SUBORDINATION RESULTS FOR A CERTAIN SUBCLASS OF NON-BAZILEVIC ANALYTIC FUNCTIONS DEFINED BY LINEAR OPERATOR

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1 italian jornal of pre and applied mathematics n ) 375 SUBORDINATION RESULTS FOR A CERTAIN SUBCLASS OF NON-BAZILEVIC ANALYTIC FUNCTIONS DEFINED BY LINEAR OPERATOR Adnan G. Alamosh Maslina Dars 1 School of Mathematical Sciences Faclty of Science and Technology Universiti Kebangsaan Malaysia 436 UKM Bangi Selangor Malaysia s: adnan omosh@yahoo.com maslina@km.ed.my Abstract. In this work, by making se of the principle of sbordination, we introdce a certain sbclass of non-bailevic analytic fnctions defined by linear operator. Sch reslts as sbordination and sperordination, sandwich theorem and ineqality properties are given. 1. Introdction Let A s denote the class of the fnctions f of the form 1) f) = + a n n, s N = {1, 2, 3,...}), n=s+1 which are analytic in the open nit disk U = { C : < 1}. If f) and F ) are analytic in U. Then we say that the fnction f) is sbordinate to F ) in U if there exists an analytic fnction w) in U sch that w) 1 and f) = F w)), denoted by f F or f) F ). Frthermore, if the fnction F ) is nivalent in U, then we have the following eqivalence see [1]): f) F ) f) = F ) and fu) F U). Let ψ : C 2 U C and h) be nivalent in U. If p) is analytic in U and satisfies the first order differential sbordination: 2) ϕ p), p ); ) h), 1 Corresponding athor.

2 376 a.g. alamosh, m. dars then p) is a soltion of the differential sbordination 2). The nivalent fnction q) is called a dominant of the soltions of the differential sbordination 2) if p) q) for all p) satisfying 2). A nivalent dominant q that satisfies q q for all dominants of 2) is called the best dominant. If p) and ϕp), p )) are nivalent in U and if p) satisfies first order differential sperordination: 3) h) ϕ p), p ); ), then p) is a soltion of the differential sperordination 3). An analytic fnction q) is called a sbordinant of the soltions of the differential sperordination 3) if q) p) for all p) satisfying 3). A nivalent sbordinant q that satisfies q q for all sbordinants of 3) is called the best sbordinant. For frther properties of sbordination and sperordination, see [1] and [11]. For fnctions f, g A s, where f is given by 1) and g is defined by g) = + n=s+1 b n n, then the Hadamard prodct or convoltion) f g of the fnctions f and g is defined by f g)) = f) g) = + n=s+1 a n b n n. For the fnctions f, g A s, we define the linear operator : A k A k for k =, 1, 2,... ), < α 1, < β 1, λ, and U by: Dα,β,λ f g)) = f g)), Dα,β,λ 1 f g)) = D α,β,λf g)) = [1 λα + β 1)]f g)) + λα + β 1)[f g))] = + [λα + β 1)n 1) + 1] a n b n n, n=s+1 and in general) Dα,β,λ k f g)) = D α,β,λ D k 1 α,β,λ f g))) 4) = + [λα + β 1)n 1) + 1] k a n b n n, λ ). n=s+1 Using 4), it is easy to verify that 5) λα + β 1)[Dα,β,λ k f g))] = D k+1 α,β,λ f g)) + [1 λα + β 1)]Dk α,β,λ f g)). Remark 1. For b n = Cδ, n), the operator Dα,β,λ k f g)) extends to Dk α,β,δ,λ f), where the operator Dα,β,δ,λ k f) was introdced and stdied by Alamosh and Dars, which generalies many other operators see [1]), where ) n + δ 1 Cδ, n) =. δ

3 sbordination reslts for a certain sbclass of non-bailevic Definition 1. A fnction f A s is said to be in the class Nα,β,λ k g, ρ, ; A, B) if it satisfies the following sbordination condition: 1 + ρ) ρ Dk+1 α,β,λ f g)) Dα,β,λ k f g)) Dα,β,λ k f g)) Dα,β,λ k f g)) 6) 1 + A 1 + B, where g A s, ρ C, < < 1, 1 B < A 1, A B, A R, and Dα,β,λ k f) as defined on 4)). Here all the powers are the principal vales. Re { Frthermore, the fnction f N k α,β,λ g, ρ, ; ϖ) if and only if f, g A s and 1 + ρ) f g)) f g)) Dα,β,λ k f g)) } > ϖ, f g)) where ϖ < 1; U). We note that: If k =, and b n = 1, then the class Nα,β,λ k g, ρ, ; A, B) redces to the class Nρ, ; A, B) which is defined by Wang el. at [5]. If k =, ρ = 1, n = 1, A = 1, B = 1 and b n = 1, then the class Nα,β,λ k g, ρ, ; A, B) redces to the class of non-bailevic fnctions which introdced by Obradovic [13]. If k =, ρ = 1, n = 1, A = 1 2ϖ, B = 1 and b n = 1, then the class Nα,β,λ k g, ρ, ; A, B) redces to the class of non-bailevic fnctions of order ϖ ϖ < 1) which was given by Tneski and Dars [12]. Other works related to non-bailevic can be fond in [2]-[7]). In the present paper, we discss and prove the sbordination and sperordination properties, sandwich theorem and ineqality properties for the class Nα,β,λ k g, ρ, ; A, B). 2. Preliminary reslts In order to establish or main reslts, we need the following definition and lemmas. Definition 2. [9]. Denote by Q the set of all fnctions f that are analytic and injective on U \ Ef), where { } Ef) = ζ U : lim f) =, ζ and sch that f ζ) for ζ U \ Ef) Lemma 1. [1] Let the fnction h) be analytic and convex nivalent) in U with h) = 1. Sppose also that the fnction g) given by 7) g) = 1 + c k k + c k+1 k

4 378 a.g. alamosh, m. dars is analytic in U. If 8) g) + g ) γ then h), g) q) = γ k γ k and q) is the best dominant of 8). Reγ) > ; γ ; U), ht)t γ k 1 dt h), Lemma 2. [8] Let q) be a convex nivalent fnction in U and let σ C, η C = C\ {} with ) { )} Re 1 + q ) σ > max, Re. q) η If the fnction g) is analytic in U and σg) + ηg ) σq) + ηq ), then g) q) and q) is the best dominant. Lemma 3. [9] Let q) be a convex nivalent fnction in U and let k C. Frther assme that Rek) >. If and is nivalent in U, then g) H[q), 1] Q, g) + kg ) q) + kq ) g) + kg ), implies q) g) and q) and q is the best sbordinant. Lemma 4. [14] Let F be analytic and convex in U. If then Lemma 5. [15] Let be analytic in U and f, g A and f, g F λf + 1 λ)g F, λ 1). f) = 1 + g) = 1 + a n n n=1 b n n n=1 be analytic and convex in U. If f) g), then a n < b 1 n N).

5 sbordination reslts for a certain sbclass of non-bailevic Main reslts We begin by presenting or first sbordination property given by Theorem 1. Theorem 1. For g A s, ρ C, < < 1, 1 B < A 1, A B, A R, and Dα,β,λ k f g) as defined by 4). Let f) N α,β,λ k g, ρ, ; A, B) with Reρ) >. Then q) Dα,β,λ k f g)) 9) 1 = λα+β 1)sρ A 1 + A d λα + β 1)sρ 1 + B 1 + B and q) is the best dominant. Proof. Define the fnction g) by 1) g) = U). f g)) Then g) is of the form 7) and analytic in U with g) = 1. Taking logarithmic differentiation of 1) with respect to and sing 5), we dedce that 11) 1 + ρ) ρ Dk+1 α,β,λ f g)) f g)) Dα,β,λ k f g)) = g) + Since f) Nα,β,λ k g, ρ, ; A, B), we have g) + Applying Lemma 1 to 11) with γ = q) = f g)) λα + β 1)ρ g ). λα + β 1)ρ g ) 1 + A 1 + B., we get λα + β 1)ρ λα + β 1)sρ λα+β 1)sρ f g)) t λα+β 1)sρ At 1 + Bt dt 12) = λα + β 1)sρ λα+β 1)sρ A 1 + A d 1 + B 1 + B, and q) is the best dominant. The proof of Theorem 1 is ths complete.

6 38 a.g. alamosh, m. dars Theorem 2. Let q) be nivalent in U, ρ C. Sppose also that q) satisfies the following ineqality: ) { )} 13) Re 1 + q ) > max, Re. q) λα + β 1)ρ If f A s satisfies the following sbordination condition: 1 + ρ) ρ Dk+1 α,β,λ f g)) Dα,β,λ 14) k f g)) Dα,β,λ k f g)) Dα,β,λ k f g)) λα + β 1)ρ q) + q ), then and q) is the best dominant. q) f g)) Proof. Let the fnction g) be defined by 1). We know that 11) holds tre. Combining 11) and 14), we find that 15) g) + λα + β 1)ρ g ) q) + λα + β 1)ρ q ). By sing Lemma 2 and 15), we easily get the assertion of Theorem 2. Taking q) = 1 + A 1 + B in Theorem 2, we get the following reslt. Corollary 1. Let ρ C and 1 B < A 1. Sppose also that ) { )} 1 B Re > max, Re. 1 + B λα + β 1)ρ If f A s satisfies the following sbordination: 1 + ρ) f g)) Dα,β,λ k f g)) Dα,β,λ k f g)) then and the fnction 1 + A 1 + B 1 + A 1 + B + λα + β 1)ρ A B) 1 + B), A f g)) 1 + B, is the best dominant. f g))

7 sbordination reslts for a certain sbclass of non-bailevic Now, by making se of Lemma 3, we now derive the following sperordination reslt. Theorem 3. Let q) be convex nivalent in U, ρ C with Reρ) >. Also let H[q), 1] Q f g)) and 1 + ρ) f g)) f g)) Dα,β,λ k f g)) f g)) be nivalent in U. If f A s satisfies the following sperordination: then 1 + ρ) λα + β 1)ρ q) + q ) f g)) f g)) Dα,β,δ,λ k f g)) q) and the fnction q) is the best sbordinant. f g)) Proof. Let the fnction g) be defined by 1). Then λα + β 1)ρ q) + q ) 1 + ρ) f g)) = g) + Dα,β,λ k f g)) Dα,β,λ k f g)) f g)), f g)) λα + β 1)ρ g ). An application of Lemma 3 yields the assertion of Theorem 3. Taking q) = 1 + A in Theorem 3, we get the following reslt. 1 + B Corollary 2. Let ρ C and 1 B < A 1 with Reρ) >. Sppose also that H[q), 1] Q, f g)) and 1 + ρ) f g)) f g)) Dα,β,λ k f g)) f g))

8 382 a.g. alamosh, m. dars be nivalent in U. If f A s satisfies the following sperordination: then 1 + ρ) 1 + A λα + β 1)ρ A B) B 1 + B) 2 f g)) f g)) Dα,β,λ k f g)) 1 + A 1 + B f g)), f g)) and the fnction 1 + A is the best sbordinant. 1 + B Combining Theorems 2 and 3, we easily get the following Sandwich-type reslt. Theorem 4. Let q 1 be convex nivalent and let q 2 be nivalent in U, ρ C with Reρ) >. Let q 2 satisfies 13). If H[q 1 ), 1] Q f g)) and 1 + ρ) f g)) f g)) Dα,β,λ k f g)) f g)) be nivalent in U, also λα + β 1)ρ q 1 ) + q 1) 1 + ρ) f g)) = q Dα,β,λ k f g)) Dα,β,λ k 2 ) + f g)) then q 1 ) q 2 ). f g)) f g)) λα + β 1)ρ q 2), and q 1 ) and q 2 ) are, respectively, the best sbordinant and dominant. Next, we consider the following: Theorem 5. If ρ > and f Nα,β,λ k g,, ϖ) ϖ < 1), then f N α,β,λ k g, ρ, ; ϖ) for < R, where 1 λα ) s 2 16) R = + β 1)sρ λα + β 1)sρ + 1. The bond R is the best possible.

9 sbordination reslts for a certain sbclass of non-bailevic Proof. We begin by writing 17) = ϖ + 1 ϖ)g) U, ϖ < 1). f g)) Then, clearly, the fnction g) is of the form 7), is analytic and has a positive real part in U. By taking the derivatives of both sides of 17), we get { ρ) f g)) ϖ} 1 ϖ Dα,β,λ k f g)) Dα,β,λ k f g)) Dα,β,λ k f g)) 18) = g) + λα + β 1)ρ g ). By making se of the following well-known estimate see [16], Theorem 1): in 18), we obtain { 1 Re 1 + ρ) 1 ϖ g ) Re {g)} 19) Re {g)} 1 2srs 1 2r 2s = r < 1) f g)) f g)) Dα,β,λ k f g)) ) 2λα + β 1)ρsrs. 1 r 2s ) ϖ}) f g)) It is seen that the right-hand side of 19) is positive, provided that r < R, where R is given by 16). In order to show that the bond R is the best possible, we consider the fnction f) A s defined by Noting that { ρ) 1 ϖ = ϖ + 1 ϖ) 1 + s f g)) U, ϖ < 1). 1 s f g)) f g)) Dα,β,λ k f g)) = 1 + s 2λα + β 1)ρss + = 1 s 1 s ) 2 ϖ} f g)) for = R, we conclde that the bond is the best possible. Theorem 5 is ths proved. Now, we give the inclsion properties:

10 384 a.g. alamosh, m. dars Theorem 6. Let ρ 2 ρ 1 and 1 B 1 B 2 < A 2 A 1 1. Then 2) N k α,β,λg, ρ 2, ; A 2, B 2 ) N k α,β,λg, ρ 1, ; A 1, B 1 ). Proof. Let f N k α,β,λ g, ρ 2, ; A 2, B 2 ). Then we have 1+ρ 2 ) D k+1 α,β,λ f g)) ρ 2 f g)) Dα,β,λ k f g)) Since 1 B 1 B 2 < A 2 A 1 1, we easily find that D k+1 α,β,λ f g)) 1 + ρ 2 ) ρ D 21) α,β,λ k 2 f g)) Dα,β,λ k f g)) 1 + A B A B 1, 1 + A 2 f g)) 1 + B 2. f g)) that is f Nα,β,λ k g, ρ 2, ; A 1, B 1 ). Ths the assertion of Theorem 6 holds for ρ 2 = ρ 1. If ρ 2 > ρ 1, by Theorem 1 and 21), we know that f Nα,β,λ k g,, ; A 1, B 1 ), that is, 22) 1 + A 1 f g)) 1 + B 1. At the same time, we have D k+1 α,β,λ f g)) 1 + ρ 1 ) ρ Dα,β,λ k 1 f g)) Dα,β,λ k f g)) 23) = + ρ 1 ρ 2 [ 1 + ρ 2 ) 1 ρ ) 1 ρ 2 Dα,β,λ k f g)) D k+1 α,β,λ f g)) ρ 2 f g)) Dα,β,δ,λ k f g)) f g)) ]. f g)) Moreover, since ρ 1 < 1 and the fnction 1 + A 1 ρ B 1 1 B 1 < A 1 1) is analytic and convex in U. Combining 21)-23) and Lemma 4, we find that 1+ρ 1 ) D k+1 α,β,λ f g)) ρ 1 f g)) Dα,β,λ k f g)) 1 + A 1 f g)) 1 + B 1, that is f N k α,β,λ g, ρ 1, ; A 1, B 1 ), which implies that the assertion 2) of Theorem 6 holds.

11 sbordination reslts for a certain sbclass of non-bailevic Theorem 7. Let f Nα,β,λ k g, ρ, ; A, B) with ρ > and 1 B < A 1. Then 24) < R { λα + β 1)sρ } < f g)) λα+β 1)sρ 1 1 A 1 B d λα + β 1)sρ λα+β 1)sρ A 1 + B d. The extremal fnction of 24) is defined by 25) F ) = Dα,β,λf k g)) = λα + β 1)sρ λα+β 1)sρ A s 1 + B d s ) 1. Proof. Let f Nα,β,λ k g, ρ, ; A, B) with ρ >. From Theorem 1, we know that 9) holds, which implies that 26) { } R Dα,β,λ k f g)) { < sp R U λα + β 1)sρ < λα + β 1)sρ λα + β 1)sρ } λα+β 1)sρ A 1 + B d ) 1 + A d 1 + B λα+β 1)sρ 1 sp U λα+β 1)sρ A 1 + B d and 27) { } R Dα,β,λ k f g)) { > inf R U λα + β 1)sρ > λα + β 1)sρ λα + β 1)sρ } λα+β 1)sρ A 1 + B d ) 1 + A d 1 + B λα+β 1)sρ 1 inf U λα+β 1)sρ 1 1 A 1 B d. Combining 26) and 27), we get 24). By noting that the fnction F ) defined by 25) belongs to the class Nα,β,λ k g, ρ, ; A, B), we obtain that eqality 24) is sharp. The proof of Theorem 7 is evidently complete. Similarly, by applying the method of proof of Theorem 7, we easily get the following reslt.

12 386 a.g. alamosh, m. dars Corollary 3. Let f Nα,β,λ k g, ρ, ; A, B) with ρ > and 1 B < A 1. Then 28) λα + β 1)sρ < R { λα+β 1)sρ A 1 + B d } f g)) < λα + β 1)sρ The extremal fnction of 28) is defined by 25). Theorem 8. Let 29) f) = + Then n=s+1 λα+β 1)sρ 1 1 A 1 B d. a n n N k α,β,λg, ρ, ; A, B), s N = {1, 2, 3,...}). 3) a s+1 [λα + β 1) + 1] k A B) + λα + β 1)ρ b s+1. The ineqality 3) is sharp, with the extremal fnction defined by 25). Proof. Combining 6) and 29), we obtain 1 + ρ) f g)) D 31) α,β,λ k f g)) Dα,β,λ k f g)) Dα,β,λ k f g)) = λα + β 1)ρ) [λα + β 1) + 1] k a s+1 b s A 1 + B. An application of Lemma 5 to 31) yields 32) + λα + β 1)ρ) [λα + β 1) + 1] k a s+1 b s+1 < A B. Ths, from 32), we easily arrive at 3) asserted by Theorem 8. Acknowledgements. The athors wold like to acknowledge and appreciate the financial spport received from Universiti Kebangsaan Malaysia nder the grant: AP

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14 388 a.g. alamosh, m. dars [14] Li, M.S., On certain sbclass of analytic fnctions, J. Soth China Normal Univ., 4 22), 15-2 in Chinese). [15] Rogosinski, W., On the coefficients of sbordinate fnctions, Proc. London Math. Soc., Ser. 2), ), [16] Bernardi, S.D., New distortion theorems for fnctions of positive real part and applications to the partial sms of nivalent convex fnctions, Proc. Amer. Math. Soc., 45 1) 1974), Accepted: