Subordination Properties for Multivalent Functions Associated with a Generalized Fractional Differintegral Operator

Using of the principle of subordination, we investigate some subordination and convolution properties for classes of multivalent functions under certain assumptions on the parameters involved, which are defined by a generalized fractional differintegral operator under certain assumptions on the parameters involved.

If ϕ(z) is analytic in U and satisfies: then ϕ is a solution of (2).The univalent function q is called dominant, if ϕ(z) ≺ q(z) for all ϕ.
Definition 1.For f (z) ∈ A(p), the fractional integral and fractional derivative operators of order λ are defined by Owa [3] (see also [4]) as: where f is an analytic function in a simply-connected region of the complex z-plane containing the origin, and the multiplicity of (z − ζ) λ−1 ((z − ζ) −λ ) is removed by requiring log(z − ζ) to be real when z − ζ > 0.
Definition 2. For f (z) ∈ A(p) and in terms of 2 F 1 , the generalized fractional integral and generalized fractional derivative operators defined by Srivastava et al. [5] (see also [6]) as: where f (z) is an analytic function in a simply-connected region of the complex z−plane containing the origin with the order f (z We note that: where D −λ z f (z) and D λ z f (z) are the fractional integral and fractional derivative operators studied by Owa [3].

Preliminaries
To prove our main results, we shall need the following lemmas.Lemma 1. [18].Let h be an analytic and convex (univalent) function in U with h(0) = 1.Additionally, let φ given by: be analytic in U. If: then: and ψ is the best dominant of (6).
Denote by P(ς) the class of functions Φ given by: which are analytic in U and satisfy the following inequality: Using the well-known growth theorem for the Carathéodory functions (cf., e.g., [19]), we may easily deduce the following result: The result is the best possible.

Properties Involving H λ,δ p,η,µ
Unless otherwise mentioned, we assume throughout this paper that p ∈ N, Then: The estimate in ( 13) is sharp. Proof.Let: Then, φ is analytic in U.After some computations, we get: Now, by using Lemma 1, we deduce that: or, equivalently, and so: Since: The inequality (13) now follows from ( 16) and (17).To prove that the result is sharp, let: Now, for f (z) defined by ( 18), we have: Letting z → −1, we obtain: which ends our proof.
Putting θ = 1 and using Lemma 1 for Equation (15) in Theorem 1, we obtain the following example.
Example 1.Let the function f (z) ∈ A(p).Then, following containment property holds, Using (4) instead of (3) in Theorem 1, one can prove the following theorem. Then: The result is sharp.
Putting θ = 1 in Theorem 2, we obtain the following example.
For a function f ∈ A(p), the generalized Bernardi-Libera-Livingston integeral operator F p,γ is defined by (see [23]): Proof.Since and: Now, the first part of this lemma follows.Furthermore, If we replace f (z) by H λ,δ p,η,µ f (z) and using the first part of this lemma, we get (21) .
Theorem 3. Suppose that p + γ > 0, f ∈ T λ,δ p,η,µ (A, B) and F p,γ defined by (20).Then: The result is sharp. Proof.Let: Then, φ is analytic in U.After some calculations, we have: Employing the same technique that was used in proving Theorem 1, the remaining part of the theorem can be proven. then: where: and: The result is possible when Proof.Suppose that f i ∈ A(p) (i = 1, 2) satisfy the condition (25).Setting: we have: Thus, by making use of the identity (3) in (29), we get: which, in view of F given by ( 27) and (30), yields: where: Since p i (z) ∈ P(ς i ) (i = 1, 2), it follows from Lemma 3 that: Now, by using (33) in (32) and then appealing to Lemma 2, we have: When , which satisfy (25), are defined by: Thus, it follows from (32) that: which evidently ends the proof.
Theorem 5. Let υ ∈ C * , and let A, B ∈ C with A = B and |B| ≤ 1. Suppose that: If f ∈ A(p) with H λ,δ p,η,µ f (z) = 0 for all z ∈ U * = U\{0}, then: where: is the best dominant. Proof.Putting: Then, ∆ is analytic in U, ∆(0) = 1 and ∆(z) = 0 for all z ∈ U. Taking the logarithmic derivatives on both sides of (34) and using (3), we have: Now, the assertions of Theorem 5 follow by Lemma 4. then: where β ∈ (1, ∞) is the positive root of the equation: Proof.Let: Then, ϕ is analytic in U, ϕ(0) = 1 and ϕ(z) = 0 for all z ∈ U. Taking the logarithmic derivatives on both sides of (37) and using the identity (3), we have: and so: zϕ (z) . Let: s, and: Then, for x, y ≤ − 1+x 2 2 , we have: where β is the positive root of Equation (36).Suppose that: ∈ Ω, and so, we obtain the required result by an application of Lemma 5.