SOME PROPERTIES OF A SUBCLASS OF ANALYTIC FUNCTIONS DEFINED BY A GENERALIZED SRIVASTAVA-ATTIYA OPERATOR. Nagat. M. Mustafa and Maslina Darus

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1 FACTA UNIVERSITATIS NIŠ) Ser. Math. Inform. Vol. 27 No ), SOME PROPERTIES OF A SUBCLASS OF ANALYTIC FUNCTIONS DEFINED BY A GENERALIZED SRIVASTAVA-ATTIYA OPERATOR Nagat. M. Mustafa and Maslina Darus Abstract. Making use of an integral operator which is defined by means of a general Hurwitz Lerch zeta function, we give some properties of the class Q α δ, β, γ). It is worth noting that the usage of Hurwitz Lerch zeta function in Geometric Function Theory was first made by Srivastava and Attiya in Indeed, in this present paper, we obtain integral means inequalities, modified Hadamard products and establish some results concerning the partial sums for functions f belonging to the class Q α δ, β, γ). Keywords: Analytic functions, Hurwitz Lerch zeta function, Srivastava Attiya operator, Integral means, Hadamard product, Partial sums. 1. Introduction Let A denote the class of all analytic functions in the open unit disk U = {z C : z < 1}, of the form 1.1) fz) = z + a k z k, z U). With a view to define the Srivastava Attiya operator, we recall here a general Hurwitz Lerch-Zeta function, which is defined in [4],[6]) by the following series: Φz, s, b) = k=0 z k k + b) s, where s C, b C Z 0 ) when z < 1), and Reb) > 1) when z = 1). Received June 10, 2012.; Accepted December 5, Mathematics Subject Classification. Primary 30C45 The authors were supported in part by LRGS/TD/2011/UKM/ICT/03/02 309

2 310 Nagat. M. Mustafa and Maslina Darus By making use of the following normalized function: G z) = 1 + b) s [Φz, s, b) b s ] ) s 1 + b = z + z k, z U). k + b Srivastava Attiya [4] introduced operator L : A A by the following: L = z + ) s 1 + b a k z k. k + b The operator L is now well-known in the literature as the Srivastava Attiya operator. Various basic properties of L are systematically investigated in [11], [12]). Owa and Srivastava [2] introduced the operator Ω α : A A, which is known as an extension of fractional derivative and fractional integral as follows: Ω α fz) = Γ2 α)z α D α z fz) = z + Γk + 1)Γ2 α) a k z k α 2, 3, 4, ), Γk + 1 α) where D α z fz) the fractional derivative of f of order α see [3]). Let: Φ z, s, b) = G 1 z) = b s z + k + b 1) s a kz k. Using the technique of Owa and Srivastava [2], we introduced the generalized integral operator Im α f) : A A by the following: Im α fz) = Γ2 α)z α Dz α Φ z, s, b), α 2, 3, 4, ) ) s Γk + 1)Γ2 α) b = z + a k z k, z U), Γk + 1 α) k 1 + b where s C, b C Z 0, and 0 α < 1. It can also be shown that this operator is the generalized Srivastava Attiya operator by taking fz) = z + Γk+1)Γ2 α) a k.

3 Some Properties of a Subclass of Analytic Functions Note that : Im 0 0,b fz) = fz). Special cases of this operator include: Im α 0,b fz) Ω α fz) is the Owa and Srivastava operator [2]. Im 0 +1 fz) L is the Srivastava and Attiya integral operator [4]. Im 0 σ,2 fz) I σ fz) is the Jung Kim Srivastava integral operator [5]. Also, the authors [1] have recently introduced a new subclass of analytic functions with negative coefficients, and stated the following: For 0 δ < 1), 0 < β 1) and 1 2 < γ 1) if δ = 0, and 1 2 < γ 1 2δ )if δ 0, we let Q α δ, β, γ) be the subclass of A consisting of functions of the form 1.1) and satisfying the inequality Im α fz)) 1 2γIm α fz)) δ) Im α fz)) 1) < β. We further let 1.2) Q α δ, β, γ) = Q α δ, β, γ) T, where T := { f A : fz) = z } a k z k, where a k 0 for all k 2, is a subclass of A introduced and studied by Silverman [9]. In [1], it was also shown that the sufficient condition for a function f to be in the class Q α δ, β, γ). Theorem 1.1. and only if Let the function f be defined by 1.2). Then f Q α δ, β, γ) if 1.3) k [1 + β2γ 1)] Γk + 1)Γ2 α) Γk + 1 α) b k 1 + b ) s a k. The result is sharp.

4 312 Nagat. M. Mustafa and Maslina Darus 2. Integral Means Inequalities In order to prove the results regarding integral means inequalities, we need the concept of subordination between analytic functions and also the following lemma. Lemma 2.1. [8] If f, g are analytic in U,such that f g, then 2π 0 Theorem π 0 fz) y dθ 2π 0 gz) y dθ, z = re iθ, 0 < r < 1, y > 0). Let f Q α δ, β, γ). Then for z = reiθ, 0 < r < 1, we have fre iθ ) y dθ 2π where the function f 2 z) defined by 2.1) f 2 z) = z 2 [1 + β2γ 1)] 0 f 2 re iθ ) y dθ, 0 < r < 1, y > 0), Γ3)Γ2 α) Γ3 α) ) b 1+b )s z 2. Proof: Let f Q α δ, β, γ) and satisfying 1.3), and f 2z) be given by 2.1). We must show that 2π y 0 1 2π a k z k 1 dθ 0 1 ) 2 [1 + β2γ 1)] Γ3)Γ2 α) z dθ. Γ3 α) 1+b )s By Lemma 2.1, it suffices to show that Setting 1 2.2) 1 a k z k [1 + β2γ 1)] Γ3)Γ2 α) Γ3 α) a k z k 1 = 1 2 [1 + β2γ 1)] From 2.2), we obtain ωz) z k [1 + β2γ 1)] This completes the proof of the theorem. Γk+1)Γ2 α) Γ3)Γ2 α) Γ3 α) ) b 1+b )s z. ) b 1+b )s ωz). ) b k 1+b )s a k z < 1.

5 Some Properties of a Subclass of Analytic Functions Modified Hadamard Products Let the functions f j z)j = 1; 2) be defined by 3.1) f j z) = z a k,j z k, for all a k,j 0, z U). The modified Hadamard product of f 1 z) and f 2 z) is defined by f 1 f 2 )z) = z a k,1 a k,2 z k. Using the techniques of Schild and Silverman [7], we prove the following results. Theorem 3.1. For functions f j z)j = 1; 2) defined by 3.1), let f 1 z) Q α δ, β, γ), f 2 z) Q α δ, µ, γ). Then f 1 f 2 )z) ξ α δ, Q α δ, β, µ, γ)), where ξ α δ, Q α δ, β, γ) = 2γ1 δ)βµ 4βµγ 2 1 δ) + 2 2γ1 δ)βµ ) Γ3)Γ2 α) Γ3 α) b 1+b ) s 1 + µ2γ 1))1 + β2γ 1)). Proof: To prove the theorem, we need to find the largest ξ = ξ α δ, Q α δ, β, µ, γ)) such that k [1 + ξ2γ 1)] Γk+1)Γ2 α) s k 1+b) a k,1 a k,2 1, 2ξγ1 δ) since and k [1 + β2γ 1)] k [1 + µ2γ 1)] Γk+1)Γ2 α) Γk+1)Γ2 α) 2µγ1 δ) By the Cauchy-Schwarz inequality, we have; k Γk+1)Γ2 α) ) 2γ1 δ) s k 1+b) ) ) 1 + β2γ 1)) β s k 1+b) s k 1+b) a k,1 1, a k, µ2γ 1)) a k,1 a k,2 1. µ

6 314 Nagat. M. Mustafa and Maslina Darus Thus, it suffices to show that Note that 1 + ξ2γ 1)) a k,1 a k,2 ξ ak,1 a k,2 k 2γ1 δ) ) Γk+1)Γ2 α) 1 + β2γ 1)) β b k 1+b Consequently, we need only to prove that k 2γ1 δ) ) Γk+1)Γ2 α) or, equivalently that b k 1+b ξ 1 + ξ2γ 1)) ξ 2γ1 δ)βµ 4βµγ 2 1 δ) + k = ψk), 1 + µ2γ 1)) a k,1 a k,2. µ β µ ) s 1 + β2γ 1)) 1 + µ2γ 1)). β µ ) s 1 + β2γ 1)) 1 + µ2γ 1)) 1 + β2γ 1)) β 2γ1 δ)βµ ) Γk+1)Γ2 α) 1 + µ2γ 1)), µ b k 1+b) s 1 + µ2γ 1))1 + β2γ 1)) is an increasing function of k, letting k = 2, we obtain ψ2) = 2γ1 δ)βµ 4βµγ 2 1 δ) + 2 which completes the proof. 2γ1 δ)βµ ) Γ3)Γ2 α) Γ3 α) b 1+b ) s 1 + µ2γ 1))1 + β2γ 1)), Theorem 3.2. For functions f j z)j = 1; 2) defined by 3.1), be in the class Q α δ, β, γ) Then the function hz) = z ) a 2 k,1 + a2 k,2 z k, belongs to the class φ α δ, Q α δ, β, γ), where φ α δ, Q α δ, β, γ) = 21 + β2γ 1))] 4β 2 γ1 δ) ) s. 1+b) 4β2 γ1 δ) Γ3)Γ2 α) Γ3 α)

7 Some Properties of a Subclass of Analytic Functions Proof: By virtue of Theorem 1.1, we obtain 3.2) k [1 + β2γ 1)] Γk+1)Γ2 α) s 2 k 1+b) a 2 k,1 1, and 3.3) k [1 + β2γ 1)] Γk+1)Γ2 α) s 2 k 1+b) a 2 k,2 1. It follows from 3.2) and3.3). 1 k [1 + β2γ 1)] Γk+1)Γ2 α) s 2 k 1+b) a 2 k,1 + a 2 2 k,2) 1. Therefore,we need to find the largest φ = φ α δ, Q α δ, β, γ) k [1 + φ2γ 1)] Γk+1)Γ2 α) ) s k 1+b) 2φγ1 δ) 1 k [1 + β2γ 1)] Γk+1)Γ2 α) s 2 k 1+b), 2 that is, 4β 2 γ1 δ) φ k1 + β2γ 1)) 2 Γk+1)Γ2 α) s = χk), k 1+b) 4β2 γ1 δ) is an increasing function of k, letting k = 2, we obtain χ2) = 21 + β2γ 1))] which completes the proof. 4β 2 γ1 δ) ) s, 1+b) 4β2 γ1 δ) Γ3)Γ2 α) Γ3 α)

8 316 Nagat. M. Mustafa and Maslina Darus 4. Partial sums By following the earlier work by Silverman[10] on partial sums of analytic functions, we study the ratio of a function of the form 1.2) to its sequence of partial sums of the form f 1 z) = z, f n z) = z + n a kz k, z U). We will determine sharp lower bounds for { } { } { fz) fn z) f } z) Re, Re, Re f n z) fz) f nz) and Re { } f n z) f. z) Theorem 4.1. { fz) 4.1) Re f n z) and 4.2) Re where c n be defined as Let f Q α δ, β, γ) and satisfying 1.3), then } 1 1, n N, z U), { } fn z), n N, z U), fz) 1 + Γn+1)Γ2 α) Γn+1 α) n [1 + β2γ 1)] c n = ) The results are sharp for every k with the function given by s n 1+b). 4.3) fz) = z zn+1, z U, n N). Proof: In order to prove 4.1), it suffices to show that 4.4) { fz) f n z) 1 1 )} = 1+ n a kz k 1 + a k z k 1 1+ n a kz k 1 = 1 + wz) 1 wz). Then wz) = a kz k n a kz k 1 + a k z. k 1

9 Notice that w0) = 0 and Some Properties of a Subclass of Analytic Functions Now wz) 1 if and only if wz) = a k 2 2 n a k z k 1 a k. 4.5) a k + a k 1. It suffices to show that the LHS of 4.5) is bounded above by the condition 1.3) which is equivalent to c k ) a k + c k 1) a k 0. To see that the function given by 4.3) gives the sharp result, we observe that for z = re πi n, fz) f n z) = 1 + zn 1 1, when z 1 ). To prove the second part of this theorem, we write 4.6) 1 + ) we find that { fn z) fz) c } n+1 = 1 + n a kz k 1 + a k z k n a kz k 1 = 1 + wz) 1 wz), wz) = 1 + ) a k z k n a kz k ) a k z. k 1 Now wz) 1 if and only if 1 + ) a k + a k 1. The equality holds in 4.2) for the extremal function f given by 4.3). This completes the proof.

10 318 Nagat. M. Mustafa and Maslina Darus Theorem ) Re and 4.8) Re Let f Q α δ, β, γ) and satisfying 1.3), then } { f z) f nz) 1 n + 1, z U), { } f n z) f. z) n The results are sharp with the function given by 4.3). Proof: To prove the result 4.7), define the function wz) by { f z) f nz) 1 n + 1 } ) = 1 + wz) 1 wz). Then wz) = Now wz) 1 if and only if n+1 ka k n ka kz k 1 + ) cn+1 k a k + n + 1 k a k 1. From the condition 1.3), it suffices to show that ) cn+1 k a k + n + 1 This is equivalent to showing that c k k) a k + n+1 k. k a k c k a k. To prove the second part of this theorem, we write n + 1)c k k a k 0. n + 1 { } f wz) = n ) n z) f z) ) n = n+1 ) ka kz k n ka, kz k 1

11 Some Properties of a Subclass of Analytic Functions yields wz) 1 wz) n+1 ) k a k 2 2 n k a k 1 + n+1 ) k a k 1, z U), if and only if 21 + n + 1 ) k a k 2 2 k a k. The bound in 4.8) is sharp for all n N with the extremal function 4.3). This completes the proof of theorem. R E F E R E N C E S 1. N.M.Mustafa and M. Darus: On a subclass of analytic functions with negative cofficients associated to an integral operator involving Hurwitz-Lerch zeta function, Vasile Alecsandri University of Bacau, Faculty of Sviences, Scientific Studies and Research, Series Mathematics and Informatics, 212) 2011), S. Owa and H. M. Srivastava: Univalent and starlike generalized hypergeometric functions, Canad. J. Math, 395) 1987), H. M. Srivastava and S. Owa: An application of the fractional derivative, Math Japon., 293) 1984), H. M. Srivastava and A. A. Attiya: An integral operator associated with the HurwitzLerch Zeta function and differential subordination, Integral Transforms Spec. Funct., 183) 2007), I.B.Jung, Y.C.Kim, and H.M.Srivastava: The Hardy space of analytic functions associated with certain one-parameter families of integral operators, J.Math. Anal. Appl, ), H.M. Srivastava and J.Choi: Series Associated with the Zeta and Related Functions, Dordrecht, Boston and London: Kluwer Academic Publishers), 2001). 7. A. Schild and H.M. Srivastava: Convolutions of Univalent functions with negative coefficients, Proc. Ann. Univ. Mariae Curie-Sk lodowska sect., A29) 1975), J. E. Littlewood: On inequalities in the theory of functions, Proc. London Math. Soc., 223)1925), H. Silverman: Univalent functions with negative coefficients, Proc. Amer. Math. Soc. one-parameter families of integral operators, ), H. Silverman: Partial sums of starlike and convex functions, J. Math. Ana. Appl ), N. E. Cho, I. H. Kim and H. M. Srivastava: Sandwich-type theorems for multivalent functions associated with the Srivastava-Attiya operator, Appl. Math. Comput., ),

12 320 Nagat. M. Mustafa and Maslina Darus 12. G. Murugusundaramoorthy: Subordination results for spiral-like functions associated with Srivastva- Attiya operator, Integral Transforms Spec. Funct., 232) 2012), Nagat.M.Mustafa, Maslina Darus School of Mathematical Sciences Faculty of Science and Technology Universiti Kebangsaan Malaysia Bangi Selangor D. Ehsan, Malaysia. corresponding author)