aa r X i v : . [ m a t h . C V ] F e b Uniqueness of p ( f ) and P [ f ] Kuldeep Singh Charak , Banarsi Lal Department of Mathematics, University of Jammu, Jammu-180 006, INDIA. E-mail: kscharak7@rediffmail.com E-mail: [email protected]
Abstract
Let f be a non-constant meromorphic function, a ( , ∞ ) be a mero-morphic function satisfying T ( r, a ) = o ( T ( r, f )) as r → ∞ , and p ( z ) bea polynomial of degree n ≥ p (0) = 0. Let P [ f ] be a non-constantdifferential polynomial of f . Under certain essential conditions, we provethe uniqueness of p ( f ) and P [ f ] when p ( f ) and P [ f ] share a with weight l ≥
0. Our result generalizes the results due to Zang and Lu, Banerjee andMajumdar, Bhoosnurmath and Kabbur and answers a question of Zangand Lu.
Keywords:
Meromorphic functions, small functions, sharing of values, differ-ential polynomials, Nevanlinna theory.
AMS subject classification: 30D35, 30D30 Introduction
Let f and g be two non constant meromorphic functions and k be a non-negativeinteger. For a ∈ C ∪ {∞} , we denote by E k ( 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 timesif m > k . If E k ( a, f ) = E k ( a, g ), we say that f and g share the value a withweight k .We write “ f and g share ( a, k )” to mean that “ f and g share the value a with weight k ”. Since E k ( a, f ) = E k ( a, g ) implies E p ( a, f ) = E p ( a, g ) forany integer p (0 ≤ p < k ), clearly if f and g share ( a, k ), then f and g share( a, p ), 0 ≤ p < k . Also we note that f and g share the value a IM(ignoringmultilicity) or CM(counting multiplicity) if and only if f and g share ( a,
0) or( a, ∞ ), respectively.A differential polynomial P [ f ] of a non-constant meromorphic function f isdefined as P [ f ] := m X i =1 M i [ f ] , where M i [ f ] = a i . Q kj =0 ( f ( j ) ) n ij with n i , n i , . . . , n ik as non-negative integersand a i (
0) are meromorphic functions satisfying T ( r, a i ) = o ( T ( r, f )) as r →∞ . The numbers d ( P ) = m ax ≤ i ≤ m P kj =0 n ij and d ( P ) = m in ≤ i ≤ m P kj =0 n ij are respectively called the degree and lower degree of P [ f ]. If d ( P ) = d ( P ) = d (say), then we say that P [ f ] is a homogeneous differential polynomial of degree d . For notational purpose, let f and g share 1 IM, and let z be a zero of f − p and a zero of g − q . We denote by N E ( r, / ( f − f − p = q = 1.By N (2 E ( r, / ( f − f − p = q ≥ N L ( r, / ( f − f − p > q ≥
1, each point in these counting functions iscounted only once; similarly, the terms N E ( r, / ( g − N (2 E ( r, / ( g − N L ( r, / ( g − N f>k ( r, / ( g − f − g − p > q = k , and similarly theterm N g>k ( r, / ( f − If f isa non-constant entire function sharing two distinct values ignoring multiplicitywith f ′ , then f ≡ f ′ ”, the study of the uniqueness of f and f ( k ) , f n and ( f m ) ( k ) , f and P [ f ] is carried out by numerous authors. For example, Zang and Lu [12]proved : Theorem A.
Let k , n be the positive integers, f be a non-constant mero-morphic function, and a ( , ∞ ) be a meromorphic function satisfying T ( r, a ) =2 ( T ( r, f )) as r → ∞ . If f n and f ( k ) share a IM and (2 k + 6)Θ( ∞ , f ) + 4Θ(0 , f ) + 2 δ k (0 , f ) > k + 12 − n, or f n and f ( k ) share a CM and ( k + 3)Θ( ∞ , f ) + 2Θ(0 , f ) + δ k (0 , f ) > k + 6 − n, then f n ≡ f ( k ) . In the same paper, T. Zhang and W. Lu asked the following question:
Question 1:
What will happen if f n and ( f ( k ) ) m share a meromorphicfunction a ( , ∞ ) satisfying T ( r, a ) = o ( T ( r, f )) as r → ∞ ? S.S.Bhoosnurmath and Kabbur [3] proved:
Theorem B.
Let f be a non-constant meromorphic function and a ( , ∞ ) be a meromorphic function satisfying T ( r, a ) = o ( T ( r, f )) as r → ∞ . Let P [ f ] be a non-constant differential polynomial of f . If f and P [ f ] share a IM and (2 Q + 6)Θ( ∞ , f ) + (2 + 3 d ( P )) δ (0 , f ) > Q + 2 d ( P ) + d ( P ) + 7 , or if f and P [ f ] share a CM and ∞ , f ) + ( d ( P ) + 1) δ (0 , f ) > , then f ≡ P [ f ] . Banerjee and Majumder [2] considered the weighted sharing of f n and ( f m ) ( k ) and proved the following result: Theorem C.
Let f be a non-constant meromorphic function, k, n, m ∈ N and l be a non negative integer. Suppose a ( , ∞ ) be a meromorphic functionsatisfying T ( r, a ) = o ( T ( r, f )) as r → ∞ such that f n and ( f m ) ( k ) share ( a, l ) .If l ≥ and ( k + 3)Θ( ∞ , f ) + ( k + 4)Θ(0 , f ) > k + 7 − n, or l = 1 and (cid:18) k + 72 (cid:19) Θ( ∞ , f ) + (cid:18) k + 92 (cid:19) Θ(0 , f ) > k + 8 − n, or l = 0 and (2 k + 6)Θ( ∞ , f ) + (2 k + 7)Θ(0 , f ) > k + 13 − n, then f n ≡ ( f m ) ( k ) . Motivated by such uniqueness investigations, it is rational to think about theproblem in more general setting:
Let f be a non-constant meromorphic function, [ f ] be a non-cnstant differential polynomial of f, p ( z ) be a polynomial of degree n ≥ and a ( , ∞ ) be a meromorphic function satisfying T ( r, a ) = o ( T ( r, f )) as r → ∞ . If p ( f ) and P [ f ] share ( a, l ) , l ≥ , then is it true that p ( f ) ≡ P [ f ] ?Generally this is not true, but under certain essential conditions, we provethe following result: Theorem 1.1.
Let f be a non-constant meromorphic function, a ( , ∞ ) be ameromorphic function satisfying T ( r, a ) = o ( T ( r, f )) as r → ∞ , and p ( z ) be apolynomial of degree n ≥ with p (0) = 0 . Let P [ f ] be a non-constant differentialpolynomial of f . Suppose p ( f ) and P [ f ] share ( a, l ) with one of the followingconditions: ( i ) l ≥ and ( Q + 3)Θ( ∞ , f ) + 2 n Θ(0 , p ( f )) + d ( P ) δ (0 , f ) > Q + 3 + 2 d ( P ) − d ( P ) + n, (1.1)( ii ) l = 1 and (cid:18) Q + 72 (cid:19) Θ( ∞ , f ) + 5 n , p ( f )) + d ( P ) δ (0 , f ) > Q + 72 + 2 d ( P ) − d ( P ) + 3 n , (1.2)( iii ) l = 0 and (2 Q + 6)Θ( ∞ , f ) + 4 n Θ(0 , p ( f )) + 2 d ( P ) δ (0 , f ) > Q + 6 + 4 d ( P ) − d ( P ) + 3 n. (1.3) Then p ( f ) ≡ P [ f ] . Example 1.2.
Consider the function f ( z ) = cosαz + 1 − /α , where α =0 , ± , ± i and p ( z ) = z . Then p ( f ) and P [ f ] ≡ f ( iv ) share (1 , l ) , l ≥ and noneof the inequalities (1.1), (1.2) and (1.3) is satisfied, and p ( f ) = P [ f ] . Thusconditions in Theorem . can not be removed. Remark 1.3.
Theorem . generalizes Theorem A , Theorem B , Theorem C (and also generalizes Theorem . and Theorem . of [2]) and provides ananswer to a question of Zhang and Lu [12]. The main tool of our investigations in this paper is Nevanlinna value distri-bution theory[5].
We shall use the following results in the proof of our main result:
Lemma 2.1. [3] Let f be a non-constant meromorphic function and P [ f ] be adifferential polynomial of f . Then m (cid:18) r, P [ f ] f d ( P ) (cid:19) ≤ ( d ( P ) − d ( P )) m (cid:18) r, f (cid:19) + S ( r, f ) , (2.1)4 (cid:18) r, P [ f ] f d ( P ) (cid:19) ≤ ( d ( P ) − d ( P )) N (cid:18) r, f (cid:19) + Q (cid:20) N ( r, f ) + N (cid:18) r, f (cid:19)(cid:21) + S ( r, f ) , (2.2) N (cid:18) r, P [ f ] (cid:19) ≤ QN ( r, f ) + ( d ( P ) − d ( P )) m (cid:18) r, f (cid:19) + N (cid:18) r, f d ( P ) (cid:19) + S ( r, f ) , (2.3) where Q = m ax ≤ i ≤ m { n i + n i + 2 n i + ... + kn ik } . Lemma 2.2. [1] Let f and g be two non-constant meromorphic functions.(i) If f and g share (1 , , then N L (cid:18) r, f − (cid:19) ≤ N (cid:18) r, f (cid:19) + N ( r, f ) + S ( r ) , (2.4) where S ( r ) = o ( T ( r )) as r → ∞ with T ( r ) = m ax { T ( r, f ); T ( r, g ) } .(ii) If f and g share (1 , , then N L (cid:18) r, f − (cid:19) + 2 N L (cid:18) r, g − (cid:19) + N (2 E (cid:18) r, f − (cid:19) − N f> (cid:18) r, g − (cid:19) ≤ N (cid:18) r, g − (cid:19) − N (cid:18) r, g − (cid:19) . (2.5) Proof of Theorem 1.1:
Let F = p ( f ) /a and G = P [ f ] /a . Then F − p ( f ) − aa and G − P [ f ] − aa . (2.6)Since p ( f ) and P [ f ] share ( a, l ), it follows that F and G share (1 , l ) except atthe zeros and poles of a . Also note that N ( r, F ) = N ( r, f ) + S ( r, f ) and N ( r, G ) = N ( r, f ) + S ( r, f ) . Define ψ = (cid:18) F ′′ F ′ − F ′ F − (cid:19) − (cid:18) G ′′ G ′ − G ′ G − (cid:19) . (2.7) Claim: ψ ≡ ψ
0. Then from (2.7), we have m ( r, ψ ) = S ( r, f ) . By the Second fundamental theorem of Nevanlinna, we have T ( r, F ) + T ( r, G ) ≤ N ( r, f ) + N (cid:18) r, F (cid:19) + N (cid:18) r, F − (cid:19) + N (cid:18) r, G (cid:19) + N (cid:18) r, G − (cid:19) − N (cid:18) r, F ′ (cid:19) − N (cid:18) r, G ′ (cid:19) + S ( r, f ) , (2.8)5here N ( r, /F ′ ) denotes the counting function of the zeros of F ′ which arenot the zeros of F ( F −
1) and N ( r, /G ′ ) denotes the counting function of thezeros of G ′ which are not the zeros of G ( G − Case 1.
When l ≥ N E (cid:18) r, F − (cid:19) ≤ N (cid:18) r, ψ (cid:19) + S ( r, f ) ≤ T ( r, ψ ) + S ( r, f )= N ( r, ψ ) + S ( r, f ) ≤ N ( r, F ) + N (2 (cid:18) r, F (cid:19) + N (2 (cid:18) r, G (cid:19) + N L (cid:18) r, F − (cid:19) + N L (cid:18) r, G − (cid:19) + N (cid:18) r, F ′ (cid:19) + N (cid:18) r, G ′ (cid:19) + S ( r, f ) . and so N (cid:18) r, F − (cid:19) + N (cid:18) r, G − (cid:19) = N E (cid:18) r, F − (cid:19) + N (2 E (cid:18) r, F − (cid:19) + N L (cid:18) r, F − (cid:19) + N L (cid:18) r, G − (cid:19) + N (cid:18) r, G − (cid:19) + S ( r, f ) ≤ N ( r, f ) + N (2 (cid:18) r, F (cid:19) + N (2 (cid:18) r, G (cid:19) + 2 N L (cid:18) r, F − (cid:19) + 2 N L (cid:18) r, G − (cid:19) + N (2 E (cid:18) r, F − (cid:19) + N (cid:18) r, G − (cid:19) + N (cid:18) r, F ′ (cid:19) + N (cid:18) r, G ′ (cid:19) + S ( r, f ) . (2.9) Subcase 1.1:
When l = 1.In this case, we have N L (cid:18) r, F − (cid:19) ≤ N (cid:18) r, F ′ | F = 0 (cid:19) ≤ N ( r, F ) + 12 N (cid:18) r, F (cid:19) , (2.10)where N (cid:0) r, F ′ | F = 0 (cid:1) denotes the zeros of F ′ , that are not the zeros of F .6rom (2.5) and (2.10), we have2 N L (cid:18) r, F − (cid:19) + 2 N L (cid:18) r, G − (cid:19) + N (2 E (cid:18) r, F − (cid:19) + N (cid:18) r, G − (cid:19) ≤ N (cid:18) r, G − (cid:19) + N L (cid:18) r, F − (cid:19) + S ( r, f ) ≤ N (cid:18) r, G − (cid:19) + 12 N ( r, F ) + 12 N (cid:18) r, F (cid:19) + S ( r, f ) ≤ N (cid:18) r, G − (cid:19) + 12 N ( r, f ) + 12 N (cid:18) r, p ( f ) (cid:19) + S ( r, f ) . (2.11)Thus, from (2.9) and (2.11), we have N (cid:18) r, F − (cid:19) + N (cid:18) r, G − (cid:19) ≤ N ( r, f ) + N (2 (cid:18) r, F (cid:19) + N (2 (cid:18) r, G (cid:19) + 12 N ( r, f ) + 12 N (cid:18) r, p ( f ) (cid:19) + N (cid:18) r, G − (cid:19) + N (cid:18) r, F ′ (cid:19) + N (cid:18) r, G ′ (cid:19) + S ( r, f ) ≤ N ( r, f ) + N (2 (cid:18) r, F (cid:19) + N (2 (cid:18) r, G (cid:19) + 12 N ( r, f ) + 12 N (cid:18) r, p ( f ) (cid:19) + T ( r, G )+ N (cid:18) r, F ′ (cid:19) + N (cid:18) r, G ′ (cid:19) + S ( r, f ) . (2.12)From (2.3), (2.8) and (2.12), we obtain T ( r, F ) ≤ N ( r, f ) + N (cid:18) r, F (cid:19) + N (2 (cid:18) r, F (cid:19) + N (cid:18) r, G (cid:19) + N (2 (cid:18) r, G (cid:19) + 12 N ( r, f ) + 12 N (cid:18) r, p ( f ) (cid:19) + S ( r, f ) ≤ N ( r, f ) + 2 N (cid:18) r, F (cid:19) + N (cid:18) r, G (cid:19) + 12 N (cid:18) r, p ( f ) (cid:19) + S ( r, f ) ≤ N ( r, f ) + 52 N (cid:18) r, p ( f ) (cid:19) + N (cid:18) r, P [ f ] (cid:19) + S ( r, f ) ≤ (cid:18) Q + 72 (cid:19) N ( r, f ) + 52 N (cid:18) r, p ( f ) (cid:19) + ( d ( P ) − d ( P )) T ( r, f ) + d ( P ) N (cid:18) r, f (cid:19) + S ( r, f ) ≤ (cid:20)(cid:18) Q + 72 (cid:19) { − Θ( ∞ , f ) } + 5 n { − Θ(0 , p ( f )) } + d ( P ) { − δ (0 , f ) } (cid:21) T ( r, f )+ ( d ( P ) − d ( P )) T ( r, f ) + S ( r, f ) . nT ( r, f ) = T ( r, F ) + S ( r, f ) ≤ (cid:20)(cid:18) Q + 72 (cid:19) { − Θ( ∞ , f ) } + 5 n { − Θ(0 , p ( f )) } + d ( P ) { − δ (0 , f ) } (cid:21) T ( r, f )+ ( d ( P ) − d ( P )) T ( r, f ) + S ( r, f ) . Thus (cid:20) { (cid:18) Q + 72 (cid:19) Θ( ∞ , f ) + 5 n , p ( f )) + d ( P ) δ (0 , f ) } − { Q + 72 + 2 d ( P ) − d ( P ) + 3 n } (cid:21) T ( r, f ) ≤ S ( r, f ) . That is, (cid:18) Q + 72 (cid:19) Θ( ∞ , f ) + 5 n , p ( f )) + d ( P ) δ (0 , f ) ≤ Q + 72 + 2 d ( P ) − d ( P ) + 3 n , which violates (1.2). Subcase 1.2:
When l ≥ N L (cid:18) r, F − (cid:19) +2 N L (cid:18) r, G − (cid:19) + N (2 E (cid:18) r, F − (cid:19) + N (cid:18) r, G − (cid:19) ≤ N (cid:18) r, G − (cid:19) + S ( r, f ) . Thus from (2.9), we obtain N (cid:18) r, F − (cid:19) + N (cid:18) r, G − (cid:19) ≤ N ( r, f ) + N (2 (cid:18) r, F (cid:19) + N (2 (cid:18) r, G (cid:19) + N (cid:18) r, G − (cid:19) + N (cid:18) r, F ′ (cid:19) + N (cid:18) r, G ′ (cid:19) + S ( r, f ) ≤ N ( r, f ) + N (2 (cid:18) r, F (cid:19) + N (2 (cid:18) r, G (cid:19) + T ( r, G )+ N (cid:18) r, F ′ (cid:19) + N (cid:18) r, G ′ (cid:19) + S ( r, f ) . (2.13)Now from (2.3), (2.8) and (2.13), we obtain T ( r, F ) ≤ N ( r, f ) + N (cid:18) r, F (cid:19) + N (2 (cid:18) r, F (cid:19) + N (cid:18) r, G (cid:19) + N (2 (cid:18) r, G (cid:19) + S ( r, f ) ≤ N ( r, f ) + 2 N (cid:18) r, F (cid:19) + N (cid:18) r, G (cid:19) + S ( r, f ) ≤ N ( r, f ) + 2 N (cid:18) r, p ( f ) (cid:19) + N (cid:18) r, P [ f ] (cid:19) + S ( r, f ) ≤ ( Q + 3) N ( r, f ) + 2 N (cid:18) r, p ( f ) (cid:19) + ( d ( P ) − d ( P )) T ( r, f ) + d ( P ) N (cid:18) r, f (cid:19) + S ( r, f ) ≤ [( Q + 3) { − Θ( ∞ , f ) } + 2 n { − Θ(0 , p ( f )) } + d ( P ) { − δ (0 , f ) } ] T ( r, f )+ ( d ( P ) − d ( P )) T ( r, f ) + S ( r, f ) . nT ( r, f ) = T ( r, F ) + S ( r, f ) ≤ [( Q + 3) { − Θ( ∞ , f ) } + 2 n { − Θ(0 , p ( f )) } + d ( P ) { − δ (0 , f ) } ] T ( r, f )+ ( d ( P ) − d ( P )) T ( r, f ) + S ( r, f ) . Thus[ { ( Q +3)Θ( ∞ , f )+2 n Θ(0 , p ( f ))+ d ( P ) δ (0 , f ) }−{ ( Q +3+2 d ( P ) − d ( P )+ n } ] T ( r, f ) ≤ S ( r, f ) . That is,( Q + 3)Θ( ∞ , f ) + 2 n Θ(0 , p ( f )) + d ( P ) δ (0 , f ) ≤ Q + 3 + 2 d ( P ) − d ( P ) + n, which violates (1.1). Case 2.
When l = 0.Then, we have N E (cid:18) r, F − (cid:19) = N E (cid:18) r, G − (cid:19) + S ( r, f ) , N (2 E (cid:18) r, F − (cid:19) = N (2 E (cid:18) r, G − (cid:19) + S ( r, f ) , and also from (2.7), we have N (cid:18) r, F − (cid:19) + N (cid:18) r, G − (cid:19) ≤ N E (cid:18) r, F − (cid:19) + N (2 E (cid:18) r, F − (cid:19) + N L (cid:18) r, F − (cid:19) + N L (cid:18) r, G − (cid:19) + N (cid:18) r, G − (cid:19) + S ( r, f ) ≤ N E (cid:18) r, F − (cid:19) + N L (cid:18) r, F − (cid:19) + N (cid:18) r, G − (cid:19) + S ( r, f ) ≤ N ( r, F ) + N (2 (cid:18) r, F (cid:19) + N (2 (cid:18) r, G (cid:19) + 2 N L (cid:18) r, F − (cid:19) + N L (cid:18) r, G − (cid:19) + N (cid:18) r, G − (cid:19) + N (cid:18) r, F ′ (cid:19) + N (cid:18) r, G ′ (cid:19) + S ( r, f ) . (2.14)9rom (2.3),(2.4),(2.8) and (2.14), we obtain T ( r, F ) ≤ N ( r, f ) + N (cid:18) r, F (cid:19) + N (2 (cid:18) r, F (cid:19) + N (cid:18) r, G (cid:19) + N (2 (cid:18) r, G (cid:19) + 2 N L (cid:18) r, F − (cid:19) + N L (cid:18) r, G − (cid:19) + S ( r, f ) ≤ N ( r, f ) + 2 N (cid:18) r, F (cid:19) + N (cid:18) r, G (cid:19) + 2 N (cid:18) r, F (cid:19) + 2 N ( r, F ) + N (cid:18) r, G (cid:19) + N ( r, G ) + S ( r, f ) ≤ N ( r, f ) + 4 N (cid:18) r, F (cid:19) + 2 N (cid:18) r, G (cid:19) + S ( r, f ) ≤ N ( r, f ) + 4 N (cid:18) r, p ( f ) (cid:19) + 2 N (cid:18) r, P [ f ] (cid:19) + S ( r, f ) ≤ (2 Q + 6) N ( r, f ) + 4 N (cid:18) r, p ( f ) (cid:19) + 2( d ( P ) − d ( P )) T ( r, f ) + 2 d ( P ) N (cid:18) r, f (cid:19) + S ( r, f ) ≤ [(2 Q + 6) { − Θ( ∞ , f ) } + 4 n { − Θ(0 , p ( f )) } + 2 d ( P ) { − δ (0 , f ) } ] T ( r, f )+ 2( d ( P ) − d ( P )) T ( r, f ) + S ( r, f ) . That is, nT ( r, f ) = T ( r, F ) + S ( r, f ) ≤ [(2 Q + 6) { − Θ( ∞ , f ) } + 4 n { − Θ(0 , p ( f )) } + 2 d ( P ) { − δ (0 , f ) } ] T ( r, f )+ 2( d ( P ) − d ( P )) T ( r, f ) + S ( r, f ) . Thus[ { (2 Q +6)Θ( ∞ , f )+4 n Θ(0 , p ( f ))+2 d ( P ) δ (0 , f ) }−{ Q +6+4 d ( P ) − d ( P )+3 n } ] T ( r, f ) ≤ S ( r, f ) . That is,(2 Q + 6)Θ( ∞ , f ) + 4 n Θ(0 , p ( f )) + 2 d ( P ) δ (0 , f ) ≤ Q + 6 + 4 d ( P ) − d ( P ) + 3 n, which violates (1.3).This proves the claim and thus ψ ≡
0. So (2.7) implies that F ′′ F ′ − F ′ F − G ′′ G ′ − G ′ G − , and so we obtain 1 F − CG − D, (2.15)where C = 0 and D are constants. 10ere, the following three cases can arise: Case ( i ) : When D = 0 , −
1. Rewriting (2.15) as G − C = F − D + 1 − DF , we have N ( r, G ) = N (cid:18) r, F − ( D + 1) /D (cid:19) . In this subcase, the Second fundamental theorem of Nevanlinna yields nT ( r, f ) = T ( r, F ) + S ( r, f ) ≤ N ( r, F ) + N (cid:18) r, F (cid:19) + N (cid:18) r, F − ( D + 1) /D (cid:19) + S ( r, f ) ≤ N ( r, F ) + N (cid:18) r, F (cid:19) + N ( r, G ) + S ( r, f ) ≤ N ( r, f ) + N (cid:18) r, p ( f ) (cid:19) + S ( r, f )= [2 { − Θ( ∞ , f ) } + n { − Θ(0 , p ( f )) } ] T ( r, f ) + S ( r, f ) . Thus [ { ∞ , f ) + n Θ(0 , p ( f )) } − T ( r, f ) ≤ S ( r, f ) . That is, 2Θ( ∞ , f ) + n Θ(0 , p ( f )) ≤ , which contradicts (1.1),(1.2) and (1.3). Case ( ii ) : When D = 0. Then from (2.15), we have G = CF − ( C − . (2.16)So if C = 1, then N (cid:18) r, G (cid:19) = N (cid:18) r, F − ( C − /C (cid:19) . nT ( r, f ) = T ( r, F ) + S ( r, f ) ≤ N ( r, F ) + N (cid:18) r, F (cid:19) + N (cid:18) r, F − ( C − /C (cid:19) + S ( r, f ) ≤ N ( r, F ) + N (cid:18) r, F (cid:19) + N (cid:18) r, G (cid:19) + S ( r, f ) ≤ N ( r, f ) + N (cid:18) r, p ( f ) (cid:19) + N (cid:18) r, P [ f ] (cid:19) + S ( r, f ) ≤ N ( r, f ) + N (cid:18) r, p ( f ) (cid:19) + QN ( r, f ) + ( d ( P ) − d ( P )) m (cid:18) r, f (cid:19) + N (cid:18) r, f d ( P ) (cid:19) + S ( r, f ) ≤ ( Q + 1) N ( r, f ) + N (cid:18) r, p ( f ) (cid:19) + ( d ( P ) − d ( P )) T ( r, f )+ d ( P ) N (cid:18) r, f (cid:19) + S ( r, f ) ≤ [( Q + 1) { − Θ( ∞ , f ) } + n { − Θ(0 , p ( f )) } + d ( P ) { − δ (0 , f ) } ] T ( r, f )+ ( d ( P ) − d ( P )) T ( r, f ) + S ( r, f ) . Thus[ { ( Q +1)Θ( ∞ , f )+ n Θ(0 , p ( f ))+ d ( P ) δ (0 , f ) }−{ Q +1+2 d ( P ) − d ( P ) } ] T ( r, f ) ≤ S ( r, f ) . That is,( Q + 1)Θ( ∞ , f ) + n Θ(0 , p ( f )) + d ( P ) δ (0 , f ) ≤ Q + 1 + 2 d ( P ) − d ( P ) , which contradicts (1.1),(1.2) and (1.3).Thus, C = 1 and so in this case from (2.16), we obtain F ≡ G and so p ( f ) ≡ P [ f ] . Case ( iii ) : When D = −
1. Then from (2.15) we have1 F − CG − − . (2.17)So if C = −
1, then N (cid:18) r, G (cid:19) = N (cid:18) r, F − C/ ( C + 1) (cid:19) , and as in the Subacase ( ii ), we find that nT ( r, f ) ≤ ( Q + 1) N ( r, f ) + N (cid:18) r, p ( f ) (cid:19) + ( d ( P ) − d ( P )) T ( r, f )+ d ( P ) N (cid:18) r, f (cid:19) + S ( r, f ) . { ( Q +1)Θ( ∞ , f )+ n Θ(0 , p ( f ))+ d ( P ) δ (0 , f ) }−{ Q +1+2 d ( P ) − d ( P ) } ] T ( r, f ) ≤ S ( r, f ) . That is,( Q + 1)Θ( ∞ , f ) + n Θ(0 , p ( f )) + d ( P ) δ (0 , f ) ≤ Q + 1 + 2 d ( P ) − d ( P ) , which contradicts (1.1),(1.2) and (1.3).Thus, C = − F G ≡ p ( f ) P [ f ] = a . Thus, in this case N ( r, f ) + N ( r, /f ) = S ( r, f ).Now, by using (2.1) and (2.2), we have( n + d ( P )) T ( r, f ) ≤ T (cid:18) r, a f n + d ( P ) (cid:19) + S ( r, f ) ≤ T (cid:18) r, (cid:20) a n − f + − − − + a f n − (cid:21) . P [ f ] f d ( P ) (cid:19) + S ( r, f ) ≤ ( n − T ( r, f ) + T (cid:18) r, P [ f ] f d ( P ) (cid:19) + S ( r, f )= ( n − T ( r, f ) + m (cid:18) r, P [ f ] f d ( P ) (cid:19) + N (cid:18) r, P [ f ] f d ( P ) (cid:19) + S ( r, f ) ≤ ( n − T ( r, f ) + ( d ( P ) − d ( P )) m (cid:18) r, f (cid:19) + ( d ( P ) − d ( P )) N (cid:18) r, f (cid:19) + Q (cid:20) N ( r, f ) + N (cid:18) r, f (cid:19)(cid:21) + S ( r, f ) ≤ ( n − T ( r, f ) + ( d ( P ) − d ( P )) T ( r, f ) + S ( r, f ) . Thus (1 + d ( P )) T ( r, f ) ≤ S ( r, f ) , which is a contradiction. (cid:3) References [1] A. Banerjee,
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