Ratio Mathematica Volume 39, 2020, pp. 213-228 Uniqueness of an entire function sharing a polynomial with its linear differential polynomial Imrul Kaish* Nasir Uddin Gazi† Abstract In this paper we consider an entire function when it shares a polyno- mial with its linear differential polynomial. Our result is an improve- ment of a result of P.Li. Keywords: Uniqueness; Entire function; Differential Polynomial; Sharing. 2010 AMS subject classifications: 30D35. 1 *Department of Mathematics and Statistics, Aliah University, Kolkata, West Bengal 700160, India; imrulksh3@gmail.com. †Department of Mathematics and Statistics, Aliah University, Kolkata, West Bengal 700160, India; nsrgazi@gmail.com 1Received on November 2nd, 2020. Accepted on December 17th, 2020. Published on Decem- ber 31st, 2020. doi: 10.23755/rm.v39i0.554. ISSN: 1592-7415. eISSN: 2282-8214. ©Kaish et al. This paper is published under the CC-BY licence agreement. 213 I. Kaish and N. Gazi 1 Introduction, Definitions and Results Let f be a non-constant meromorphic function defined in the open complex plane C and a = a(z) be a polynomial. We denote by E(a; f) the set of zeros of f−a, counted with multiplicities and by E(a; f) the set of distinct zeros of f−a. If for two non-constant meromorphic functions f and g, we have E(a; f) = E(a; g), we say that f and g share a CM and if E(a; f) = E(a; g), we say that f and g share a IM. We denote by S(r,f) any function satisfying S(r,f) = o{T(r,f)}, as r → ∞, possibly outside of a set with finite measure. For an entire function f, we define deg(f) in the following way: deg(f) = ∞, if f is a transcendental entire function and deg(f) is the degree of the polynomial, if f is a polynomial. The investigation of uniqueness of an entire function sharing two values intro- duced by L. A. Rubel and C. C. Yang [Rubel and Yang, 1977] in 1977. Following is their result. Theorem A. [Rubel and Yang, 1977] Let f be a non-constant entire function. If E(a; f) = E(a; f(1)) and E(b; f) = E(b; f(1)), for distinct finite complex num- bers a and b, then f ≡ f(1). In 1979 E. Mues and N. Steinmetz [Mues and Steinmetz, 1979] tried to im- prove Theorem A by considering IM sharing of values. They proved the following theorem. Theorem B. [Mues and Steinmetz, 1979]. Let f be a non-constant entire func- tion and a, b be two distinct finite complex values. If E(a; f) = E(a; f(1)) and E(b; f) = E(b; f(1)), then f ≡ f(1). In 1986 G. Jank, E. Mues and L. Volkmann [Jank et al., 1986] considered an entire function sharing a nonzero value with its derivatives and they proved the following result. Theorem C. [Jank et al., 1986] Let f be a non-constant entire function and a be a non-zero finite value. If E(a; f) = E(a; f(1)) ⊂ E(a; f(2)), then f ≡ f(1). H. Zhong [Zhong, 1995] tried to improve Theorem C by taking higher order derivatives. By the following example he concluded that in Theorem C the second derivative cannot be straight way replaced by any higher order derivatives. Example 1.1. [Zhong, 1995] Let k(≥ 3) be a positive integer and ω( 6= 1) be a (k − 1)th root of unity. If f = eωz + ω − 1, then f, f(1), and f(k) share the value ω CM, but f 6≡ f(1). 214 Uniqueness of an entire function sharing a polynomial with its linear differential polynomial Considering two consecutive higher order derivatives H. Zhong [Zhong, 1995] improved Theorem C in another direction. The following is the improved result. Theorem D. [Zhong, 1995] Let f be a non-constant entire function and a be a non-zero finite value. If E(a; f) = E(a; f(1)) and E(a; f) ⊂ E(a; f(n)) ∩ E(a; f(n+1)) for n(≥ 1), then f ≡ f(n). For further discussion we need the following notation. Let f be a non-constant meromorphic function, a = a(z) be a polynomial and A be a set of complex num- bers. We denote by nA(t,a; f), the number of zeros of f−a, counted according to their multiplicities which lie in A∩{z : |z|≤r}. The integrated counting function NA(r,a; f) of the zeros of f −a which lie in A∩{z : |z|≤r} is defined as NA(r,a; f) = ∫ r 0 nA(t,a; f) −nA(0,a; f) t dt + nA(0,a; f) log r, where nA(0,a; f) denotes the multiplicity of zeros of f−a at origin. NA(r,a; f) be the reduced counting function of zeros of f −a in A∩{z : |z|≤r}. Clearly if A = C then NA(r,a; f) = N(r,a; f) and NA(r,a; f) = N(r,a; f). For standard definitions and notations of the value distribution theory we refer the reader to [Hayman, 1964] and [Yang and Yi, 2003]. Recently I. Lahiri and I. Kaish [Lahiri and Kaish, 2017] improved Theorem D by considering a shared polynomial. They proved the following result. Theorem E. [Lahiri and Kaish, 2017] Let f be a non-constant entire function and a = a(z)(6≡ 0) be a polynomial with deg(a) 6= deg(f). Suppose that A = E(a; f)∆E(a; f(1)) and B = E(a,f(1))\{E(a,f(n)) ∩ E(a,f(n+1))}, where 4 denotes the symmetric difference of sets and n(≥ 1) is an integer. If (i) NA(r,a; f) + NA(r,a; f(1)) = O{logT(r,f)}, (ii) NB(r,a; f(1)) = S(r,f) and (iii) each common zero of f −a and f(1) −a has the same multiplicity, then f = λez, where λ( 6= 0) is a constant. Throughout the paper we denote by L = L(f) a nonconstant linear differential polynomial generated by f of the form L = L(f) = a1f (1) + a2f (2) + ......... + anf (n), (1) where a1,a2, .......,an(6= 0) are constants. Considering Linear differential polynomial P.Li [Li, 1999] improved Theorem D in the following way. 215 I. Kaish and N. Gazi Theorem F. [Li, 1999]. Let f be a non-constant entire function and L be defined in (1) and a be a non-zero finite complex number. If E(a; f) = E(a; f(1)) ⊂ E(a; L) ∩E(a; L(1)) then f = f(1) = L. In this paper we extend Theorem D and Theorem F in the following way Theorem 1.1. Let f be a non-constant entire function, L be defined in (1) and a = a(z)(6≡ 0) be a polynomial with deg(a) 6= deg(f). Suppose that A = E(a; f)∆E(a; f(1)) and B = E(a,f(1))\{E(a,L(p))∩E(a,L(q))} where p,q are integers satisfying q > p ≥ deg(a). If (i) NA(r,a; f) + NA(r,a; f(1)) = O{log T(r,f)}, (ii) NB(r,a; f(1)) = S(r,f) and (iii) each common zero of f −a and f(1) −a has the same multiplicity, then f = L = λez, where λ( 6= 0) is a constant. Putting A = B = ∅ we get the following corollary. Corolary 1.1. Let f be a non-constant entire function, L be defined in (1) and a = a(z)(6≡ 0) be a polynomial with deg(a) 6= deg(f). If E(a; f) = E(a; f(1)) and E(a,f(1)) ⊂ E(a,L(p)) ∩ E(a,L(q)) where p,q are integers satisfying q > p ≥ deg(a), then f = L = λez, where λ( 6= 0) is a constant. Remark 1.1. If in Corollary 1.1, a is a non-zero constant and p = deg(a) = 0,q = p + 1 then it is a particular form of Theorem F. Remark 1.2. If in (1), a1 = a2 = .......an−1 = 0 and an = 1 then L = f(n) and if in Corollary 1.1, a is a non-zero constant and p = deg(a),q = p + 1, then Corollary 1.1 is the Theorem D. Remark 1.3. It is an open problem whether the Theorem 1.1 is valid or not if we omit the condition p ≥ deg(a). 2 Lemmas In this section we present some necessary lemmas. Lemma 2.1. [Lahiri and Kaish, 2017]. Let f be a transcendental entire function of finite order and a = a(z)(6≡ 0) be a polynomial and A = E(a; f)∆E(a; f(1)). If 216 Uniqueness of an entire function sharing a polynomial with its linear differential polynomial (i) NA(r,a; f) + NA(r,a; f(1)) = O{log T(r,f)}, (ii) each common zero of f −a and f(1) −a has the same multiplicity, then m(r,a; f) = S(r,f). Lemma 2.2. [Lain, 1993]. Suppose f be an entire function, a0,a1, .....an are polynomials and a0,an are not identically zero. Then each solution of the linear differential equation anf(n) + an−1f(n−1) + ...... + a0f = 0 is of finite order. Lemma 2.3. [Hayman, 1964]. Let f be a non-constant meromorphic function and a1,a2,a3 be three distinct meromorphic functions satisfying T(r,aν) = S(r,f) for ν = 1, 2, 3 then T(r,f) ≤ N(r, 0; f −a1) + N(r, 0; f −a2) + N(r, 0; f −a3) + S(r,f). Lemma 2.4. Let f be a transcendental entire function and a = a(z)(6≡ 0) be a polynomial. Also let L(f), L(a) be the linear differential polynomials generated by f and a respectively. Suppose h = (a−a(1))(L(p)(f) −L(p)(a)) − (a−L(p)(a))(f(1) −a(1)) f −a , A = E(a; f)\E(a; f(1)) and B = E(a,f(1))\{E(a,L(p)) ∩ E(a,L(q))}, where p,q are integers satisfying 0 ≤ p < q. If (i) NA(r,a; f) + NB(r,a; f(1)) = S(r,f), (ii) each common zero of f −a and f(1) −a has the same multiplicity, (iii) h is a transcendental entire or meromorphic, then m(r,a,f(1)) = S(r,f). Proof. Since a−a(1) = (f(1)−a(1))−(f(1)−a), if z0 be a common zero of f−a and f(1)−a with multiplicity r(≥ 2), then z0 is a zero of a−a(1) with multiplicity r − 1. So N(2(r,a; f) ≤ 2N(r, 0; a−a(1)) + NA(r,a; f) = S(r,f), (2) where N(2(r,a; f) be the counting function of multiple zeros of f −a. Using (2) and from the hypothesis we get N(r,h) ≤ NA(r,a; f) + NB(r,a; f(1)) + N(2(r,a; f) + S(r,f) = S(r,f) 217 I. Kaish and N. Gazi Since m(r,h) = S(r,f), we have T(r,h) = S(r,f) From h = (a−a(1))(L(p)(f) −L(p)(a)) − (a−L(p)(a))(f(1) −a(1)) f −a , we get f = a + 1 h {(a−a(1))(L(p)(f) −L(p)(a)) − (a−L(p)(a))(f(1) −a(1))} = a + 1 h {(a−a(1))(L(p)(f) −a) − (a−L(p)(a))(f(1) −a)}. (3) Case 1. Let p > 0 . Differentiating (3) we get f(1) = a(1) + ( 1 h )(1){(a−a(1))(L(p)(f) −a) − (a−L(p)(a))(f(1) −a)} + 1 h {(a(1) −a(2))(L(p)(f) −a) + (a−a(1))(L(p+1) −a(1))}− 1 h {(a(1) −L(p+1)(a))(f(1) −a) + (a−L(p)(a))(f(2) −a(1))}. This implies (f(1) −a){1 + ( 1 h )(1)(a−L(p)(a)) + 1 h (a(1) −L(p+1)(a))} = a(1) −a + ( 1 h )(1)(a−a(1))(L(p)(f)−a) + 1 h (a(1) −a(2))(L(p)(f)−a) + 1 h (a− a(1))(L(p+1)(f) −a(1)) − 1 h (a−L(p)(a))(f(2) −a(1)) = a(1) − a + (a−a (1) h )(1)(L(p)(f) − L(p−1)(a)) + (a−a (1) h )(1)(L(p−1)(a) − a) + a−a(1) h (L(p+1)(f)−L(p)(a)) + a−a (1) h (L(p)(a)−a(1))− 1 h (a−L(p)(a))(f(2)−a(1)), or, (f(1) −a){1 + (a−L (p)(a) h ))(1)} = (a(1) −a) + {(a−a (1) h )(L(p−1)(a) −a)}(1) + (a−a (1) h )(1)(L(p)(f)−L(p−1)(a))+a−a (1) h (L(p+1)(f)−L(p)(a))−1 h (a−L(p)(a))(f(2)− a(1)), or 1 f(1) −a = h1 h2 − 1 h2 ( a−a(1) h )(1)( L(p)(f) −L(p−1)(a) f(1) −a ) +( a−a(1) hh2 )( L(p+1)(f) −L(p)(a) f(1) −a ) − 1 hh2 (a−L(p)(a))( f(2) −a(1) f(1) −a ), (4) where h1 = 1 + ( a−L(p)(a) h )(1), h2 = a (1) −a + {(a−a (1) h )(L(p−1)(a) −a)}(1). We now verify that h1 6≡ 0,h2 6≡ 0. 218 Uniqueness of an entire function sharing a polynomial with its linear differential polynomial If h1 ≡ 0, then 1 + ( a−L(p)(a) h )(1) ≡ 0. Integrating we get 1 h = c1−z a−L(p)(a) , where c1 is a constant. This is a contradiction, because h is transcendental. If h2 ≡ 0, then a(1) − a + {(a−a (1) h )(L(p−1)(a) −a)}(1) ≡ 0. Integrating we get h = (a−a (1))(L(p−1)(a)−a) P(z) , where P(z) is a polynomial. This is again a contradiction. Therefore h1 6≡ 0,h2 6≡ 0. Again T(r,h1) + T(r,h1) = S(r,f), since T(r,h) = S(r,f). Now from (4) and using Lemma of logarithmic derivative we get m(r,a; f(1)) = m(r, 1 f(1)−a) = S(r,f). Case 2. Let p = 0. Then L(p)(f) = L(f). Suppose L(f) = a1f(1) + a2f(2) + ......... + anf(n) and L(a) = a1a(1) + a2a(2) + ......... + ana(n), where a1,a2, .......,an( 6= 0) are constant, n(≥ 1) be an integer. From the definition of h we get f = a + 1 h {(a−a(1))(L(f) −a) − (a−L(a))(f(1) −a)} Differentiating we get f(1) = a(1) + ( 1 h )(1){(a−a(1))(L(f) −a) − (a−L(a))(f(1) −a)} + 1 h {(a(1) −a(2))(L(f) −a) + (a−a(1))(L(1)(f) −a(1))} − 1 h {(a(1) −L(1)(a))(f(1) −a) − (a−L(a))(f(2) −a(1))}. This implies (f(1)−a){1 + (a−L(a) h )(1)} = (a(1)−a)+(a−a (1) h )(1)(L(f)−a)+a−a (1) h (L(1)(f)− a(1))−a−L(a) h (f(2)−a(1)) = (a(1)−a)+(a−a (1) h )(1)(L(f)−L1(a))+(a−a (1) h )(1)(L1(a)− a)+(a−a (1) h )(L(1)(f)−L(a))+(a−a (1) h )(L(a)−a(1))−a−L(a) h (f(2)−a(1)) = (a(1)− a) + {(a−a (1) h )(L1(a) −a)}(1) + (a−a (1) h )(1)(L(f) − L1(a)) + (a−a (1) h )(L(1)(f) − L(a)) − a−L(a) h (f(2) −a(1)) Or, 1 f(1) −a = h3 h4 − 1 h4 ( a−a(1) h )(1)( L(f) −L1(a) f(1) −a ) +( a−a(1) hh4 )( L(1)(f) −L(a) f(1) −a ) − ( a−L(a) hh4 )( f(2) −a(1) f(1) −a ), (5) where L1(a) = a1a + a2a (1) + ..... + ana (n−1), h3 = 1 + ( a−L(a) h )(1) and h4 = a (1) −a + {(a−a (1) h )(L1(a) −a)}(1) 219 I. Kaish and N. Gazi Similarly as in Case 1, h3 6≡ 0, h4 6≡ 0. Also T(r,h3) + T(r,h4) = S(r,f). Therefore from (5) and using Lemma of logarithmic derivative we get m(r,a; f(1)) = m(r, 1 f(1)−a) = S(r,f). This completes the proof of the lemma. Lemma 2.5. Let f be a transcendental entire function, a = a(z)(6≡ 0) be a polynomial and L = L(f) be define in (1). Suppose (i) NA(r,a; f) + NA(r,a; f(1)) = S(r,f), where A = E(a; f)∆E(a; f(1)) (ii) NB(r,a; f(1))) = S(r,f), where B = E(a,f(1))\{E(a,L(p)) ∩E(a,L(q))} p,q are integers satisfying q > p ≥ deg(a), (iii) each common zero of f −a and f(1) −a has the same multiplicity, (iv) m(r, a; f) = S(r, f), then f = L = λez, where λ(6= 0) is a constant. Proof. Let α = f(1) −a f −a , (6) From the hypothesis we get, N(r,α) ≤ NA(r,a; f) + S(r,f) = S(r,f) and m(r,α) = m(r, f(1) −a f −a ) = m(r, f(1) −a(1) + a(1) −a f −a ) ≤ m(r,a; f) + S(r,f) = S(r,f). Therefore T(r,α) = S(r,f). From (6) we get f(1) = αf + a(1 −α) = α1f + β1, where α1 = α and β1 = a(1 −α) Differentiating we get, f(2) = α2f + β2, 220 Uniqueness of an entire function sharing a polynomial with its linear differential polynomial where α2 = α (1) 1 + α1α1 and β2 = β (1) 1 + α1β1. Similarly, f(k) = αkf + βk, where αk+1 = α (1) k + α1αk and βk+1 = β (1) k + αkβ1. Clearly T(r,αk) + T(r,βk) = S(r,f), because T(r,α) = S(r,f). Now L(p) = n∑ k=1 akf (p+k) = ( n∑ k=1 akαp+k)f + ( n∑ k=1 akβp+k) = µ1f + ν1, (7) where µ1 = n∑ k=1 akαp+k, ν1 = n∑ k=1 akβp+k L(q) = n∑ k=1 akf (q+k) = ( n∑ k=1 akαq+k)f + ( n∑ k=1 akβq+k) = µ2f + ν2, (8) where µ2 = n∑ k=1 akαq+k, ν2 = n∑ k=1 akβq+k. Clearly T(r,µi) + T(r,νi) = S(r,f), i = 1, 2. Let D = E(a; f) ∩E(a; f(1)) ∩E(a; L(p)) ∩E(a; L(q)). Note that D 6= ∅, because otherwise, N(r,a; f) = S(r,f). Then from the hypothesis T(r,f) = S(r,f), a contradiction. Let z1 ∈ D then f(z1) = f(1)(z1) = L(p)(z1) = L(q)(z1) = a(z1). Now from (7) and (8) we get a(z1) = µ1(z1)a(z1) + ν1(z1) and a(z1) = µ2(z1)a(z1) + ν2(z1) If µ1a + ν1 −a 6≡ 0, then N(r,a; f) ≤ NA(r,a; f) + NB(r,a; f(1)) + ND(r,a; f) + S(r,f) ≤ NA(r, 0; µ1a + ν1 −a) + S(r,f) = S(r,f), a contradiction. Therefore µ1a + ν1 −a ≡ 0. (9) 221 I. Kaish and N. Gazi Similarly µ2a + ν2 −a ≡ 0. (10) From (9) and (10) we get µ1 ≡ µ2 ≡ 1 and ν1 ≡ 0 ≡ ν2. Then from (7) L(p) ≡ f. (11) Also µ1 ≡ 1 implies n∑ k=1 akαp+k ≡ 1. (12) From (12) we see that α has no pole. Because if α has a pole of order d(≥ 1) then the left hand side of (12) has a pole of order (p + k)d but the right hand side is a constant. Again by simple calculation from (12) we get anα n+p + P [α] ≡ 0. (13) where P [α] is a differential polynomial in α with degree not exceeding (n + p− 1). If α is transcendental entire, then by Clunie’s Lemma we have m(r,α) = S(r,α), a contradiction. If α is a nonconstant polynomial then left hand side of (13) is also a noncon- stant polynomial, which is again a contradiction. Therefore α is a constant. Now from f (1)−a f−a = α, we get f (1) −αf = a(1 −α). Integrating we get e−αzf = (1 −α) ∫ ae−αzdz = (1 −α)P(z)e−αz + λ, where λ(6= 0) is a constant and P(z) is a polynomial of degree atmost deg(a), or, f = (1 −α)P(z) + λeαz. Now f(r+1) = λαr+1eαz, if r = deg(a) 222 Uniqueness of an entire function sharing a polynomial with its linear differential polynomial Therefore L(p) = n∑ k=1 akf (p+k) = ( n∑ k=1 akα p+k)λeαz = λeαz = f(1) α − 1 −α α p(1)(z), (14) Suppose α 6= 1. Since D = E(a; f) ∩E(a; f(1)) ∩E(a; L(p)) ∩E(a; L(q)) 6= ∅, we have f(z2) = f(1)(z2) = L(p)(z2) = L(q)(z2) = a(z2), for some z2 ∈ D. From (14) we get a(z2) = a(z2) α − 1 −α α P(1)(z2) or, a(z2)(1 − 1 α ) + 1 −α α P(1)(z2) = 0 or, (α− 1){a(z2) −P(1)(z2)} = 0 or, a(z2) −P(1)(z2) = 0. Clearly a(z) −P(1)(z) 6≡ 0, because deg(P(1)(z)) is less than deg(a). N(r,a; f) ≤ NA(r,a; f) + NB(r,a; f(1)) + ND(r,a; f) + S(r,f) ≤ N(r, 0; a−P(1)) + S(r,f) = S(r,f). Then from the hypothesis T(r,f) = S(r,f), a contradiction. Therefore α = 1, so f = λez. Again L = n∑ k=1 akf (k) = ( n∑ k=1 akα k)λeαz = λez. 223 I. Kaish and N. Gazi Therefore f = L = λez. This completes the lemma. 3 Proof of the Main Theorem Proof. First we claim that f is a transcendental entire function. If f is a polynomial, then T(r,f) = O(log r) and NA(r,a; f) + NA(r,a; f(1)) = O(log r). Then from the hypothesis we get O(log r) = O(log T(r,f)) = S(r,f), which implies T(r,f) = S(r,f), a contradiction. Therefore A = ∅. Similarly NB(r,a; f(1)) = S(r,f) implies B = ∅. Therefore E(a; f) = E(a; f(1)) and E(a; f(1)) ⊂ E(a; L(p)) ∩E(a; L(q)). Let deg(f) = m and deg(a) = r . If m ≥ r + 1 then deg(f − a) = m and deg(f(1) −a) ≤ m− 1 which contradicts that E(a,f) = E(a,f(1)). If m ≤ r − 1 , then deg(f − a) = deg(f(1) − a) = r. Since E(a,f) = E(a,f(1)), (f −a) = t(f(1) −a), where t(6= 0) is a constant. If t = 1, then f = f(1), which is a contradiction because f is a polynomial. If t 6= 1 then tf(1)−f ≡ (t−1)a, which is impossible because deg((t−1)a) = r and deg(tf(1) −f) = m and m < r. Therefore our claim ” f is transcendental entire function ” is established. Now we prove the result into two cases. Case 1. Let f ≡ L(p). Then m(r,a; f) = m(r, a f −a 1 a ) ≤ m(r, a f −a ) + S(r,f) = m(r, a f −a + 1 − 1) + S(r,f) ≤ m(r, a f −a + 1) + S(r,f) ≤ m(r, f f −a ) + S(r,f) = m(r, L(p) f −a ) + S(r,f), (15) since p ≥ deg(a), by Lemma of logarithmic derivative, m(r, L (p) f−a) = S(r,f). So from (15) m(r,a; f) = S(r,f). Therefore by Lemma 5, f = L = λez,λ(6= 0) is a constant. Case 2. Let f 6≡ L(p). This case can be divided into two subcases. 224 Uniqueness of an entire function sharing a polynomial with its linear differential polynomial Subcase 2.1. Let f(1) 6≡ L(p). Since a−a(1) = (f(1)−a(1))−(f(1)−a), a common zero of f−a and f(1)−a of multiplicity s(≥ 2) is a zero of a−a(1) with multiplicity s− 1(≥ 1). Therefore N(2(r,a; f(1) | f = a) ≤ 2N(r, 0; a−a(1)) = S(r,f), where N(2(r,a; f(1) | f = a) denotes the counting function (counted with multiplicities) of those multiple zeros of f(1) −a which are also zeros of f −a. Now N(2(r,a; f (1)) ≤ NA(r,a; f(1)) + NB(r,a; f(1)) + N(2(r,a; f(1) | f = a) + S(r,f) = S(r,f). (16) Using (16) and from the hypothesis we get N(r,a; f(1)) ≤ NB(r,a; f(1)) + N(r, a−L(p)(a) a−a(1) ; L(p)(f) −L(p)(a) f(1) −a(1) ) + S(r,f) ≤ T(r, a−L(p)(a) a−a(1) ; L(p)(f) −L(p)(a) f(1) −a(1) ) + S(r,f) = N(r, L(p)(f) −L(p)(a) f(1) −a(1) ) + S(r,f) ≤ N(r,a(1); f(1)) + S(r,f). (17) Again m(r,a; f) = m(r, f(1) −a(1) f −a 1 f(1) −a(1) ) ≤ m(r,a(1); f(1)) + S(r,f) = T(r,f(1)) −N(r,a(1); f(1)) + S(r,f) = m(r,f(1)) −N(r,a(1); f(1)) + S(r,f) ≤ m(r,f) −N(r,a(1); f(1)) + S(r,f) = T(r,f) −N(r,a(1); f(1)) + S(r,f), i.e N(r,a(1); f(1)) ≤ N(r,a; f) + S(r,f). So from (17) we get N(r,a; f(1)) ≤ N(r,a; f) + S(r,f). (18) Also N(r,a; f) ≤ NA(r,a; f) + N(r,a; f | f(1) = a) ≤ N(r,a; f(1)) + S(r,f). (19) 225 I. Kaish and N. Gazi From (18) and (19) we get N(r,a; f(1)) = N(r,a; f) + S(r,f). (20) Let h = (a−a(1))(L(p)(f) −L(p)(a)) − (a−L(p)(a))(f(1) −a(1)) f −a , which is de- fined in Lemma 2.4. Clearly T(r,h) = S(r,h). Now T(r,f) = m(r,f) = m(r,a + 1 h {(a−a(1))(L(p)(f) −L(p)(a)) − (a−L(p))(f(1) −a(1))} ≤ m(r, (a−a(1))L(p)(f) − (a−L(p))f(1)) + S(r,f) ≤ m(r,f(1)) + S(r,f) = T(r,f(1)) + S(r,f) = m(r,f(1)) + S(r,f) ≤ m(r,f) + S(r,f) = T(r,f) + S(r,f). Therefore T(r,f(1)) = T(r,f) + S(r,f). (21) If h is transcendental, then by Lemma 2.4, m(r,a; f(1)) = S(r,f) and from (20) and (21) m(r,a; f) = S(r,f). So from Lemma 2.5, f = L = λez, λ( 6= 0), is a constant. If h is rational, then by Lemma 2.2 we see that f is of finite order. So by Lemma 2.1 we get m(r,a; f) = S(r,f). Therefore from Lemma 2.5, f = L = λez, λ(6= 0) is a constant. 226 Uniqueness of an entire function sharing a polynomial with its linear differential polynomial Subcase 2.2. Let f(1) ≡ L(p). Now m(r,a; f) = m(r, a(1) f −a 1 a(1) ) ≤ m(r, a(1) f −a ) + S(r,f) = m(r, f(1) − (f(1) −a(1)) f −a + S(r,f) ≤ m(r, f(1) f −a ) + S(r,f) = m(r, L(p) f −a ) + S(r,f). (22) Since p ≥ deg(a), by Lemma of logarithmic derivative, m(r, L (p) f−a) = S(r,f), so from (22) m(r,a; f) = S(r,f). Therefore from Lemma 2.5, we get f = L = λez, λ(6= 0), is a constant. This completes the proof of the Main Theorem. 4 Conclusions Finally we arrive at the conclusion that a non-constant entire function sharing a polynomial with its linear differential polynomial with some conditions defined in Theorem (1.1) belongs to the class of functions F = {λez : λ ∈ C\{0}}. 5 Acknowledgements Authors are thankful to all the referees for their remarkable suggestions and all the authors of various papers and books which have been consulted to built the work. References W. K. Hayman. Meromorphic Functions. The Clarendon Press, Oxford, 1964. G. Jank, E. Mues, and L. Volkmann. Meromorphe funktionen, die mit ihrer ersten und zweiten ableitung einen endlichen wert teilen. Complex Var. Theory Appl., 6:51–71, 1986. 227 I. Kaish and N. Gazi I. Lahiri and I. Kaish. An entire function sharing a polynomial with its derivatives. Boll. Unione Mat. Ital., 10:229–240, 2017. I. Lain. Nevanlinna theory and complex differential equations. Watter de Gruyter, New York, 1993. P. Li. Entire functions that share one value with their linear differential polynomi- als. Kodai Math. j., 22:446–457, 1999. E. Mues and N. Steinmetz. Meromorphe functionen, die mit ihrer ableitung werte teilen. Manuscripta Math., 29:195–206, 1979. L. A. Rubel and C. C. Yang. Values shared by an entire function and its derivative. Lecture Notes in Math.(Springer), 599:101–103, 1977. C. C. Yang and H. X. Yi. Uniqueness Theory of Meromorphic Functions. Science Press and Kluwer Academic Publishers, New York, 2003. H. Zhong. Entire functions that share one value with their derivatives. Kodai Math. J., 18:250–259, 1995. 228