CUBO, A Mathematical Journal Vol. 24, no. 01, pp. 167–186, April 2022 DOI: 10.4067/S0719-06462022000100167 Uniqueness of entire functions whose difference polynomials share a polynomial with finite weight Goutam Haldar 1 1Department of Mathematics, Malda College, Rabindra Avenue, Malda, West Bengal 732101, India. goutamiit1986@gmail.com goutamiitm@gmail.com ABSTRACT In this paper, we use the concept of weighted sharing of val- ues to investigate the uniqueness results when two difference polynomials of entire functions share a nonzero polynomial with finite weight. Our result improves and extends some re- cent results due to Sahoo-Karmakar [J. Cont. Math. Anal. 52(2) (2017), 102–110] and that of Li et al. [Bull. Malays. Math. Sci. Soc., 39 (2016), 499–515]. Some examples have been exhibited which are relevant to the content of the paper. RESUMEN En este art́ıculo, usamos el concepto de intercambio pesado de valores para investigar los resultados de unicidad cuando dos polinomios de diferencia de funciones enteras comparten un polinomio no cero con peso finito. Nuestro resultado mejora y extiende algunos resultados de Sahoo-Karmakar [J. Cont. Math. Anal. 52(2) (2017), 102–110] y los de Li et al. [Bull. Malays. Math. Sci. Soc., 39 (2016), 499–515]. Se ex- hiben algunos ejemplos que son relevantes para el contenido del art́ıculo. Keywords and Phrases: Entire function, difference polynomial, shift and difference operator, weighted sharing. 2020 AMS Mathematics Subject Classification: 30D35, 39A10. Accepted: 28 February, 2022 Received: 05 June, 2021 c©2022 G. Haldar. This open access article is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License. http://cubo.ufro.cl/ http://dx.doi.org/10.4067/S0719-06462022000100167 https://orcid.org/0000-0001-8169-6190 mailto:goutamiit1986@gmail.com mailto:goutamiitm@gmail.com 168 G. Haldar CUBO 24, 1 (2022) 1 Introduction Let f and g be two non-constant meromorphic functions defined in the open complex plane C. If for some a ∈ C ∪ {∞}, the zero of f − a and g − a have the same locations as well as same multiplicities, we say that f and g share the value a CM (counting multiplicities), and if we do not consider the multiplicities into account, then f and g are said to share the value a IM (ignoring multiplicities)(see [37]). We adopt the standard notations of the Nevanlinna theory of meromorphic functions (see [14, 22, 41]). For a non-constant meromorphic function f, we denote by T (r, f) the Nevanlinna characteristic function of f and by S(r, f) any quantity satisfying S(r, f) = o{T (r, f)} as r → ∞ outside of an exceptional set of finite linear measure. We define shift and difference operators of f(z) by f(z + c) and ∆cf(z) = f(z + c) − f(z), respectively. Note that ∆nc f(z) = ∆ n−1 c (∆cf(z)), where c is a nonzero complex number and n ≥ 2 is a positive integer. For further generalization of ∆cf, we now define the linear difference operator of an entire (mero- morphic) function f as Lc(f) = f(z + c) + c0f(z), where c0 is a finite complex constant. Clearly, for the particular choice of the constant c0 = −1, we get Lc(f) = ∆cf. In 1959, Hayman [13] proved the following result. Theorem A ([13]). Let f be a transcendental entire function and let n be an integer such that n ≥ 1. Then fnf′ = 1 has infinitely many solutions. A number of authors have shown their interest to find the uniqueness of entire and meromorphic functions whose differential polynomials share certain values or fixed points, and obtained some remarkable results (see [3, 9, 10, 26, 33, 34, 36, 37, 39, 42]). In recent years, the difference variant of the Nevanlinna theory has been established in [8, 11, 12]. Using these theories, some mathematicians in the world began to study the uniqueness questions of meromorphic functions sharing values with their shifts, and study the value distribution of the nonlinear difference polynomials, and produced many fine works, for example, see [1, 5, 6, 7, 11, 15, 16, 23, 27, 29, 30, 31, 40, 44]. We recall the following result from Laine-Yang [23]. Theorem B ([23]). Let f be a transcendental entire function of finite order, and c be a non-zero complex constant. Then, for n ≥ 2, f(z)nf(z + c) assumes every non-zero value a ∈ C infinitely often. Later on, Liu-Yang [28] extended Theorem B, and proved the following result: Theorem C ([28]). Let f be a transcendental entire function of finite order, and let η be a nonzero complex constant. Then for n ≥ 2 the function f(z)nf(z + η) − P0(z) has infinitely many zeros, where P0 is any given polynomial such that P0 6≡ 0. CUBO 24, 1 (2022) Uniqueness of entire functions 169 Regarding uniqueness corresponding to Theorem C, Li et al. [24] obtained the following result. Theorem D ([24]). Let f and g be two distinct transcendental entire functions of finite order, and let P0 6≡ 0 be a polynomial. Let η is a nonzero complex constant and n ≥ 4 is an integer such that 2 deg(P0) < n + 1. Also, suppose that f(z) nf(z + η) − P0(z) and g(z) ng(z + η) − P0(z) share 0 CM. Then one of the following assertions holds. (I) If n ≥ 4 and f(z)nf(z + η)/P0(z) is a Mobius transformation of g(z) ng(z + η)/P0(z), then either (i) f = tg, where t is a constant satisfying tn+1 = 1 (ii) f = eQ and g = te−Q, where P0 reduces to a nonzero constant c, t is a constant such that tn+1 = c2, and Q is a non-constant polynomial. (II) If n ≥ 6, then (I)(i) or (I)(ii) holds. In 2016, Li-Li [25] obtained the IM analogues of the above Theorem D as follows. Theorem E ([25]). Let f and g be two distinct transcendental entire functions of finite order, and let P0 6≡ 0 be a polynomial. Let η is a nonzero complex constant and n ≥ 4 is an integer such that 2 deg(P0) < n + 1. Also, suppose that f(z) nf(z + η) − P0(z) and g(z) ng(z + η) − P0(z) share 0 IM. Then one of the following assertions holds. (I) If n ≥ 4 and f(z)nf(z + η)/P0(z) is a Mobius transformation of g(z) ng(z + η)/P0(z), then either (i) f = tg, where t is a constant satisfying tn+1 = 1, (ii) f = eQ and g = te−Q, where P0 reduces to a nonzero constant c, t is a constant such that tn+1 = c2, and Q is a non-constant polynomial. (II) If n ≥ 12, then (I)(i) or (I)(ii) holds. In 2001, the notion of weighted sharing was originally defined in the literature ([18, 19]), which is the gradual change of shared values from CM to IM. Below we recall the definition. Definition 1.1 ([18, 19]). Let k be a non-negative integer or infinity. For a ∈ C∪{∞}, we denote by Ek(a; f) the set of all a-points of f, where an a-point of multiplicity m is counted m times if m ≤ k and k + 1 times if m > k. If Ek(a; f) = Ek(a; g), we say that f, g share the value a with weight k. Clearly, if f, g share (a, k) then f, g share (a, p) for any integer p, 0 ≤ p < k. Also we note that f, g share a value a IM or CM if and only if f, g share (a, 0) or (a, ∞), respectively. Using the notion of weighted sharing, Sahoo-Karmakar [35] further improved Theorem D as follows. 170 G. Haldar CUBO 24, 1 (2022) Theorem F ([35]). Let f, g, P0 and n be defined as in Theorem D. Suppose that f(z) nf(z + η) − P0(z) and g(z) ng(z + η) − P0(z) share (0, 2). (I) If n ≥ 4 and f(z)nf(z + η)/P0(z) is a Mobius transformation of g(z) ng(z + η)/P0(z), then either (i) f = tg, where t is a constant satisfying tn+1 = 1 (ii) f = eQ and g = te−Q, where P0 reduces to a nonzero constant c, t is a constant such that tn+1 = c2, and Q is a non-constant polynomial. (II) If n ≥ 6, then (I)(i) or (I)(ii) holds. Observing the above results, it is natural to ask the following questions. Question 1.2. What can be said about the relationship of two finite order non-constant mero- morphic functions f and g if their more general nonlinear difference polynomials f(z)nLc(f) and g(z)nLc(g) share a polynomial P(z) 6≡ 0, where Lc(f) = f(z +c)+c0f(z) with c and c0 being finite nonzero complex constants, and n ≥ 2 being a positive integer? Question 1.3. Is it possible to further reduce the nature of sharing from (0, 2) to (0, 1) in Theorem F? Question 1.4. Can the lower bound of n be further reduced in Theorems E and F? Question 1.5. What can be said about the uniqueness of f and g if we consider the difference polynomial of the form f(z)n∆cf and g(z) n∆cg in Theorems E and F? The purpose of this paper is to answer all the questions raised above. In fact we have been successfully able to reduce the nature of sharing of f(z)nf(z +η)−P0(z) and g(z) ng(z +η)−P0(z) in Theorem F. We have also reduced the lower bound of n in Theorems E and F successfully. 2 Main results Now we state our main result. Theorem 2.1. Let f and g be two transcendental entire functions of finite order, P 6≡ 0 be a polynomial. Let c be a non-zero complex constant, and n be a positive integer such that 2 deg(P) < n + 1. Let l be a non-negative integer such that f(z)nLc(f) − P(z) and g(z) nLc(g) − P(z) share (0, l) and g(z), g(z + c) share 0 CM. If n ≥ 4 and f(z)nLc(f)/P(z) is a Mobius transformation of g(z)nLc(g)/P(z), or one of the following conditions holds: (i) l ≥ 2 and n ≥ 5; CUBO 24, 1 (2022) Uniqueness of entire functions 171 (ii) l = 1 and n ≥ 6; (iii) l = 0 and n ≥ 11, then one of the following conclusions can be realized: (a) f = tg, where t is a constant satisfying tn+1 = 1; (b) When c0 = 0, f = e U and g = te−U, where P(z) reduces to a nonzero constant d, t is a constant such that tn+1 = d2 and U is a non-constant polynomial; (c) When c0 6= 0, f = c1e az, g(z) = c2e −az, where a, c1, c2 and d are non-zero constants satisfying (c1c2) n+1(eac + c0)(e −ac + c0) = d 2. If Lc(f) = ∆cf, then one can easily get the following corollary from Theorem 2.1 which answers Question 1.5. Corollary 2.2. Let f and g be two transcendental entire functions of finite order, P 6≡ 0 be a polynomial. Let c be a non-zero complex constant, and n be a positive integer such that 2 deg(P) < n + 1. Let l be a non-negative integer such that f(z)n∆cf − P(z) and g(z) n∆cg − P(z) share (0, l) and g(z), g(z + c) share 0 CM. If n ≥ 4 and f(z)n∆c(f)/P(z) is a Mobius transformation of g(z)n∆c(g)/P(z), or one of the following conditions holds: (i) l ≥ 2 and n ≥ 5; (ii) l = 1 and n ≥ 6; (iii) l = 0 and n ≥ 11, then one of the following conclusions can be realized: (a) f = tg, where t is a constant satisfying tn+1 = 1; (b) f = c1e az, g(z) = c2e −az, where a, c1, c2 and d are non-zero constants satisfying (c1c2) n+1(eac + c0)(e −ac + c0) = d 2. The following examples show that both the conclusions of Theorem 2.1 actually holds. Example 2.3. Let f(z) = ez and g = tf, where t is a constant such that tn+1 = 1, and η be any non-zero complex constant. Then for any given polynomial p such that p 6≡ 0 with 2 deg(p) < n+1, f(z)nf(z + η)− p(z) and g(z)ng(z + η)− p(z) share (0, ∞). Also f(z)n(f(z + η)− f(z))− p(z) and g(z)n(g(z + η) − g(z)) − p(z)share (0, ∞). Here f and g satisfy the conclusion (a) of Theorem 2.1. Example 2.4. Let f(z) = e2πiz/η and g(z) = te−2πiz/η, where t is a constant such that tn+1 = 1, η is a non-zero complex constant. Then f(z)nf(z + η) and g(z)ng(z + η) share (1, ∞). Here f and g satisfy the conclusion (b) of Theorem 2.1. Example 2.5. Let f(z) = ez, g(z) = e−z, η = − log(−1) and P(z) = 2. Then one can easily verify that f(z)n(f(z + η) − f(z)) and g(z)n(g(z + η) − g(z)) share (2, ∞). Here f and g satisfy the conclusion (b) of Theorem 2.1. 172 G. Haldar CUBO 24, 1 (2022) The following example shows that Theorem 2.1 is not true for infinite order entire functions. Example 2.6. Let f(z) = e2πiz/η ee 2πiz/η and g(z) = 1 ee 2πiz/η , where η is a non-zero constant. Then it is easy to verify that f(z)nf(z + η) and g(z)ng(z + η) share (1, ∞). But there does not exist a non-zero constant t such that f = tg or fg = t, where tn+1 = 1. 3 Auxiliary definitions Throughout the paper we have used the following definitions and notations. Definition 3.1 ([17]). Let a ∈ C ∪ {∞}. We denote by N(r, a; f |= 1) the counting function of simple a points of f. For p ∈ N we denote by N(r, a; f |≤ p) the counting function of those a-points of f (counted with multiplicities) whose multiplicities are not greater than p. By N(r, a; f |≤ p) we denote the corresponding reduced counting function. In a similar manner we can define N(r, a; f |≥ p) and N(r, a; f |≥ p). Definition 3.2 ([19]). Let p ∈ N∪{∞}. We denote by Np(r, a; f) the counting function of a-points of f, where an a-point of multiplicity m is counted m times if m ≤ p and p times if m > p. Then Np(r, a; f) = N(r, a; f) + N(r, a; f |≥ 2) + · · · + N(r, a; f |≥ p). Clearly N1(r, a; f) = N(r, a; f). Definition 3.3 ([43]). Let f and g be two non-constant meromorphic functions such that f and g share (a, 0). Let z0 be an a-point of f with multiplicity p, an a-point of g with multiplicity q. We denote by NL(r, a; f) the reduced counting function of those a-points of f and g where p > q, by N 1) E (r, a; f) the counting function of those a-points of f and g where p = q = 1, by N (2 E (r, a; f) the reduced counting function of those a-points of f and g where p = q ≥ 2. In the same way we can define NL(r, a; g), N 1) E (r, a; g), N (2 E (r, a; g). In a similar manner we can define NL(r, a; f) and NL(r, a; g) for a ∈ C ∪ {∞}. When f and g share (a, m), m ≥ 1, then N 1) E (r, a; f) = N(r, a; f |= 1). Definition 3.4 ([19]). Let f, g share a value (a, 0). We denote by N∗(r, a; f, g) the reduced counting function of those a-points of f whose multiplicities differ from the multiplicities of the corresponding a-points of g. Clearly N∗(r, a; f, g) = N∗(r, a; g, f) and N∗(r, a; f, g) = NL(r, a; f)+ NL(r, a; g). 4 Some lemmas We now prove several lemmas which will play key roles in proving the main results of the paper. Let F and G be two non-constant meromorphic functions. Henceforth we shall denote by H the CUBO 24, 1 (2022) Uniqueness of entire functions 173 following function H = ( F ′′ F ′ − 2F ′ F − 1 ) − ( G′′ G′ − 2G′ G − 1 ) . (4.1) Lemma 4.1 ([8]). Let f(z) be a meromorphic function of finite order ρ, and let c be a fixed non-zero complex constant. Then for each ǫ > 0, we have T (r, f(z + c)) = T (r, f) + O(rρ−1+ǫ) + O{log r}. Lemma 4.2 ([8]). Let f(z) be a meromorphic function of finite order ρ and let c be a non-zero complex number. Then for each ǫ > 0, we have m ( r, f(z + c) f(z) ) + m ( r, f(z) f(z + c) ) = O(rρ−1+ǫ). Lemma 4.3 ([32]). Let f be a non-constant meromorphic function and let R(f) = n ∑ i=0 aif i/ m ∑ j=0 bjf j be an irreducible rational function in f with constant coefficients {ai} and {bj} where an 6= 0 and bm 6= 0. Then T (r, R(f)) = d T (r, f) + S(r, f), where d = max{n, m}. Lemma 4.4 ([25]). Let f and g be two transcendental entire functions of finite order, c 6= 0 be a complex constant, α(z) be a small function of f and g, P(z) = anz n + an−1z n−1 + · · · + a1z + a0 be a nonzero polynomial, where a0, a1, . . . , an(6= 0) are complex constants, and let n > Γ1 be an integer. If P(f)f(z + c) and P(g)g(z + c) share α(z) IM, then ρ(f) = ρ(g). Lemma 4.5. Let f be a transcendental entire function of finite order, and Lc(f) = f(z+c)+c0f(z), where c, c0 ∈ C − {0}. Then for n ∈ N, nT (r, f) + S(r, f) ≤ T (r, f(z)nLc(f)) ≤ (n + 1)T (r, f) + S(r, f). Proof. This lemma can be proved in a similar manner as done in the proof of Lemma 2.4 and Remark 2.1 of [30]. Remark 4.6. If c0 = 0, then Lc(f) = f(z + c) and therefore by Lemma 2.3 of [30], we can get T (r, f(z)nLc(f)) = (n + 1)T (r, f) + S(r, f). (4.2) Remark 4.7. If c0 6= 1, then the following example shows that one can not get equality just like (4.2). Example 4.8 ([30]). If f(z) = ez, ec = 2, c0 = −1, then T (r, f(z) nLc(f)) = T (r, e (n+1)z) = (n + 1)T (r, f) + S(r, f). If f(z) = ez + z, c = 2πi, c0 = −1, then T (r, f(z) nLc(f)) = T (r, 2πi(e z + z)n) = nT (r, f) + S(r, f). Remark 4.9. From the above example, it can be easily seen that f(z) and f(z + c) share 0 CM for the first one, but for the second one f(z) and f(z + c) do not share 0 CM. Regarding this one may ask, in order to get equality just like (4.2), is it sufficient to assume that f(z) and f(z + c) share 0 CM? In this direction, we prove the following lemma. 174 G. Haldar CUBO 24, 1 (2022) Lemma 4.10. Let F = f(z)nLc(f), where f(z) is an entire function of finite order, and f(z), f(z + c) share 0 CM. Then T (r, F) = (n + 1)T (r, f) + S(r, f). Proof. Keeping in view of Lemmas 4.1 and 4.3, we have T (r, F) = T (r, f(z)nLc(f)) = m(r, f nLc(f)) ≤ m(r, f(z)n) + m(r, Lc(f)) + S(r, f) ≤ T (f(z)n) + m ( r, Lc(f) f(z) ) + m(r, f(z)) + S(r, f) ≤ (n + 1)T (r, f) + S(r, f). Since f(z) and f(z + c) share 0 CM, we must have N ( r, ∞; Lc(f) f(z) ) = S(r, f). So, keeping in view of Lemmas 4.2 and 4.3, we obtain (n + 1)T (r, f) = T (r, f(z)n+1) = m(r, f(z)n+1P(f(z))) = m ( r, F f(z) Lc(f) ) ≤ m(r, F) + m ( r, f(z) Lc(f) ) + S(r, f) ≤ T (r, F) + T ( r, Lc(f) f(z) ) + S(r, f) = T (r, F) + N ( r, ∞; Lc(f) f(z) ) + m ( r, Lc(f) f(z) ) + S(r, f) = T (r, F) + S(r, f). From the above two inequalities, we must have T (r, F) = (n + 1)T (r, f) + S(r, f). Lemma 4.11 ([37]). Let F and G be non-constant meromorphic functions such that G is a Mobius transformation of F. Suppose that there exists a subset I ⊂ R+ with linear measure mesI = +∞ such that for r ∈ I and r −→ ∞ N(r, 0; F) + N(r, 0; G) + N(r, ∞; F) + N(r, ∞; G) < (λ + o(1))T (r, G), where λ < 1. If there exists a point z0 ∈ C satisfying F(z0) = G(z0) = 1, then either F = G or FG = 1. Lemma 4.12 ([38]). Let f(z) and g(z) be two non-constant meromorphic functions. Then N ( r, ∞; f g ) − N ( r, ∞; g f ) = N(r, ∞; f) + N(r, 0; g) − N(r, ∞; g) − N(r, 0; f). Lemma 4.13. Let f(z) be a transcendental entire function of finite order, c ∈ C–{0} be finite complex constant and n ∈ N. Let F(z) = f(z)nLc(f), where Lc(f) 6≡ 0. Then nT (r, f) ≤ T (r, F) − N(r, 0; Lc(f)) + S(r, f). CUBO 24, 1 (2022) Uniqueness of entire functions 175 Proof. Using Lemmas 4.2 and 4.12, and the first fundamental theorem of Nevanlinna, we obtain m(r, f(z)n+1) = m ( r, f(z)F Lc(f) ) ≤ m(r, F) + m ( r, f(z) Lc(f) ) + S(r, f) ≤ m(r, F) + T ( r, f(z) Lc(f) ) − N ( r, ∞; f(z) Lc(f) ) + S(r, f) ≤ m(r, F) + T ( r, Lc(f) f(z) ) − N ( r, ∞; f(z) Lc(f) ) + S(r, f) ≤ m(r, F) + N ( r, ∞; Lc(f) f(z) ) + m ( r, Lc(f) f(z) ) − N ( r, ∞; f(z) Lc(f) ) + S(r, f) ≤ m(r, F) + N(r, 0; f) − N(r, 0; Lc(f)) + S(r, f). i.e., m(f(z)n+1) ≤ T (r, F) + T (r, f) − N(r, 0; Lc(f)) + S(r, f). By Lemma 4.3, we obtain (n + 1)T (r, f) = m(r, fn+1) ≤ T (r, F) + T (r, f) − N(r, 0; Lc(f)) + S(r, f), i.e., nT (r, f) ≤ T (r, F) − N(r, 0; Lc(f)) + S(r, f). Lemma 4.14 ([2]). If f, g be two non-constant meromorphic functions sharing (1, 1), then 2NL(r, 1; f) + 2NL(r, 1; g) + N (2 E (r, 1; f) − Nf>2(r, 1; g) ≤ N(r, 1; g) − N(r, 1; g). Lemma 4.15 ([4]). If f, g be two non-constant meromorphic functions sharing (1, 1), then Nf>2(r, 1; g) ≤ 1 2 N(r, 0; f) + 1 2 N(r, ∞; f) − 1 2 N0(r, 0; f ′) + S(r, f), where N0(r, 0; f ′) is the counting function of those zeros of f′ which are not the zeros of f(f − 1). Lemma 4.16 ([43]). If f, g be two non-constant meromorphic functions sharing (1, 0) and H 6≡ 0, then N 1) E (r, 1; f) ≤ N(r, 0; H) + S(r, f) ≤ N(r, ∞; H) + S(r, f) + S(r, g). Lemma 4.17 ([4]). If f, g be two non-constant meromorphic functions such that they share (1, 0), then NL(r, 1; f) + 2NL(r, 1; g) + N (2 E (r, 1; f) − Nf>1(r, 1; g) − Ng>1(r, 1; f) ≤ N(r, 1; g) − N(r, 1; g). 176 G. Haldar CUBO 24, 1 (2022) Lemma 4.18 ([4]). If f, g be share (1, 0), then (i) NL(r, 1; f) ≤ N(r, 0; f) + N(r, ∞; f) + S(r, f). (ii) Nf>1(r, 1; g) ≤ N(r, 0; f) + N(r, ∞; f) − N0(r, 0; f ′) + S(r, f). (iii) Ng>1(r, 1; f) ≤ N(r, 0; g) + N(r, ∞; g) − N0(r, 0; g ′) + S(r, g). Lemma 4.19 ([20]). If f, g be be two non-constant meromorphic functions that share (1, 0), (∞, 0) and H 6≡ 0, then N(r, ∞; H) ≤ N(r, 0; f |≥ 2) + N(r, 0; g |≥ 2) + N∗(r, 1; f, g) + N∗(r, ∞; f, g) + N0(r, 0; f ′) + N0(r, 0; g ′) + S(r, f) + S(r, g), where N0(r, 0; f ′) is the reduced counting function of those zeros of f′ which are not the zeros of f(f − 1) and N0(r, 0; g ′) is similarly defined. Lemma 4.20 ([21]). If N(r, 0; f(k) | f 6= 0) denotes the counting function of those zeros of f(k) which are not the zeros of f, where a zero of f(k) is counted according to its multiplicity, then N ( r, 0; f(k) | f 6= 0 ) ≤ kN(r, ∞; f) + N (r, 0; f |< k) + kN (r, 0f |≥ k) + S(r, f). 5 Proofs of the theorems Proof of Theorem 2.1. Let F = f(z)nLc(f)/P(z) and G = g(z) nLc(g)/P(z). Then F and G are two transcendental meromorphic functions that share (1, l) except the zeros and poles of P(z). Since g(z) and g(z + c) share 0 CM, from Lemma 4.10, we obtain T (r, G) = (n + 1)T (r, g) + O{rρ(f)−1+ǫ} + O{log r}. (5.1) Since f and g are of finite order, it follows from Lemma (4.5) and (5.1) that F and G are also of finite order. Moreover, from Lemma 4.4 we deduce that ρ(f) = ρ(g) = ρ(F) = ρ(G). We consider the following two cases separately. Case 1: Suppose that F is a Mobius transformation of G, i.e., F = AG + B CG + D , (5.2) where A, B, C, D are complex constants satisfying AD−BC 6= 0. Let z0 be a 1-point such that F . Since F, G share (1, 2), z0 is also a 1-point of G. Therefore, from (5.2), we obtain A + B = C + D, and hence (5.2) can be written as F − 1 = G − 1 αG + β , CUBO 24, 1 (2022) Uniqueness of entire functions 177 where α = C/(A − C) and β = D/(A − C). From this we can say that F , G share (1, ∞). Now using the standard Valiron-Mohon’ko Lemma 4.3, we obtain from (5.2) that T (r, F) = T (r, G) + O(log r). Then using Lemmas 4.5 and 4.10 and the fact that f and g are transcendental entire functions of finite order, we deduce T (r, f) ≤ n + 1 n T (r, g) + S(r, f) + S(r, g) and T (r, G) T (r, g) −→ n + 1 (5.3) as r −→ ∞, r ∈ I. Now keeping in view of (5.3), Lemma 4.2 and the condition that f and g are transcendental entire functions, we obtain N(r, 0; F) + N(r, ∞; F) = N(r, 0; f(z)nLc(f)) + O(log r) ≤ N(r, 0; f(z)) + N(r, 0; Lc(f)) + O(log r) ≤ N(r, 0; f(z)) + T (r, Lc(f)) + O(log r) ≤ N(r, 0; f(z)) + m(r, Lc(f)) + O(log r) ≤ N(r, 0; f(z)) + m ( r, Lc(f) f(z) ) + m(r, f(z)) + O(log r) ≤ 2T (r, f) + S(r, f) ≤ 2n + 2 n T (r, g) + S(r, g). Similarly, we obtain N(r, 0; G) + N(r, ∞; G) ≤ 2T (r, g) + S(r, g). Thus using (5.3), we obtain N(r, 0; F) + N(r, ∞; F) + N(r, 0; G) + N(r, ∞; G) ≤ 2(2n + 1) n(n + 1) T (r, G) + S(r, g). (5.4) Since, g(z) and g(z + c) share 0 CM, we get that N(r, 0; Lc(g)/g(z)) = S(r, g). Thus, keeping in view of this, Lemmas 4.2, 4.10 and applying the second fundamental theorem of Nevanlinna on G, we obtain (n + 1)T (r, g) = T (r, G) ≤ N(r, ∞; G) + N(r, 0; G) + N(r, 1; G) + S(r, g) ≤ N(r, 0; g) + N(r, 0; Lc(g)) + N(r, 1; G) + S(r, g) ≤ N(r, 0; g) + T (r, Lc(g)) + N(r, 1; G) + S(r, g) ≤ N(r, 0; g) + T ( r, Lc(g) g(z) ) + T (r, g) + S(r, g) ≤ 2T (r, g) + N(r, 1; G) + S(r, g), i.e., (n − 1)T (r, g) ≤ 2T (r, g) + N(r, 1; G) + S(r, g). From this and the fact that F and G share (1, 2), we conclude that there exists a point z0 ∈ C such that F(z0) = G(z0) = 1. Hence from (5.4), Lemma 4.11 and the condition n ≥ 4, we conclude that either FG = 1 or F = G. Now we consider the following sub-cases. 178 G. Haldar CUBO 24, 1 (2022) Subcase 1.1: F ≡ G. Then we get f(z)n(f(z + c) + c0f(z)) ≡ g(z) n(g(z + c) + c0g(z)). Let h(z) = f(z)/g(z). Then we deduce that (h(z)nh(z + c) − 1)g(z + c) = −c0(h n+1(z) − 1)g(z). (5.5) Suppose h is not constant. Then from (5.5), we obtain g(z) g(z + c) = h(z)nh(z + c) − 1 c0(h(z)n+1 − 1) . As g(z) and g(z+c) share 0 CM, from the above equation we can say that h(z)n+1 and h(z)nh(z+c) share (1, ∞). Let z0 be a zero of h n+1−1. Then we must have h(z)n+10 = 1 and h(z0) nh(z0+c) = 1. Hence h(z0 + c) = h(z0), and therefore by Lemma 4.1, we obtain N(r, 1; hn+1) ≤ N(r, 0; h(z + c) − h(z)) ≤ 2T (r, h) + S(r, h). Keeping in mind the above inequality and Lemma 4.3 and applying the second fundamental theo- rem of Nevanlinna to hn+1, we obtain (n + 1)T (r, h) = T (r, hn+1) ≤ N(r, ∞; hn+1) + N(r, 0; hn+1) + N(r, 1; hn+1) + S(r, h) ≤ 4T (r, h) + S(r, h), i.e., (n − 3)T (r, h) ≤ S(r, h), which is not possible since n ≥ 4. Hence h is constant. Then (5.5) reduces to (hn+1 − 1)Lc(g) = 0. As Lc(g) 6≡ 0, we must have h n+1 = 1 and thus f = tg, for a constant t such that tn+1 = 1, which is the conclusion (a). Subcase 1.2: Suppose FG ≡ 1. Then we have f(z)nLc(f)g(z) nLc(g) = P0(z) 2. (5.6) From (5.6) and the condition that f and g are transcendental entire functions, one can immediately say that both f and g have at most finitely many zeros. So, we may write f(z) = P1(z)e Q1(z), g(z) = P1(z)e Q2(z), (5.7) where P1, P2, Q1, Q2 are polynomials, and Q1, Q2 are non-constants. Substituting (5.7) in (5.6), we obtain (P1P2) nen(Q1+Q2)[P1(z + c)P2(z + c)e Q1(z+c)+Q2(z+c) + c20P1P2e Q1+Q2 +c0P1P2(z + c)e Q1+Q2(z+c) + c0P1(z + c)P2e Q1(z+c)+Q2] = P(z)2. (5.8) CUBO 24, 1 (2022) Uniqueness of entire functions 179 Keeping in view of (5.7), we must have n(Q1(z) + Q2(z)) + Q1(z + c) + Q2(z + c) = A1, (5.9) n(Q1(z) + Q2(z)) + Q1(z) + Q2(z + c) = A2, (5.10) n(Q1(z) + Q2(z)) + Q1(z + c) + Q2(z) = A3, (5.11) (n + 1)(Q1(z) + Q2(z)) = A4, (5.12) where A1, A2, A3, A4 are constants. Let Q1(z) + Q2(z) = W(z). Then (5.9) can be written as nW(z) + W(z + c) = A1, (5.13) for all z ∈ C. Therefore, from (5.13), we must have W = B, where B is a constant, and therefore, we have Q2 = B − Q1. (5.14) Keeping in view of (5.14), (5.7) can be written as f(z) = P1e Q1(z), g(z) = P2e Be−Q1(z). (5.15) Now (5.8) can be written as (P1P2) n[P1(z + c)P2(z + c)e A4 + c0P1(z + c)P2e A3 + c0P1P2(z + c)e A2 +c20P1P2e A4] = P(z)2. (5.16) If P1P2 is not a constant, then the degree of the left side of (5.16) is at least n + 1. But the condition 2 deg(P) < n + 1 implies that the degree of the right side of (5.16) is less than n + 1, which is a contradiction. Thus P1P2 and P reduce to non-zero constants. Since P1, P2 are both polynomials and their product is constant, each of them must be constant. Therefore, (5.15) can be written as f(z) = eU, g(z) = eBe−U, (5.17) where U is a non-constant polynomial. Using the above forms of f and g and keeping in mind that P is a constant, say d, (5.6) reduces to e(n+1)B(eU(z+c)−U(z) + c0)(e −(U(z+c)−U(z)) + c0) = d 2. (5.18) If c0 = 0, (5.18) reduces to e (n+1)B = d2. Set eB = t. Then (5.17) can be written as f(z) = eU , g(z) = te−U, where t is a constant such that tn+1 = 1, 180 G. Haldar CUBO 24, 1 (2022) which is the conclusion (b). If c0 6= 0, then from (5.18), one can say that e U(z+c)−U(z) + c0 has no zeros. Then φ(z) = eU(z+c)−U(z) 6= 0, −c0, ∞. By Picard’s theorem, φ is constant and so deg(U(z)) = 1. Therefore, from (5.17), one may obtain f(z) = c1e az, g(z) = c2e −az, where a, c1 and c2 are non-zero constants. Using these in (5.6), we obtain (c1c2) n+1(eac + c0)(e −ac + c0) = d 2, which is the conclusion (c). Case 2: Suppose n ≥ 5. Since f(z)nLc(f) − P(z) and g(z) nLc(g) − P(z) share (0, l), it follows that F and G share (1, l). Let H 6≡ 0. First suppose l ≥ 2. Using Lemmas 4.16 and 4.19, we obtain N(r, 1; F) = N(r, 1; F |= 1) + N(r, 1; F |≥ 2) ≤ N(r, ∞; H) + N(r, 1; F |≥ 2) ≤ N(r, 0; F |≥ 2) + N(r, 0; G |≥ 2) + N∗(r, 1; F, G) + N(r, 1; F |≥ 2) + N0(r, 0; F ′) + N0(r, 0; G ′) + S(r, F) + S(r, G). (5.19) Keping in view of the above observation and Lemma 4.20, we see that N0(r, 0; G ′) + N(r, 1; F |≥ 2) + N∗(r, 1; F, G) ≤ N0(r, 0; G ′) + N(r, 1; F |≥ 2) + N(r, 1; F |≥ 3) + S(r, F) ≤ N0(r, 0; G ′) + N(r, 1; G |≥ 2) + N(r, 1; G |≥ 3) + S(r, F) + S(r, G) ≤ N0(r, 0; G ′) + N(r, 1; G) − N(r, 1; G) + S(r, F) + S(r, G) ≤ N(r, 0; G′ | G 6= 0) ≤ N(r, 0; G) + S(r, G). (5.20) Since g(z) and g(z + c) share 0 CM, we must have N(r, ∞, Lc(g)/g(z)) = 0. Hence using (5.19), (5.20), Lemmas 4.2, 4.13 and applying second fundamental theorem of Nevan- linna to F , we obtain nT (r, f) ≤ T (r, F) − N(r, 0; Lc(f)) + S(r, f) ≤ N(r, 0; F) + N(r, ∞; F) + N(r, 1; F) − N(r, 0; F ′) − N(r, 0; Lc(f)) + S(r, f) ≤ N2(r, 0; F) + N2(r, 0; G) − N(r, 0; Lc(f)) + S(r, f) + S(r, g) ≤ N2(r, 0; f nLc(f)) + N2 ( r, 0; gn+1 Lc(g) g(z) ) − N(r, 0; Lc(f)) + S(r, f) + S(r, g) ≤ 2N(r, 0; f) + 2N(r, 0; g) + N ( r, 0; Lc(g) g(z) ) + S(r, f) + S(r, g) CUBO 24, 1 (2022) Uniqueness of entire functions 181 ≤ 2(T (r, f) + T (r, g)) + T ( r, Lc(g) g(z) ) + S(r, f) + S(r, g) ≤ 2(T (r, f) + T (r, g)) + N ( r, ∞; Lc(g) g(z) ) + m ( r, Lc(g) g(z) ) + S(r, f) + S(r, g) ≤ 2(T (r, f) + T (r, g)) + S(r, f) + S(r, g). (5.21) Similarly, using Lemmas 4.2, 4.13 and applying second fundamental theorem of Nevanlinna to G, we obtain nT (r, g) ≤ T (r, G) − N(r, 0; Lc(g)) + S(r, g) ≤ N(r, 0; G) + N(r, ∞; G) + N(r, 1; G) − N(r, 0; G′) − N(r, 0; Lc(g)) + S(r, g) ≤ N2(r, 0; F) + N2(r, 0; G) − N(r, 0; Lc(g)) + S(r, f) + S(r, g) ≤ N2(r, 0; f(z) nLc(f)) + N2 (r, 0; g nLc(g)) − N(r, 0; Lc(g)) + S(r, f) + S(r, g) ≤ 2N(r, 0; f) + 2N(r, 0; g) + N (r, 0; Lc(f)) + S(r, f) + S(r, g) ≤ 2(T (r, f) + T (r, g)) + T (r, Lc(f)) + S(r, f) + S(r, g) ≤ 2(T (r, f) + T (r, g)) + m ( r, Lc(f) f(z) ) + m(r, f(z)) + S(r, f) + S(r, g) ≤ 2(T (r, f) + T (r, g)) + T (r, f) + S(r, f) + S(r, g). (5.22) Combining (5.21) and (5.22), we get (n − 5)T (r, f) + (n − 4)T (r, g) ≤ S(r, f) + S(r, g), which contradicts with n ≥ 5. When l = 1, Keeping in view of Lemmas 4.14, 4.15, 4.16, 4.19 and 4.20, we obtain N(r, 1; F) = N(r, 1; F |= 1) + NL(r, 1; F) + NL(r, 1; G) + N (2 E (r, 1; F) ≤ N(r, 0; F |≥ 2) + N(r, 0; G |≥ 2) + N∗(r, 1; F, G) + NL(r, 1; F) + NL(r, 1; G) + N (2 E (r, 1; F) + N0(r, 0; F ′) + N0(r, 0; G ′) + S(r, F) + S(r, G) ≤ N(r, 0; F |≥ 2) + N(r, 0; G |≥ 2) + 2NL(r, 1; F) + 2NL(r, 1; G) + N (2 E (r, 1; F) + N0(r, 0; F ′) + N0(r, 0; G ′) + S(r, F) + S(r, G) ≤ N(r, 0; F |≥ 2) + N(r, 0; G |≥ 2) + NF >2(r, 1; G) + N(r, 1; G) − N(r, 1; G) + N0(r, 0; F ′) + N0(r, 0; G ′) + S(r, F) + S(r, G) ≤ N(r, 0; F |≥ 2) + N(r, 0; G |≥ 2) + N(r, 0; G′ | G 6= 0) + 1 2 N(r, 0; F) + N0(r, 0; F ′) + S(r, F) + S(r, G) ≤ N(r, 0; F |≥ 2) + 1 2 N(r, 0; F) + N2(r, 0; G) + N0(r, 0; F ′) + S(r, F) + S(r, G). (5.23) Since g(z), g(z + c) share 0 CM, N(r, ∞; g(z + c)/g(z)) = 0, and therefore, using Lemma 4.2, we obtain T (r, g(z + c)/g(z)) = 0. 182 G. Haldar CUBO 24, 1 (2022) Hence using (5.23), Lemmas 4.2, 4.13 and applying second fundamental theorem of Nevanlinna to F , we obtain nT (r, f) ≤ T (r, F) − N(r, 0; Lc(f)) + S(r, f) ≤ N(r, 0; F) + N(r, 1; F) − N(r, 0; F ′) − N(r, 0; Lc(f)) + S(r, f) ≤ N2(r, 0; F) + N2(r, 0; G) + 1 2 N(r, 0; F) − N(r, 0; Lc(f)) + S(r, f) + S(r, g) ≤ 2N(r, 0; f) + N2 ( r, 0; gn+1 Lc(g) g ) + 1 2 N(r, 0; F) + S(r, f) + S(r, g) ≤ 2N(r, 0; f) + 2N(r, 0; g) + 1 2 (N(r, 0; f) + N(r, 0, Lc(f))) + N ( r, 0; Lc(g) g ) + S(r, f) + S(r, g) ≤ 5 2 T (r, f) + 2T (r, g) + 1 2 T (r, Lc(f)) + T ( r, Lc(g) g ) + S(r, f) + S(r, g) ≤ 5 2 T (r, f) + 2T (r, g) + 1 2 m ( r, Lc(f) f ) + 1 2 m(r, f(z)) + S(r, f) + S(r, g) ≤ 3T (r, f) + 2T (r, g) + S(r, f) + S(r, g). (5.24) In a similar manner, we may obtain nT (r, g) ≤ 3T (r, f) + 5 2 T (r, g) + S(r, f) + S(r, g). (5.25) Combining (5.24) and (5.25), we obtain (n − 6)T (r, f) + ( n − 5 2 ) T (r, g) ≤ S(r, f) + S(r, g), which is a contradiction since n ≥ 6. When l = 0, using Lemmas 4.16, 4.17, 4.18, 4.19 and 4.20, we obtain N(r, 1; F) = N(r, 1; F |= 1) + NL(r, 1; F) + NL(r, 1; G) + N (2 E (r, 1; F) ≤ N(r, 0; F |≥ 2) + N(r, 0; G |≥ 2) + N∗(r, 1; F, G) + NL(r, 1; F) + NL(r, 1; G) + N (2 E (r, 1; F) + N0(r, 0; F ′) + N0(r, 0; G ′) + S(r, F) + S(r, G) ≤ N(r, 0; F |≥ 2) + N(r, 0; G |≥ 2) + 2NL(r, 1; F) + 2NL(r, 1; G) + N (2 E (r, 1; F) + N0(r, 0; F ′) + N0(r, 0; G ′) + S(r, F) + S(r, G) ≤ N(r, 0; F |≥ 2) + N(r, 0; G |≥ 2) + NL(r, 1; F) + NF >1(r, 1; G) + NG>1(r, 1; F) + N(r, 1; G) − N(r, 1; G) + N0(r, 0; F ′) + N0(r, 0; G ′) + S(r, F) + S(r, G) ≤ N(r, 0; F |≥ 2) + N(r, 0; G |≥ 2) + N(r, 0; G′ | G 6= 0) + 2N(r, 0; F) + N(r, 0; G) + N0(r, 0; F ′) + S(r, F) + S(r, G) ≤ N2(r, 0; F) + N(r, 0; F) + N2(r, 0; G) + N(r, 0; G) + N0(r, 0; F ′) + S(r, F) + S(r, G). (5.26) CUBO 24, 1 (2022) Uniqueness of entire functions 183 Hence using (5.26), Lemmas 4.2, 4.13 and applying second fundamental theorem of Nevanlinna to F , we obtain nT (r, f) ≤ T (r, F) − N(r, 0; Lc(f)) + S(r, f) ≤ N(r, 0; F) + N(r, 1; F) − N(r, 0; F ′) − N(r, 0; Lc(f)) + S(r, f) ≤ N2(r, 0; F) + N2(r, 0; G) + 2N(r, 0; F) + N(r, 0; G) − N(r, 0; Lc(f)) + S(r, f) + S(r, g) ≤ 2N(r, 0; f) + N2 ( r, 0; gn+1(z) Lc(g) g(z) ) + N ( r, 0; gn+1(z) Lc(g) g(z) ) + 2N(r, 0; fn(z)Lc(f)) + S(r, f) + S(r, g) ≤ 4N(r, 0; f) + 3N(r, 0; g) + N ( r, 0; Lc(g) g(z) ) + N ( r, 0; Lc(g) g(z) ) + 2N(r, 0; Lc(f)) + S(r, f) + S(r, g) ≤ 4T (r, f) + 3T (r, g) + 2T ( r, Lc(g) g(z) ) + 2T (r, Lc(f)) + S(r, f) + S(r, g) ≤ 4T (r, f) + 3T (r, g) + 2m(r, Lc(f)) + S(r, f) + S(r, g) ≤ 4T (r, f) + 3T (r, g) + 2m ( r, Lc(f) f(z) ) + 2m(r, f(z)) + S(r, f) + S(r, g) ≤ 6T (r, f) + 3T (r, g) + S(r, f) + S(r, g). (5.27) In a similar manner, we obtain nT (r, g) ≤ 5T (r, f) + 6T (r, g) + S(r, f) + S(r, g). 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