Research Article Uniqueness Theorems of Difference Operator on Entire Functions

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1 Discrete Dynamics in Nature and Society Volume 203, Article ID 3649, 6 pages Research Article Uniqueness Theorems of Difference Operator on Entire Functions Jie Ding School of Mathematics, TaiYuan University of Technology, TaiYuan, ShanXi , China Correspondence should be addressed to Jie Ding; jieding984@gmail.com Received 22 October 202; Accepted 6 December 202 Academic Editor: Yanbin Sang Copyright 203 Jie Ding. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We investigate the uniqueness questions of the difference operator on entire functions and obtain three uniqueness theorems using the idea of weight sharing.. Introduction Afunctionf(z) is called meromorphic, if it is analytic in the complex plane except at poles. It is assumed that the reader is familiar with the standard symbols and fundamental results of Nevanlinna theory such as the characteristic function T(r, f), and proximity function m(r, f), counting function f) (see [, 2]). In addition we use S(r, f) denotes any quantity that satisfies the condition: S(r, f) = o(t(r, f)) as r possibly outside an exceptional set of finite logarithmic measure. Let f and g be two nonconstant meromorphic functions, a C { }, wesaythatf and g share the value a IM (ignoring multiplicities) if f aand g ahavethesamezeros, they share the value a CM (counting multiplicities) if f a and g ahave the same zeros with the same multiplicities. When a= the zeros of f amean the poles of f (see [2]). Let p be a positive integer and a C { }. Weuse N p) (r, /(f a)) to denote the counting function of the zeros of f a(counted with proper multiplicities) whose multiplicities are not bigger than p, N (p+ (r, /(f a)) to denote the counting function of the zeros of f awhose multiplicities are not less than p+. N p) (r, /(f a)) and N (p+ (r, /(f a)) denote their corresponding reduced counting functions (ignoring multiplicities), respectively. We denote by E(a, f) the set of zeros of f awith multiplicity, E p) (a, f) the set of zeros of f a(counted with proper multiplicities) whose multiplicities are not greater than p. In 997, Yang and Hua(see[3]) studied the uniqueness of the differential monomials and obtained the following result. Theorem A. Let f and g be nonconstant entire functions, and let n 3 be an integer. If f n f and g n g share CM, then either f(z) = c exp cz, g(z) = c 2 exp cz,wherec, c 2,andc are constants satisfying (c c 2 ) n+ c 2 = or f=tgfor a constant such that t n+ =. Recently, a number of papers (including [, 3 7]) have focused on complex difference equations and differences analogues of Nevanlinna theory. In particular, Qi et al. (see [6]) proved Theorem B, which can be considered as a difference counterpart of Theorem A. Theorem B. Let f and g be transcendental entire functions with finite order, c be a nonzero complex constant and n 6 be an integer. If f n f(z + c) and g n g(z + c) share CM, then fg = t or f=t 2 g for some constant t and t 2 which satisfy t n+ =and t n+ 2 =. In 20, Zhang et al. (see [7]) investigated the distribution of zeros and shared values of the difference operator on meromorphic functions and uniqueness of difference polynomials with the same points or fixed points. They obtained the following results. Theorem C. Let f and g be nonconstant entire functions of finite order, and let n 5 be an integer. Suppose that c is a

2 2 Discrete Dynamics in Nature and Society nonzero complex constant such that f(z + c) f(z) 0and g(z + c) g(z) 0.Iff n [f(z + c) f(z)] and g n [g(z + c) g(z)] share CM, and g(z + c) and g(z) share 0 CM, then f(z) = c exp az and g(z) = c 2 exp az,wherec, c 2,anda are constants satisfying (c c 2 ) n+ (exp az +exp az 2) = or f=tg for a constant such that t n+ =. Theorem D. Let f and g be nonconstant entire functions of finite order, and let n 5 be an integer. Suppose that c is a nonzero complex constant such that f(z + c) f(z) 0and g(z + c) g(z) 0.Iff n [f(z + c) f(z)] and g n [g(z + c) g(z)] share z CM, and g(z+c) and g(z) share 0 CM, then f=tg,wheret is a constant satisfying t n+ =. We investigate the uniqueness theorem of another differences polynomial and prove Theorem. Theorem. Let f and g be nonconstant transcendental entire functions of finite order, and let n 9be an integer. Suppose that c is a nonzero real constant such that f(z+2c)+f(z+c)+ f(z) = g(z+2c)+g(z+c)+g(z) and f(z+2c)+f(z+c)+f(z) 0, g(z + 2c) + g(z + c) + g(z) 0.IfE 3) (z, f n [f(z + 2c) + f(z+c)+f(z)]) = E 3) (z, g n [g(z+2c)+g(z+c)+g(z)]),then f(z) = tg(z), wheret is a constant satisfying t n+ =except that t=. In paper [5], Wang et al. improved the Theorem B and proved the following result. Theorem E. Let f and g be transcendental entire functions with finite order, c be a nonzero complex constant and n 6 be an integer. If E 3) (, f n f(z + c)) = E 3) (, g n g(z + c)), then fg = t or f=t 2 g for some constant t and t 2 which satisfy t n+ =and t n+ 2 =. The purpose of this paper is to induce the idea of weight sharingtotheoremscandd,theresultsasfollow. 2. Some Lemmas In order to prove our theorems, we need the following Lemmas. Lemma 5 is a difference analogue of the logarithmic derivative lemma, given by Halburd and Korhonen [9] and Chiang and Feng [7] independently. Lemma 5 (see [9]). Let f(z) be a meromorphic function of finite order, and let c C and δ (0, ).Then m (r, f (z+c) ) + m (r, f (z+c) ) =O( T(r,f) )=S(r,f), r δ for all r outside of a possibly exceptional set with finite logarithmic measure. Lemma 6 (see [2]). Let f(z) be a nonconstant meromorphic function, and let P(f) = a 0 f n +a f n + a n,wherea 0 ( =0), a,...,a n are small functions of f.then () T (r, P(f)) = nt (r, f) + S (r, f). (2) Lemma 7 (see [8]). Let f and g be two nonconstant meromorphicfunctionsatisfyinge k) (, f) = E k) (, g) for some positive integer k N.DefineH as follow If H 0,then H=( f f 2f f ) (g g 2g g ). (3) N (r, H) N (2 (r, f) + N (2 (r, f ) Theorem 2. Let f and g be nonconstant entire functions of finite order, and let n 6 be an integer. Suppose that c is a nonzero complex constant such that f(z + c) f(z) 0 and g(z + c) g(z) 0.IfE 3) (, f n [f(z + c) f(z)]) = E 3) (, g n [g(z + c) g(z)]) share CM, g(z + c) and g(z) share 0CM,thenf(z) = c exp az and g(z) = c 2 exp az,wherec, c 2, and a are constants satisfying (c c 2 ) n+ (exp az +exp az 2) = or f=tgfor a constant such that t n+ =. + N (2 (r, g) + N (2 (r, g )+N 0 (r, + N (k+ (r, g )+N (k+ (r, f ) f ) ) + S (r, f) + S (r, g), g (4) Theorem 3. Let f and g be nonconstant entire functions of finite order, and let n 6 be an integer. Suppose that c is a non-zero complex constant such that f(z + c) f(z) 0 and g(z + c) g(z) 0.IfE 3) (, f n [f(z + c) f(z)]) = E 3) (, g n [g(z + c) g(z)]) share z CM, g(z + c) and g(z) share 0 CM, then f = tg,wheret is a constant satisfying t n+ =. Remark 4. Some ideas of this paper are based on [5, 7]. where N 0 (r, /f ) denotes the counting function of zeros of f but not zeros of f(f ),andn 0 (r, /g ) is similarly defined. Lemma 8 (see [9]). Under the conditions of Lemma 7, we have N ) (r, f )=N ) (r, g ) H) +S(r,f)+S(r,g). (5)

3 Discrete Dynamics in Nature and Society 3 Lemma 9 (see [9]). If H=(f /f 2f /(f )) (g /g 2g /(g )) 0, then either f gor fg provided that f)+g)+/f)+/g) lim sup, r,r I T (r) (6) where T(r) := max{t(r, f), T(r, g)} and I is a set with infinite linear measure. Lemma 0. Let f(z) be a meromorphic function of finite order, c C.Then N (r, f (z+c)) = ) +S(r, ). (7) Proof. Using Lemma5 andtheformula(2)in[2] N (r, f (z+c)) ) +S(r, ). (8) Replacing f(z) with f(z c),wehave f(z)) f(z c))+s(r,f(z c)) for every c C,sowededucethat =f(z c))+s(r,f(z)), (9) f(z)) f(z+c))+s(r,f(z)). (0) From (8)and(0), we obtain that f(z+c)) =f(z))+s(r,f(z)). () Thus we completed the proof. Lemma (see [9]). Let T : (0, + ) (0, + ) be a nondecreasing continuous function, s > 0, 0 < α <, and let F R + be the set of all r satisfy T (r) αt(r+s). (2) If the logarithmic measure of F is infinite, then 3. Proof of Theorems Proof of Theorem. We define log T(r,f) lim sup =. (3) r log r F (z) = fn (z) [f (z+2c) +f(z+c) +f(z)], z G (z) = gn (z) [g (z+2c) +g(z+c) +g(z)]. z (4) In Lemma 7,wereplacefand g,byfand G respectively, we claim that H 0.Ifitisnottrue,thenH 0.From Lemma 8 we have that rln ) (r, F ) =N ) (r, ) H) G +S(r,f)+S(r,g) N (2 (r, F )+N (2 (r, G ) F )+N 0 (r, F )+N (4 (r, G ) +S(r,f)+S(r,g). (5) From the Nevanlinna second foundational theorem, we can get that F )+ G ) F )+N 0 (r, + F )+ G ) +S(r,f)+S(r,g). (6) From the definitions of N k) and N (k, the following inequalities are obvious: F ) 2 N ) (r, 2 F ), G ) 2 N ) (r, 2 G ). F )+N (4 (r, G )+N (4 (r, F ) Combining (5), (6), and (7), we deduce that 2 F )+2N (2 (r, F ) G ) (7)

4 4 Discrete Dynamics in Nature and Society +2 G )+2N (2 (r, G )+S(r,f)+S(r,g) 4N (r, f )+4N (r, g ) + 2N (r, f (z+c) ) + 2N (r, g (z+c) ) + 2N (r, f (z+2c) ) + 2N (r, g (z+2c) ) +S(r,f)+S(r,g) 8N (r, f ) + 8N (r, g )+S(r,f)+S(r,g) 8T (r, f ) + 8T (r, g )+S(r,f)+S(r,g). (8) We can apply Lemma 5, Lemma 6,andLemma 0 to show that (n+) T(r,f) =T(r,f n+ ) T(r, F (z)) + T (r, f (z+2) +f(z+) +f(z) )+S(r,f(z)) =T(r, F (z)) + N (r, +S(r,f(z)) f (z+2) +f(z+) +f(z) ) T(r, F (z)) +T(r,f(z))+S(r,f(z)), which implies (9) T (r, F) nt(r,f)+s(r,f). (20) The same augment as above, we have that T (r, G) nt(r,g)+s(r,g). (2) Because of Lemma 9,wehavethatF Gor FG.We will consider the following two cases. Case. Suppose that F(z) = G(z).Then f n (z) [f (z+2c) +f(z+c) +f(z)] =g n (z) [g (z+2c) +g(z+c) +g(z)]. Let h(z) = f(z)/g(z),wededucethat h n (z) [h (z+2c) g (z+2c) +h(z+c) g (z+c) +h(z) g (z)] =g(z+2c) +g(z+c) +g(z). (24) (25) If h(z + c) h(z), bythehypothesisf(z + 2c) + f(z + c) + f(z) =g(z+2c)+g(z+c)+g(z),wegetthath(z) =.So T(r,h n ) + h n )+hn ) h n ) 2T(r, h) +S(r, h), (26) which means h is a constant, because of n 0. Then h(z) = t and t is a constant satisfying t n+ =except that t=. Case 2. Suppose that F(z) G(z).Then f n (z) [f (z+2c) +f(z+c) +f(z)] g n (z) [g (z+2c) +g(z+c) +g(z)] =z 2. (27) Note that zero is a Picard exceptional value of f and g, then f(z) = e P(z) and g(z) = e Q(z),whereP(z) and Q(z) are polynomials. In (27), we let z=0,then e np(0) [e P(2c) +e P(c) +e P(0) ]e nq(0) [e Q(2c) +e Q(c) +e Q(0) ]=0. (28) It is impossible, because of c is a real number. From (8), (20), and (2), we can deduce that (n 8) [T(r,f)+T(r,g)] S(r,f)+S(r,g). (22) which is a contraction. Therefore, H 0. Noting that Proof of Theorem 2. Denoting F (z) =f n (z) [f (z+c) f(z)], G (z) =g n (z) [g (z+c) g(z)]. (29) F )+ G ) 4[T(r,f)+T(r,g)] +S(r,f)+S(r,g) T(r), where T(r) = max T(r, F), T(r, G). (23) In Lemma 7,wereplacefand g,byfand G respectively. If H 0,byLemma8we deduce that N ) (r, F ) =N ) (r, G )

5 Discrete Dynamics in Nature and Society 5 H) +S(r,f)+S(r,g) + 6N (r, g )+S(r,f)+S(r,g) N (2 (r, F )+N (2 (r, G ) F )+N 0 (r, F )+N (4 (r, G ) +S(r,f)+S(r,g). (30) 6T (r, f ) + 6T (r, g )+S(r,f)+S(r,g). We can apply Lemmas 5, 6,and0 to show that (n+) T(r,f) =T(r,f n+ ) T(r, F (z)) (32) The same reasons as in the proof of Theorem, wehave that F )+ G ) F )+N 0 (r, + F )+ G ) +S(r,f)+S(r,g), F ) 2 N ) (r, F ) F ) 2 F ), G ) 2 N ) (r, G ) G ) 2 G ). Combining (30)and(3), we deduce that (3) + T (r, f (z+c) f(z) )+S(r,f(z)) f (z+c) f(z) =T(r, F (z)) + N (r, )+S(r,f(z)) T(r, F (z)) +T(r,f(z))+S(r,f(z)), (33) which implies T (r, F) nt(r, f) +S(r, f). (34) We have that (g(z + c) g(z))/g(z)),sinceg(z + c) and g(z) share 0 CM, then T (r, G) (n+) T (r, g) +S(r, g). (35) From (32), (34), and (35), we can deduce that (n 6) T(r,f)+(n 5) T(r,g) S(r,f)+S(r,g). which is a contraction. Therefore, H 0. Noting that F )+ G ) (36) 2 F )+2N (2 (r, F ) 3 [T (r, f) + T (r, g)] + S (r, f) (37) +2 G )+2N (2 (r, G )+S(r,f)+S(r,g) 4N (r, f )+4N (r, g ) + 2N (r, f (z+c) f(z) ) + 2N (r, g (z+c) g(z) )+S(r,f)+S(r,g) 4N (r, )+2T(r,f(z+c) f(z)) f + 6N (r, g )+S(r,f)+S(r,g) =4N (r, )+2m(r,f(z+c) f(z)) f +S(r,g) T(r), where T(r) = max T(r, F), T(r, G). Because of Lemma 9,wehavethatF Gor FG. By using the same methods as in the proof of Theorem.0 in [7], we can complete the proof of Theorem 2. Proof of Theorem 3. Theproofisalmostliterallythesameas the proof of Theorem 2, with the methods as in the proof of Theorem.9 in [7] replacing the methods as in the proof of Theorem.0 in [7]. Acknowledgment This research was supported by the funds of Taiyuan University of Technology.

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