On certain subclasses of analytic functions

Transcription

1 Stud. Univ. Babeş-Bolyai Math. 58(013), No. 1, 9 14 On certain subclasses of analytic functions Imran Faisal, Zahid Shareef and Maslina Darus Abstract. In the present paper, we introduce and study certain new subclasses of analytic functions in the open unit disk U. Some inclusion relationships and integral preserving properties have also discussed in particular with reference to a new integral operator. Mathematics Subject Classification (010): 30C45. Keywords: Analytic functions, starlike functions, inclusion relationship. 1. Introduction Let A be the class of functions of the form f = z + a k z k which are analytic and normalized in the open unit disk U = {z : z < 1}. Next we define some well known subclasses such as starlike, convex, close-to-convex and quasi-convex functions of A, denoted by S (ξ), C(ξ), K(ρ, ξ) and K (ρ, ξ) respectively as follow(cf.[1]-[3]): { ( zf S ) } (ξ) = f A : R > ξ, z U, 0 ξ < 1. f { ( C(ξ) = f A : R 1 + zf ) } f > ξ, z U, 0 ξ < 1. { ( zf K(ρ, ξ) = f A : g S ) } (ξ) R > ρ, z U, 0 ρ < 1. g { ( (zf K ) ) } (ρ, ξ) = f A : g C(ξ) R g > ρ, z U. Note that f C(ξ) zf S (ξ) f K (ρ, ξ) zf K(ρ, ξ).

2 10 Imran Faisal, Zahid Shareef and Maslina Darus For f A and β, γ 0, we define a new differential operator as follows: Θ 0 (β, γ)f = f (γ + β + 1)Θ 1 (β, γ)f = βf + (γ + 1)(zf ) Θ n (β, γ)f = z +. ( ) n β + k(γ + 1) a k z k. (1.1) γ + β + 1 This operator is closely related to the following operators: 1. Θ n (λ, 0)f = Θ n (λ)f = z + ( ) n k+λ 1+λ ak z k,(see [4, 5]);. Θ n (1, 0)f = Θ n f = z + ( k+1 ) n ak z k,(see[6]); 3. Θ n (0, 0)f = Θ n f = z + (k) n a k z k,(see[7]). (1.1) (γ + 1)z(Θ n (β, γ)f) = (γ β)θ n+1 (β, γ)f βθ n (β, γ)f. Now for linear operator Θ n (β, γ) we define the following subclasses of A: S n(ξ, β, γ) = {f A : Θ n (β, γ)f S (ξ)}; C n (ξ, β, γ) = {f A : Θ n (β, γ)f C(ξ)}; K n (ρ, ξ, β, γ) = {f A : Θ n (β, γ)f K(ρ, ξ)}; K n(ρ, ξ, β, γ) = {f A : Θ n (β, γ)f K (ρ, ξ)}.. Inclusion relationships Lemma.1. [8, 9] Let ϕ(µ, υ) be a complex function such that ϕ : D C, D C C, and let µ = µ 1 + iµ, υ = υ 1 + iυ. Suppose that ϕ(µ, υ) satisfies the following conditions: 1. ϕ(µ, υ) is continuous in D;. (1, 0) D and Rϕ(1, 0) > 0; 3. Rϕ(iµ, υ 1 ) 0 for all (iµ, υ 1 ) D such that υ 1 1 (1 + µ ). Let h = 1 + c 1 z + c z + be analytic in U, such that (h, zh ) D for all z U. If R{ϕ(h, zh )} > 0(z U), then R{h} > 0. Theorem.. Let f A, 0 ξ < 1, β, γ 0, n N then S n+1(ξ, β, γ) S n(ξ, β, γ) S n 1(ξ, β, γ). Proof. Let f S n+1(ξ, β, γ), and suppose that Since z(θ n (β, γ)f) Θ n (β, γ)f (1 + β )Θn+1 (β, γ)f γ + 1 Θ n (β, γ)f = ξ + (1 ξ)h. = ξ + (1 ξ)h + β γ + 1,

3 On certain subclasses of analytic functions 11 therefore z(θ n+1 (β, γ)f) Θ n+1 (β, γ)f ξ = (1 ξ)h + (γ + 1)(1 ξ)zh β + (γ + 1)ξ + (1 ξ)h. Taking h = µ = µ 1 + iµ and zh = υ = υ 1 + iυ, we define ϕ(µ, υ) by: ϕ(µ, υ) = (1 ξ)µ + (γ + 1)(1 ξ)υ β + (γ + 1)ξ + (1 ξ)µ. R{ϕ(iµ, υ 1 )} = [1 + (γ + 1)ξ](γ + 1)(1 ξ)υ 1 (1 + (γ + 1)ξ) + (1 ξ) µ, R{ϕ(iµ, υ 1 )} [1 + (γ + 1)ξ](γ + 1)(1 ξ)(1 + µ ) (1 + (γ + 1)ξ) + (1 ξ) µ < 0. Clearly ϕ(µ, υ) satisfies the conditions of Lemma.1. Hence R{h} > 0(z U), implies f Sn(ξ, β, γ). Theorem.3. Let f A, 0 ξ < 1, β, γ 0, n N 0 then C n+1 (ξ, β, γ) C n (ξ, β, γ) C n 1 (ξ, β, γ). Proof. Let f C n+1 (ξ, β, γ) Θ n+1 (β, γ)f C(ξ) z(θ n+1 (β, γ)f) S (ξ) Θ n+1 (β, γ)(zf ) S (ξ) zf Sn+1(ξ, β, γ) Sn(ξ, β, γ) zf Sn(ξ, β, γ) Θ n (β, γ)(zf ) S (ξ) z(θ n (β, γ)f) S (ξ) Θ n (β, γ)f C(ξ) f C n (ξ, β, γ). Theorem.4. Let f A, 0 ξ < 1, β, γ 0, 0 ρ < 1, n N 0 then K n+1 (ρ, ξ, β, γ) K n (ρ, ξ, β, γ) K n 1 (ρ, ξ, β, γ). Proof. Let f K n+1 (ρ, ξ, β, γ) and suppose that Using (1.1) we have Since (Θn (β,γ)zf ) Θ n (β,γ)g ( z(θn (β, γ)f) Θ n ) = ρ + (1 ρ)h, z U. (β, γ)g z(θ n+1 (β, γ)f) Θ n+1 (β, γ)g z(θ n+1 (β, γ)f) Θ n+1 (β, γ)g = (γ+1)z(θ n (β,γ)f ) Θ n (β,γ)g (γ+1)z(θ n (β,γ)g) Θ n (β,γ)g + βz(θn (β,γ)zf ) Θ n (β,γ)g + β = ρ+(1 ρ)h and g S n+1(ξ, β, γ) S n(ξ, β, γ). Therefore ρ = (1 ρ)h + (γ + 1)(1 ρ)zh (γ + 1)(ξ + (1 ξ)h) + β. Taking h = µ = µ 1 + iµ and zh = υ = υ 1 + iυ, we define ϕ(µ, υ) by This implies ϕ(µ, υ) = (1 ρ)µ + (γ + 1)(1 ρ)υ (γ + 1)(ξ + (1 ξ)h) + β. R[ϕ(iµ, υ 1 )] = [β+ξ(γ+1)+(γ+1)(1 ξ)h 1(x 1,y 1)] +[(γ+1)(1 ξ)h (x 1,y 1)] < 0. (γ+1)(1+µ )(1 ρ)[β+ξ(γ+1)+(γ+1)(1 ξ)h1(x1,y1)].

4 1 Imran Faisal, Zahid Shareef and Maslina Darus Hence, the function ϕ(µ, υ) satisfies the conditions of Lemma.1. Implies R{h} > 0(z U) gives f K n (ρ, ξ, β, γ). Similarly we proved the following theorem. Theorem.5. Let f A, 0 ξ < 1, 0 ρ < 1, β, γ 0, n N 0 then K n+1(ρ, ξ, β, γ) K n(ρ, ξ, β, γ) K n 1(ρ, ξ, β, γ). 3. Integral operator For c > 1 and f A, the integral operator L c (f) : A A is defined by L c (f) = c + 1 z c z The operator L c (f) was introduced by Bernardi [10]. 0 t c 1 f(t)dt. (3.1) Theorem 3.1. Let c > 1, 0 ξ < 1. If f S n(ξ, β, γ), then L c (f) S n(ξ, β, γ). Proof. By using (3.1) we get z(θ n (β, γ)l c f) Θ n (β, γ)l c f Let z(θn (β, γ)l c f) Θ n (β, γ)l c f z(θ n (β, γ)l c f) Θ n (β, γ)l c f This implies and ϕ(µ, υ) = (1 ξ)µ + R[ϕ(iµ, υ 1 )] = = (c + 1) Θn (β, γ)f Θ n (β, γ)l c f c. = ξ + (1 ξ)h, h = 1 + c 1 z + c z +. ξ = (1 ξ)h + (1 ξ)zh ξ + (1 ξ)h + c. (1 ξ)υ, (same as Theorem.), ξ + c + (1 ξ)µ (ξ + c)(1 ξ)υ 1 [ξ + c] + [(1 ξ)µ ] (ξ + c)(1 ξ)(1 + µ ) [ξ + c] + [(1 ξ)µ ] < 0. After using Theorem.1 and Lemma.1., we have L c (f) S n(ξ, λ, α, β, µ). Theorem 3.. Let c > 1, 0 ξ < 1. If f C n (ξ, β, γ), then L c (f) C n (ξ, β, γ). Proof. Proof is same as that of Theorem.3. Theorem 3.3. Let c > 1, 0 ξ < 1, 0 ρ < 1. If f K n (ρ, ξ, β, γ), then L c (f) K n (ρ, ξ, β, γ). Proof. Since f K n (ξ, β, γ) Θ n (β, γ)f K(ρ, ξ). Let z(θ n (β, γ)l c f) Θ n (β, γ)l c g = ρ + (1 ρ).

5 On certain subclasses of analytic functions 13 Using (3.1) we have ( z(θn (β, γ)f) Θ n ) = (β, γ)g z(θ n (β,γ)l c(zf ) Θ n (β,γ)l cg z(θ n (β,γ)l c(g) Θ n (β,γ)l cg + c (Θn (β,γ)l c(zf ) Θ n (β,γ)l cg + c. Since g Sn(ξ, β, γ) L c (g) Sn(ξ, β, γ). Let z(θ n (β, γ)l c (g) Θ n (β, γ)l c g = ξ + (1 ξ)h, R(H) > 0. ( z(θn (β, γ)f) (1 ρ)zh Θ n ) ρ = (1 ρ)h + (β, γ)g ξ + (1 ξ)h + c. Using method of Theorem.3, we get ϕ(µ, υ) = (1 ρ)µ + (1 ρ)υ ξ + (1 ξ)h + c. Taking h = µ = µ 1 + iµ and zh = υ = υ 1 + iυ, we define the function ϕ(µ, υ) by: R[ϕ(iµ, υ 1 )] = 1 (1 + µ )(1 ρ)[(ξ + c) + (1 ξ)h 1 (x 1, y 1 )] [(ξ + c) + (1 ξ)h 1 (x 1, y 1 )] + [(1 ξ)h (x, y )] < 0. Hence, by using Lemma.1. we have L c (f) K n (ρ, ξ, β, γ). Similarly we proved the following theorem. Theorem 3.4. Let c > 1, 0 ξ < 1, and 0 ρ < 1. If f K n(ρ, ξ, β, γ), then L c (f) K n(ρ, ξ, β, γ). Acknowledgement. The work is supported by LRGS/TD/011/UKM/ICT/03/0 and GUP References [1] Robertson, M.S., On the theory of univalent functions, Ann. Math., 37(1936), [] Noor, K.I., On quasi-convex functions and related topics, Internat. J. Math. Math. Sci., 10(1987), [3] Libera, R.J., Some radius of convexity problems, Duke Math. J., 31(1964), [4] Cho, N.E., Kim, T.H., Multiplier transformations and strongly close to convex functions, Bull. Korean Math. Soc., 40(003), [5] Cho, N.E., Srivastava, H.M., Argument estimates of certain analytic functions defined by a class of multiplier transformations, Math. Comput. Modeling, 37(003), [6] Uralegaddi, B.A., Somanatha, C., Certain classes of univalent functions, In: current topics in analytic functions theory, eds., H.M. Srivastava and S. Owa, World Scientific Company, Singapore, 199, [7] Salagean, G.S., Subclasses of univalent functions, Lecture Notes in Math., 1013, Springer-Verlag, Heideberg, 1983,

6 14 Imran Faisal, Zahid Shareef and Maslina Darus [8] Miller, S.S., Differential inequalities and Caratheodory function, Bull. Amer. Math. Soc., 8(1975), [9] Miller, S.S., Mocanu, P.T., Second order differential inequalities in the complex plane, J. Math. Anal. Appl., 65(1978), [10] Bernardi, S.D., Convex and starlike univalent functions, Trans. Amer. Math. Soc., 135(1969), Imran Faisal Department of Mathematics COMSATS Institute of Information Technology Attock Campus, Pakistan Zahid Shareef School of Mathematical Sciences University Kebangsaan Malaysia, Bangi, Selangor, Malaysia Maslina Darus School of Mathematical Sciences University Kebangsaan Malaysia, Bangi, Selangor, Malaysia