (r, F ) ≤ N(r, F ) +N(r, 1F) +N(r,1F − (1 + 1D ))≤ N(r, 1F) + S(r, f)i.e.,mT (r, F ) ≤ N(r, 1F) + S(r, f)≤ mT (r, f) + S(r, f).So, we have mT (r, f) = N(r, 1f ) and so Θ(0, f) = 1−m. Which contradicts (3.15)−(3.17).If D = −1, thenFF − 1 ≡ C1G− 1 (3.18)and from which we knowN(r, 1F ) = N(r,G) = S(r, f) and hence, N(r,1F ) = S(r, f).If C 6= −1,we know from (3.18) thatN(r,1G− (1 + C))= N(r, F ) = S(r, f).So from Lemma 2.1 and the Second fundamental theorem we getT (r, (fn)(k)) ≤ N(r,G) +N(r, 1G) +N(r,1G− (1 + C))+ S(r, f)≤ N(r,1(fn)(k))+ S(r, f)mT (r, f) ≤ (k + 1)N(r, 1f) + kN(r, f) + S(r, f),which is absurd. So C = −1 and we get from (3.18) that FG ≡ 1, which implies[(fn)(k)fn]= a2fn+m .In view of the first fundamental theorem, we get from above(n+m)T (r, f) ≤ k[N(r, f) +N(r, 1f)] + S(r, f) = S(r, f),which is impossible.Sub case 2.2. D = 0 and so from (3.14) we getG− 1 ≡ C(F − 1).If C 6= 1, thenG ≡ C(F − 1 + 1C)and N(r,1G) = N(r,1F − (1− 1C )).ON MEROMORPHIC FUNCTIONS THAT SHARE ... 35By the second fundamental theorem and Lemma 2.1 for p = 1 and Lemma 2.3 wehavemT (r, f) + S(r, f) = T (r, F )≤ N(r, F ) +N(r, 1F) +(r,1F − (1− 1C ))+ S(r,G)≤ N(r, fm) +N(r, 1fm) +N(r,1(fn)(k))+ S(r, f)≤ N(r, f) +N(r, 1f) + (k + 1)N(r,1f) + kN(r, f) + S(r, f)≤ (k + 2)N(r, 1f) + (k + 1)N(r, f) + S(r, f).Hence(k + 1)Θ(∞, f) + (k + 2)Θ(0, f) ≤ 2k + 3−m.So, it follows that(k + 4)Θ(∞, f) + (k + 5)Θ(0, f) ≤ 3Θ(∞, f) + (k + 1)Θ(∞, f)+ (k + 3)Θ(0, f) + 2Θ(0, f)≤ 2k + 9−m(k +92)Θ(∞, f) + (k + 112)Θ(0, f) ≤ 2k + 10−m,and(2k + 7)Θ(∞, f) + (2k + 8)Θ(0, f) ≤ 4k + 15−m.This contradicts (1.7) − (1.9). Hence C = 1 and so F ≡ G, that is fm ≡ (fn)(k).This completes the proof of the theorem.Acknowledgments. The authors would like to thank the anonymous referee forhis/her comments that helped us improve this article.References[1] W. K. Hayman, Meromorphic functions, Clarendon Press, Oxford (1964).[2] L. Yang, Distribution Theory, Springer Verlag, Berlin (1993).[3] I.Lahiri, Weighted sharing and uniqueness of meromorphic function, Nagoya Math. J., 161,193-206(2001).[4] R.Bruck, On entire functions which share one value CM with their first derivative, Result.Math. 30, 21-24(1996).[5] L.Z.Yang, Solution of a differential equation and its applications, Kodai. Math. J. 22, 458-464(1999).[6] K.W.Yu, On entire and meromorphic functions that share small functions with their deriva-tives, J. Inequal. Pure Appl. Math. 4(1)(2003) Art.21(Online:http://jipam.vu.edu.au/).[7] L.P.Liu and Y.X.Gu, Uniqueness of meromorphic functions that share one small functionwith their derivatives, Kodai. Math. J. 27, 272-279(2004).[8] I.Lahiri and A.Sarkar, Uniqueness of meromorphic function and its derivative, J. Inequal.Pure Appl. Math. 5(1)(2004) Art.20 (Online:http://jipam.vu.edu.au/).[9] Q.C. Zhang, Meromorphic function that shares one small function with their derivatives, J.Inequal. Pure. Appl. Math. 6(4)(2005) Art.116 (Online:http://jipam.vu.edu.au/).[10] Jin-Dong Li and Guang-Xin Huang, On meromorphic functions that share one small functionwith their derivatives, Palestine Journal of Mathematics, Vol. 4(1) (2015), 91-96.36 HARINA P. WAGHAMORE AND RAJESHWARI S.[11] A.Z. Mohon’ko, On the Nevanlinna charecteristics of some meromorphic functions, Theoryof Functions, Funct. Anal. Appl. 14 (1971), 83-87.HARINA P.WAGHAMOREDepartment of Mathematics, Central College Campus, Bangalore University, Bangalore-560 001, INDIAE-mail address: harinapw@gmail.comRAJESHWARI S.Department of Mathematics, Central College Campus, Bangalore University, Bangalore-560 001, INDIAE-mail address: rajeshwaripreetham@gmail.com

On meromorphic functions that share a small function with its derivatives.

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On meromorphic functions that share a small function with its derivatives.

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