Volume 8 (2007), Issue 1, Article 16, 4 pp. ON A GENERALIZATION OF ALPHA CONVEXITY KHALIDA INAYAT NOOR M ATHEMATICS D EPARTMENT, COMSATS I NSTITUTE OF I NFORMATION T ECHNOLOGY, I SLAMABAD , PAKISTAN . khalidanoor@hotmail.com Received 28 November, 2006; accepted 06 February, 2007 Communicated by N.E. Cho A BSTRACT. In this paper, we introduce and study a class M̃k (α, β, γ), k ≥ 2 of analytic functions defined in the unit disc. This class generalizes the concept of alpha-convexity and include several other known classes of analytic functions. Inclusion results, an integral representation and a radius problem is discussed for this class. Key words and phrases: Starlike, Convex, Strongly alpha-convex, Bounded boundary rotation. 2000 Mathematics Subject Classification. 30C45, 30C50. 1. I NTRODUCTION Let P̃ denote the class of functions of the form (1.1) p(z) = 1 + c1 z + c2 z 2 + · · · , which are analytic in the unit disc E = {z : |z| < 1}. Let P̃ (γ) be the subclass of P̃ consisting of functions p which satisfy the condition πγ (1.2) |arg p(z)| ≤ , for some γ(γ > 0), z ∈ E. 2 We note that P̃ (1) = P is the class of analytic functions with positive real part. We introduce the class P˜k (γ) as follows: An analytic function p given by (1.1) belongs to P˜k (γ), for z ∈ E, if and only if there exist p1 , p2 ∈ P̃ (γ) such that k 1 k 1 (1.3) p(z) = + p1 (z) − − p2 (z), k ≥ 2. 4 2 4 2 We now define the class M̃k (α, β, γ) as follows: This research is supported by the Higher Education Commission, Pakistan, through research grant No: 1-28/HEC/HRD/2005/90. 306-06 2 K HALIDA I NAYAT N OOR Definition 1.1. Let α ≥ 0, β ≥ 0 (α+β 6= 0) and let f be analytic in E with f (0) = 0, f 0 (0) = 0 (z) 1 and f (z)f 6= 0. Then f ∈ M̃k (α, β, γ) if and only if , for z ∈ E, z β (zf 0 (z))0 α zf 0 (z) + ∈ P˜k (γ). α + β f (z) α + β f 0 (z) We note that, for k = 2, β = (1 − α), we have the class M̃2 (α, 1 − α, γ) = M̃α (γ) of strongly alpha-convex functions introduced and studied in [4]. We also have the following special cases. (i) M̃2 (α, 0, 1) = S ? , M̃2 (0, β, 1) = C, where S ? and C are respectively the wellknown classes of starlike and convex functions. It is known [3] that M̃α (γ) ⊂ S ? and M̃2 (α, 0, γ) coincides with the class of strongly starlike functions of order γ, see [1, 7, 8]. (ii) M̃k (α, 0, 1) = Rk , M̃k (0, β, 1) = Vk , where Rk is the class of functions of bounded radius rotation and Vk is the class of functions of bounded boundary rotation. Also M̃k (0, β, γ) = V˜k (γ) ⊂ Vk and M̃k (α, 0, γ) = R̃k (γ) ⊂ Rk . 2. M AIN R ESULTS Theorem 2.1. A function f ∈ M̃k (α, β, γ), α, β > 0, if and only if, there exists a function F ∈ R̃k (γ) such that β # α+β " α+β Z z β (F (t)) α+β (2.1) f (z) = dt . α t 0 Proof. A simple calculation yields α zf 0 (z) β (zf 0 (z))0 zF 0 (z) + = . α + β f (z) α + β f 0 (z) F (z) If the right hand side belongs to P˜k (γ) so does the left and conversely, and the result follows. Theorem 2.2. Let f ∈ M̃k (α, β, γ). Then the function β 0 α+β zf (z) (2.2) g(z) = f (z) f (z) belongs to R̃k (γ) for z ∈ E. Proof. Differentiating (2.2) logarithmically, we have zg 0 (z) α zf 0 (z) β (zf 0 (z))0 = + , g(z) α + β f (z) α + β f 0 (z) and, since f ∈ M̃k (α, β, γ), we obtain the required result. Theorem 2.3. Let f ∈ M̃k (α, β, γ), Proof. Let zf 0 (z) f (z) β > 0, 0 < γ ≤ 1. Then f ∈ R̃k (γ) for z ∈ E. = p(z). Then (zf 0 (z))0 zp0 (z) = p(z) + . f 0 (z) p(z) J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 16, 4 pp. http://jipam.vu.edu.au/ A LPHA C ONVEXITY 3 Therefore, for z ∈ E, (2.3) β (zf 0 (z))0 α zf 0 (z) + = α + β f (z) α + β f 0 (z) β zp0 (z) p(z) + α + β p(z) ∈ P˜k (γ). Let φ(α, β) = (2.4) α z β z + . α + β 1 − z α + β (1 − z)2 Then, using (1.3) and (2.4), we have k 1 k 1 φ(α, β) φ(α, β) φ(α, β) p? = + p1 ? − − p2 ? , z 4 2 z 4 2 z where ? denotes the convolution (Hadamard product). This gives us β zp0 (z) k 1 β zp01 (z) p(z) + = + p1 (z) + α + β p(z) 4 2 α + β p1 (z) k 1 β zp02 (z) − − p2 (z) + . 4 2 α + β p2 (z) From (2.3), it follows that pi + β zp0i α + β pi ∈ P̃ (γ), i = 1, 2, and, using a result due to Nunokawa and Owa [6], we conclude that pi ∈ P̃ (γ) in E, i = 1, 2. Consequently p ∈ P˜k (γ) and hence f ∈ R̃k (γ) for z ∈ E. Theorem 2.4. Let, for (α1 + β1 ) 6= 0, α1 α < , α1 + β1 α+β β1 β < α1 + β1 α+β and 0 ≤ γ < 1. Then M̃k (α, β, γ) ⊂ M˜K (α1 , β1 , γ), z ∈ E. Proof. We can write α1 zf 0 (z) β1 (zf 0 (z))0 + = α1 + β1 f (z) α1 + β1 f 0 (z) β1 (α + β) zf 0 (z) 1− β(α1 + β1 ) f (z) β1 (α + β) α zf 0 (z) β (zf 0 (z))0 + + β(α1 + β1 ) α + β f (z) α + β f 0 (z) β1 (α + β) β1 (α + β) = 1− H1 (z) + H2 (z), β(α1 + β1 ) β(α1 + β1 ) where H1 , H2 ∈ P˜k (γ) by using Definition 1.1 and Theorem 2.3. Since 0 < γ ≤ 1, the class P̃ (γ) is a convex set and consequently, by (1.3), the class P˜k (γ) is a convex set. This implies H ∈ P˜k (γ) and therefore f ∈ M̃k (α1 , β1 , γ). This completes the proof. Theorem 2.5. Let f ∈ M̃k (α, β, γ). Then α Z z β f (t) α+β 0 α+β dt belongs to (2.5) h(z) = (f (t)) t 0 J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 16, 4 pp. V˜k (γ) for z ∈ E. http://jipam.vu.edu.au/ 4 K HALIDA I NAYAT N OOR Proof. From (2.5), we have α f (z) α+β h (z) = (f (z)) . z Now the proof is immediate when we differentiate both sides logarithmically and use the fact that f ∈ M̃k (α, β, γ). 0 0 β α+β In the following we study the converse case of Theorem 2.3 with γ = 1. Theorem 2.6. Let f ∈ R̃k (1). Then f ∈ M̃k (α, β, 1), β > 0 for |z| < r(α, β), where 1 β (2.6) r(α, β) = 1 − ρ2 2 − ρ, with ρ = . α+β This result is best possible. Proof. Since f ∈ R̃k (1), zf 0 (z) f (z) ∈ P˜k (1) = Pk , and α zf 0 (z) β (zf 0 (z))0 β zp0 (z) + = p(z) + . α + β f (z) α + β f 0 (z) α + β p(z) Let φ(α, β) be as given by (2.4). Now using (1.3) and convolution techniques, we have φ(α, β) β zp0 (z) = p(z) ? p(z) + α + β p(z) z k 1 φ(α, β) k 1 φ(α, β) = + p1 (z) ? − − p2 (z) ? . 4 2 z 4 2 z n o φ(α,β) ˜ Since pi ∈ P2 (1) = P and it is known [2] that Re > 12 for |z| < r(α, β), it follows i z h ∈ P for |z| < r(α, β), i = 1, 2. with from a well known result, see [5] that pi ? φ(α,β) z r(α, β) given by (2.6). 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