Filomat 31:1 (2017), 113 123 DOI 10.2298/FIL1701113L Published by Faculty of Sciences and Mathematics, University of Niš, Serbia Available at: htt://www.mf.ni.ac.rs/filomat Convolution Proerties for Certain Meromorhically Multivalent Functions Jin-Lin Liu a, Reha Srivastava b a Deartment of Mathematics, Yangzhou University, Yangzhou 225002, P. R. China b Deartment of Mathematics and Statistics, University of Victoria, Victoria, British Columbia V8W 3R4, Canada Abstract. Let Σ denote the class of functions of the form f (z) z + a n z n ( N). Two new subclasses H, (λ, A, B) and Q, (λ, A, B) of meromorhically multivalent functions starlie resect to -symmetric oints in Σ are investigated. Certain convolution roerties for these subclasses are obtained. 1. Introduction and Preliminaries Throughout this aer we assume that N {1, 2, 3, }, N \ {1}, 1 B < 0, B < A B and 1 λ < +. (1.1) For functions f and analytic in the oen unit dis U {z : z < 1}, the function f is said to be subordinate to, written f (z) (z) (z U), if there exists an analytic function w in U w(0) 0 and w(z) < 1 such that f (z) (w(z)) in U. Let Σ denote the class of functions of the form f (z) z + a n z n ( N), (1.2) which are analytic in the unctured oen unit dis U 0 U \ {0}. Let f j (z) z + a n,j z n Σ (j 1, 2). The Hadamard roduct (or convolution) of f 1 and f 2 is defined by ( f 1 f 2 )(z) z + a n,1 a n,2 z n. 2010 Mathematics Subject Classification. Primary 30C45, 30C80. Keywords. Meromorhically multivalent function; Hadamard roduct (or convolution); subordination; -symmetric oint. Received: 09 January 2017; Acceted: 13 January 2017 Communicated by Hari M. Srivastava This wor is suorted by National Natural Science Foundation of China(Grant No.11571299) and Natural Science Foundation of Jiangsu Province(Grant No. BK20151304). Email addresses: jlliu@yzu.edu.cn (Jin-Lin Liu), rehas@math.uvic.ca (Reha Srivastava)
J.-L. Liu, R. Srivastava / Filomat 31:1 (2017), 113 123 114 In [3] (see also [4]) Dzio obtained the following result. Lemma. Let f Σ defined by (1.2) satisfy [(1 A)δ n,, + ()(nλ + (λ 1))] a n (A B), (1.3) where Then δ n,, 0 1 ( n+ N ), ( n+ N ). (1.4) (1 λ) f (z) λz f (z) f, (z) 1 + Az 1 + Bz (z U), (1.5) where f, (z) 1 1 j0 ε j f (εj z) and ε ex ( ) 2πi. (1.6) Recently, Liu and Srivastava [9] introduced and investigated two new subclasses H, (λ, A, B) and Q, (λ, A, B) of Σ as follows. A function f Σ is said to be in the class H, (λ, A, B) if and only if it satisfies the coefficient inequality (1.3). Also, a function f Σ is said to be in the class Q, (λ, A, B) if and only if it satisfies the coefficient inequality n[(1 A)δ n,, + ()(nλ + (λ 1))] a n 2 (A B). (1.7) For f Σ, one can see that f Q, (λ, A, B) if and only if 2z + z f (z) H, (λ, A, B). (1.8) In [9] the authors ointed out that each function in the classes H, (λ, A, B) and Q, (λ, A, B) is meromorhically starlie resect to -symmetric oints. The subject of meromorhically univalent and multivalent functions continue to receive a great deal of attention. Very recently, Srivastava et al. (see, e.g., [14, 15, 16, 17, 18, 19, 20, 22]) investigated various subclasses of meromorhic functions. Certain roerties such as distortion bounds, inclusion relations and integral transforms for these subclasses are studied. Motivated essentially by these wors and some other wors (see also, e.g., [1], [2], [5-13], [21] and [23-26]), we aim at investigating here convolution roerties of the subclasses H, (λ, A, B) and Q, (λ, A, B).
J.-L. Liu, R. Srivastava / Filomat 31:1 (2017), 113 123 115 2. Convolution Proerties for the Classes H, (λ, A, B) and Q, (λ, A, B) In this section we assume that 1 B j < 0 and B j < A j B j (j 1, 2). (2.1) Furthermore, we denote by λ 1 the root in (1, + ) of the equation: h(λ) aλ 2 + bλ + c 0, where We also denote and a ( 1 )( 2 ), b [( 1 )( 2 ) ( 1 )(A 2 B 2 ) ( 2 )(A 1 )], c (A 1 )(A 2 B 2 ). A(B) B + Ã(B) B + Theorem 1. Let 2λ 2 () (2 + 1)λ 2 (2.2) (2.3) j. (2.4) f j H, (λ, A j, B j ) (j 1, 2) 2 N and 1 B max{b 1, B 2 }. Then we have the following: (i) If (1 A 1 )(1 A 2 ) ( 1 )( 2 ) and λ 1, then f 1 f 2 H, (λ, A(B), B). (ii) If (1 A 1 )(1 A 2 ) > ( 1 )( 2 ) and λ λ 1, then f 1 f 2 H, (λ, A(B), B). (iii) If (1 A 1 )(1 A 2 ) > ( 1 )( 2 ) and 1 λ < λ 1, then f 1 f 2 H, (λ, Ã(B), B). In all cases (i)-(iii) the numbers A(B) and Ã(B) are otimal in the sense that they cannot be decreased for each B. Proof. Suose that 1 B max{b 1, B 2 } B j (j 1 or 2). It follows from (2.1) and (2.3) that A(B) B 2λ j 2λ j j j (2λ 1) j λ 1 λ j 2B > 0, 1 A j + 1
J.-L. Liu, R. Srivastava / Filomat 31:1 (2017), 113 123 116 which imlies that B < A(B) B. Also, (2.1) and (2.4) give that Ã(B) B (2 + 1)λ j j 2B > 0, which imlies that B < Ã(B) B. Let 2 N and f j (z) z + a n,j z n H, (λ, A j, B j ) (z U 0 ; j 1, 2). Then (1 A j )δ n,, + ( j )(nλ + (λ 1)) ( ) a n,1a n,2 (1 A j )δ n,, + ( j )(nλ + (λ 1)) a n,j 1. (2.5) ( ) Also, f 1 f 2 H, (λ, A, B) if and only if (1 A)δ n,, + ()(nλ + (λ 1)) a n,1 a n,2 1. (2.6) (A B) In order to rove Theorem 1, it follows from (2.5) and (2.6) that we only need to find the smallest A such that (1 A)δ n,, + ()(nλ + (λ 1)) (A B) (1 A j )δ n,, + ( j )(nλ + (λ 1)) ( ) (n ). (2.7) For n and n+ N, (2.7) is equivalent to λ(n+) 2 2 + 2 λ(n+) ϕ 1 (n). (2.8) It can be verified that ϕ 1 (n) (n, λ 1) is decreasing in n and so, in view of 2 N, ϕ 1 (n) ϕ 1 () B + For n and n+ 2λ 2 N, (2.7) becomes 2. (2.9) nλ+(λ1) 2 ψ 1B 1 (n) (2.10) j and we have ψ 1 (n) ψ 1 ( + 1) B + (2+1)λ 2. (2.11)
J.-L. Liu, R. Srivastava / Filomat 31:1 (2017), 113 123 117 Now 2λ j h(λ) λ(a 1 )(A 2 B 2 ), j + 1 (2 + 1)λ λ j (2.12) where h(λ) λ( λ)( 1 )( 2 ) λ[( 1 )(A 2 B 2 ) + ( 2 )(A 1 )] + (A 1 )(A 2 B 2 ) aλ 2 + bλ + c (2.13) a ( 1 )( 2 ), b [( 1 )( 2 ) ( 1 )(A 2 B 2 ) ( 2 )(A 1 )], c (A 1 )(A 2 B 2 ). Note that a < 0, h(0) c > 0 and h(1) ( 1)( 1 )( 2 ) [( 1 )(A 2 B 2 ) + ( 2 )(A 1 )] + (A 1 )(A 2 B 2 ) (1 A 1 )(1 A 2 ) ( 1 )( 2 ). (2.14) Therefore, if (i) or (ii) is satisfied, then it follows from (2.7) to (2.14) that h(λ) 0, ψ 1 ( + 1) ϕ 1 () A(B), and f 1 f 2 H, (λ, A(B), B). Furthermore, for B < A 0 < A(B), we have 1 A + ()(2λ 1) A 0 B > 1 A(B) + ()(2λ 1) A(B) B Hence for functions f j (z) z + we have f 1 f 2 H, (λ, A 0, B). 1 A j + ( j )(2λ 1) 1 A j + ( j )(2λ 1) 1. 1 A j + ( j )(2λ 1) z H, (λ, A j, B j ) (j 1, 2) (iii) If (1 A 1 )(1 A 2 ) > ( 1 )( 2 ) and 1 λ < λ 1, then we have h(λ) > 0, ϕ 1 () < ψ 1 ( + 1) Ã(B), and f 1 f 2 H, (λ, Ã(B), B). Furthermore, the number Ã(B) cannot be decreased as can be seen from the functions f j defined by f j (z) z + ( ) ((2 + 1)λ )( j ) z+1 H, (λ, A j, B j ) (j 1, 2). Theorem 2. Let f 1 H, (λ, A 1, B 1 ) and f 2 Q, (λ, A 2, B 2 )
J.-L. Liu, R. Srivastava / Filomat 31:1 (2017), 113 123 118 2 N and 1 B max{b 1, B 2 }. Also let A(B), Ã(B) and λ 1 be given as in Theorem 1. Then we have the following: (i) If (1 A 1 )(1 A 2 ) ( 1 )( 2 ) and λ 1, then f 1 f 2 Q, (λ, A(B), B). (ii) If (1 A 1 )(1 A 2 ) > ( 1 )( 2 ) and λ λ 1, then f 1 f 2 Q, (λ, A(B), B). (iii) If (1 A 1 )(1 A 2 ) > ( 1 )( 2 ) and 1 λ < λ 1, then f 1 f 2 Q, (λ, Ã(B), B). In all cases (i)-(iii) the numbers A(B) and Ã(B) are otimal in the sense that they cannot be decreased for each B. Proof. Since f 1 H, (λ, A 1, B 1 ), 2z + z f 2 (z) H, (λ, A 2, B 2 ) (see (1.8)), and ( f 1 (z) 2z + z f 2 (z) ) 2z + z( f 1 f 2 ) (z) the assertion of the theorem follows from Theorem 1. Next, we denote by λ 2 the root in (1, + ) of the equation: h 1 (λ) a 1 λ 2 + b 1 λ + c 1 0, (z U 0 ), where a 1 (3 + 1)( 1 )( 2 ), b 1 ( + 1)( 1 )( 2 ) 2 [( 1 )(A 2 B 2 ) + ( 2 )(A 1 )], c 1 2 (A 1 )(A 2 B 2 ). (2.15) We also denote à 1 (B) B + Theorem 3. Let 2 () ( + 1)((2 + 1)λ ) j. (2.16) f 1 H, (λ, A 1, B 1 ) and f 2 Q, (λ, A 2, B 2 ) 2 N and 1 B max{b 1, B 2 }. Then we have the following: (i) If 2 (1 A 1 )(1 A 2 ) (2 + 1)( 1 )( 2 ) and λ 1, then f 1 f 2 H, (λ, A(B), B). (ii) If 2 (1 A 1 )(1 A 2 ) > (2 + 1)( 1 )( 2 ) and λ λ 2, then f 1 f 2 H, (λ, A(B), B).
J.-L. Liu, R. Srivastava / Filomat 31:1 (2017), 113 123 119 (iii) If 2 (1 A 1 )(1 A 2 ) > (2 + 1)( 1 )( 2 ) and 1 λ < λ 2, then f 1 f 2 H, (λ, Ã1(B), B). In all cases (i)-(iii) the numbers A(B) and Ã1(B) are otimal in the sense that they cannot be decreased for each B. Proof. It can be verified that à 1 (B) B ( + 1)((2 + 1)λ ) 2 j > and so B < Ã1(B) < B. In order to rove Theorem 3, we only need to find the smallest A such that j 2B > 0 (1 A)δ n,, + ()(nλ + (λ 1)) n (A B) (1 A j )δ n,, + ( j )(nλ + (λ 1)) ( ) (2.17) for all n. For n and n+ N, (2.17) is equivalent to λn(n+) 2 2 n 2 ϕ 2 (n). (2.18) Define Then (λ, x) x λx(x + ) (λ, x) λ(2x + ) 3λ 2 3λ 1 3λ 1 2 j 1 j 1 j 1 + 1 j x j j j 1 A j + 1 1 > 0 (x ; λ 1), 1 A j (x ; λ 1). which imlies that ϕ 2 (n) defined by (2.18) is decreasing in n (n ). Hence, in view of 2 ϕ 2 (n) ϕ 2 () B + For n and n+ 2λ 2 2 N, (2.17) reduces to A(B). n(nλ+(λ1)) 2 2 ψ 2 (n) N, we have
and, in view of 2 N, we have J.-L. Liu, R. Srivastava / Filomat 31:1 (2017), 113 123 120 ψ 2 (n) ψ 2 ( + 1) B + (+1)((2+1)λ). 2 2 Now 2λ j h 1 (λ) λ 2 (A 1 )(A 2 B 2 ), j + 1 ( + 1)((2 + 1)λ ) λ 2 j (2.19) where h 1 (λ) a 1 λ 2 + b 1 λ + c 1 and a 1, b 1, c 1 are given by (2.15). Note that a 1 < 0, h 1 (0) c 1 > 0 and h 1 (1) 2 2 ( 1 )( 2 ) 2 [( 1 )(A 2 B 2 ) + ( 2 )(A 1 )] + 2 (A 1 )(A 2 B 2 ) ( + 1) 2 ( 1 )( 2 ) 2 (1 A 1 )(1 A 2 ) (2 + 1)( 1 )( 2 ). The remaining art of the roof of Theorem 3 is similar to that as in Theorem 1 and hence we omit it. The roof of the Theorem is comleted. From Theorem 3 we have the following theorem at once. Theorem 4. Let f j Q, (λ, A j, B j ) (j 1, 2) 2 N and 1 B max{b 1, B 2 }. Also let A(B), Ã1(B) and λ 2 be given as in Theorem 3. Then we have the following: (i) If 2 (1 A 1 )(1 A 2 ) (2 + 1)( 1 )( 2 ) and λ 1, then f 1 f 2 Q, (λ, A(B), B). (ii) If 2 (1 A 1 )(1 A 2 ) > (2 + 1)( 1 )( 2 ) and λ λ 2, then f 1 f 2 Q, (λ, A(B), B). (iii) If 2 (1 A 1 )(1 A 2 ) > (2 + 1)( 1 )( 2 ) and 1 λ < λ 2, then f 1 f 2 Q, (λ, Ã1(B), B). In all cases (i)-(iii) the numbers A(B) and Ã1(B) are otimal in the sense that they cannot be decreased for each B. Finally, we denote by λ 3 the root in (1, + ) of the equation: h 2 (λ) a 2 λ 2 + b 2 λ + c 2 0, where a 2 [( + 1) 2 (2 + 1) 2 3 ]( 1 )( 2 ), b 2 ( + 1) 2 ( 1 )( 2 ) 3 [( 1 )(A 2 B 2 ) + ( 2 )(A 1 )], c 2 3 (A 1 )(A 2 B 2 ). (2.20)
J.-L. Liu, R. Srivastava / Filomat 31:1 (2017), 113 123 121 We also denote 3 () Ã 2 (B) B + ( + 1) 2 ((2 + 1)λ ) j. (2.21) Theorem 5. Let f j Q, (λ, A j, B j ) (j 1, 2) 2 N and 1 B max{b 1, B 2 }. Then we have the following: (i) If 3 (1 A 1 )(1 A 2 ) (3 2 + 3 + 1)( 1 )( 2 ) and λ 1, then f 1 f 2 H, (λ, A(B), B). (ii) If 3 (1 A 1 )(1 A 2 ) > (3 2 + 3 + 1)( 1 )( 2 ) and λ λ 3, then f 1 f 2 H, (λ, A(B), B). (iii) If 3 (1 A 1 )(1 A 2 ) > (3 2 + 3 + 1)( 1 )( 2 ) and 1 λ < λ 3, then f 1 f 2 H, (λ, Ã2(B), B). In all cases (i)-(iii) the numbers A(B) and Ã2(B) are otimal in the sense that they cannot be decreased for each B. Proof. It can be seen that B < Ã2(B) < B. In order to rove Theorem 5, we only need to find the smallest A such that ( ) (1 A)δ n,, + ()(nλ + (λ 1)) 2 n (1 A j )δ n,, + ( j )(nλ + (λ 1)) (2.22) (A B) ( ) for n. For n and n+ N, (2.22) can be written as λn 2 (n+) 3 2 n2 2 2 + n2 + 2 λ(n+) A simle calculation shows that ϕ 3 (n) (n, λ 1) is decreasing in n. Therefore ϕ 3 (n) ϕ 3 () B + A(B). 2λ 2 2 ϕ 3 (n). (2.23) For n and n+ N, (2.22) becomes n 2 (nλ+(λ1)) 2 3 ψ 3 (n) and we have ψ 3 (n) ψ 3 ( + 1) B + (+1) 2 ((2+1)λ). 2 3
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