IJMMSInternational Journal of Mathematics and Mathematical Sciences1687-04250161-1712Hindawi Publishing Corporation95203610.1155/2010/952036952036Research ArticleOn Differential Subordinations of Multivalent Functions Involving a Certain Fractional Derivative OperatorChoiJae HoAoufMohamed KamalDepartment of Mathematics EducationDaegu National University of Education1797-6 Daemyong 2 dongNamgu, Daegu 705-715South Koreadnue.ac.kr201008032010201018122009280220102010Copyright © 2010This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We investigate several results concerning the differential subordination of analytic and multivalent functions which is defined by using a certain fractional derivative operator. Some special cases are also considered.

1. Introduction and Definitions

Let 𝒜(p) denote the class of functions f(z) of the form

f(z)=zp+k=1ap+kzp+k(p:={1,2,3,}), which are analytic in the open unit disk 𝕌={z:z,|z|<1}. Also let 𝒜0 denote the class of all analytic functions p(z) with p(0)=1 which are defined on 𝕌. If f and g are analytic in 𝕌 with f(0)=g(0), then we say that f is said to be subordinate to g in 𝕌, written fg or f(z)g(z), if there exists the Schwarz function w, analytic in 𝕌 such that w(0)=0, |w(z)|<1(z𝕌), and f(z)=g(w(z))(z𝕌). In particular, if the function g is univalent, then the above subordination is equivalent to f(0)=g(0) and f(𝕌)g(𝕌).

Let a, b, and c be complex numbers with c0,-1,-2,. Then the Gaussian hypergeometric function 2F1(a,b;c;z) is defined by

2F1(a,b;c;z)=k=0(a)k(b)k(c)kzkk!, where (η)k is the Pochhammer symbol defined, in terms of the Gamma function, by

(η)k=Γ(η+k)Γ(η)={1(k=0),η(η+1)(η+k-1)(k). The hypergeometric function 2F1(a,b;c;z) is analytic in 𝕌, and if a or b is a negative integer, then it reduces to a polynomial.

There are a number of definitions for fractional calculus operators in the literature (cf., e.g., [1, 2]). We use here the Saigo-type fractional derivative operator defined as follows (see ; see also ).

Definition 1.1.

Let 0λ<1 and μ,ν. Then the generalized fractional derivative operator 𝒥0,zλ,μ,ν of a function f(z) is defined by 𝒥0,zλ,μ,νf(z)=ddz(zλ-μΓ(1-λ)0z(z-ζ)-λ2F1(μ-λ,1-ν;1-λ;1-ζz)f(ζ)dζ). The function f(z) is an analytic function in a simply-connected region of the z-plane containing the origin, with the order f(z)=O(|z|ϵ)(z0) for ϵ>max{0,μ-ν}-1, and the multiplicity of (z-ζ)-λ is removed by requiring that log(z-ζ) be real when z-ζ>0.

Definition 1.2.

Under the hypotheses of Definition 1.1, the fractional derivative operator 𝒥0,zλ+m,μ+m,ν+m of a function f(z) is defined by 𝒥0,zλ+m,μ+m,ν+mf(z)=dmdzm𝒥0,zλ,μ,νf(z)(z𝕌;m0:={0}).

With the aid of the above definitions, we define a modification of the fractional derivative operator Δz,pλ,μ,ν by

Δz,pλ,μ,νf(z)=Γ(p+1-μ)Γ(p+1-λ+ν)Γ(p+1)Γ(p+1-μ+ν)zμ𝒥0,zλ,μ,νf(z), for f(z)𝒜(p) and μ-ν-p<1. Then it is observed that Δz,pλ,μ,ν also maps 𝒜(p) onto itself as follows:

Δz,pλ,μ,νf(z)=zp+k=1(p+1)k(p+1-μ+ν)k(p+1-μ)k(p+1-λ+ν)kak+pzk+p(z𝕌;  0λ<1;  μ-ν-p<1;  f𝒜(p)). It is easily verified from (1.8) that

z(Δz,pλ,μ,νf(z))=(p-μ)Δz,pλ+1,μ+1,ν+1f(z)+μΔz,pλ,μ,νf(z). Note that Δz,p0,0,νf=f, Δz,p1,1,νf=zf/p, and Δz,pλ,λ,νf=Ωz(λ,p)f, where Ωz(λ,p) is the fractional derivative operator defined by Srivastava and Aouf [5, 6].

In this manuscript, we will use the method of differential subordination to derive certain properties of multivalent functions defined by fractional derivative operator Δz,pλ,μ,ν.

2. Main Results

In order to establish our results, we require the following lemma due to Miller and Mocanu .

Lemma 2.1.

Let q(z) be univalent in 𝕌 and let θ(w) and ϕ(w) be analytic in a domain 𝒟 containing q(𝕌) with ϕ(w)0 when wq(𝕌). Set Q(z)=zq(z)ϕ(q(z)), h(z)=θ(q(z))+Q(z) and suppose that

Q(z) is starlike (univalent) in 𝕌,

Re{zh(z)/Q(z)}=Re{θ(q(z))/ϕ(q(z))+zQ(z)/Q(z)}>0(z𝕌).

If p(z) is analytic in 𝕌, with p(0)=q(0), p(𝕌)𝒟, and θ(p(z))+zp(z)ϕ(p(z))θ(q(z))+zq(z)ϕ(q(z))=h(z), then p(z)q(z) and q(z) is the best dominant.

We begin by proving the following

Theorem 2.2.

Let α,β,γ and β0, and let 0λ<1, μ,ν, μp-1, μ-ν-p<1, and γ(p-μ-1)/β<2. Suppose that q(z)𝒜0 is univalent in 𝕌 and satisfies Re(1+zq′′(z)q(z))>{γ(p-μ-1)-ββifγ(p-μ-1)β1,0ifγ(p-μ-1)β1. If f(z)𝒜(p) and Δz,pλ,μ,νf(z)Δz,pλ+1,μ+1,ν+1f(z){αΔz,pλ+1,μ+1,ν+1f(z)Δz,pλ,μ,νf(z)+βΔz,pλ+2,μ+2,ν+2f(z)Δz,pλ+1,μ+1,ν+1f(z)+γ}1p-μ-1{(p-μ)(α+β)-α+[γ(p-μ-1)-β]q(z)-βzq(z)}, then Δz,pλ,μ,νf(z)Δz,pλ+1,μ+1,ν+1f(z)q(z) and q(z) is the best dominant.

Proof.

Define the function p(z) by p(z)=Δz,pλ,μ,νf(z)Δz,pλ+1,μ+1,ν+1f(z)(z𝕌). Then p(z) is analytic in 𝕌 with p(0)=1. A simple computation using (2.5) gives zp(z)p(z)=z(Δz,pλ,μ,νf(z))Δz,pλ,μ,νf(z)-z(Δz,pλ+1,μ+1,ν+1f(z))Δz,pλ+1,μ+1,ν+1f(z). By applying the identity (1.9) in (2.6), we obtain Δz,pλ+2,μ+2,ν+2f(z)Δz,pλ+1,μ+1,ν+1f(z)=1p-μ-1{p-μp(z)-1-zp(z)p(z)}. Making use of (2.5) and (2.7), we have {αΔz,pλ+1,μ+1,ν+1f(z)Δz,pλ,μ,νf(z)+βΔz,pλ+2,μ+2,ν+2f(z)Δz,pλ+1,μ+1,ν+1f(z)+γ}Δz,pλ,μ,νf(z)Δz,pλ+1,μ+1,ν+1f(z)={αp(z)+βp-μ-1(p-μp(z)-1-zp(z)p(z))+γ}p(z)=1p-μ-1{(p-μ)(α+β)-α+[γ(p-μ-1)-β]p(z)-βzp(z)}. In view of (2.8), the subordination (2.3) becomes [γ(p-μ-1)-β]p(z)-βzp(z)[γ(p-μ-1)-β]q(z)-βzq(z) and this can be written as (2.1), where θ(w)=[γ(p-μ-1)-β]w,ϕ(w)=-β. Since β0, we find from (2.10) that θ(w) and ϕ(w) are analytic in with ϕ(w)0. Let the functions Q(z) and h(z) be defined by Q(z)=zq(z)ϕ(q(z))=-βzq(z),h(z)=θ(q(z))+Q(z)=[γ(p-μ-1)-β]q(z)-βzq(z). Then, by virtue of (2.2), we see that Q(z) is starlike and Re{zh(z)Q(z)}=Re{β-γ(p-μ-1)β+(1+zq′′(z)q(z))}>0. Hence, by using Lemma 2.1, we conclude that p(z)q(z), which completes the proof of Theorem 2.2.

Remark 2.3.

If we put λ=μ in Theorem 2.2, then we get new subordination result for the fractional derivative operator Ωz(λ,p) due to Srivastava and Aouf [5, 6].

Theorem 2.4.

Let α,β,γ,δ and α,δ0, and let 0λ<1, μ,ν, μp, μ-ν-p<1, and 1+δ(p-μ)(α+γ)/α>0. Suppose that q(z)𝒜0 is univalent in 𝕌 and satisfies Re(1+zq′′(z)q(z))>{δ(μ-p)(α+γ)αifδ(p-μ)(α+γ)α0,0ifδ(p-μ)(α+γ)α0. If f(z)𝒜(p) and {αΔz,pλ+1,μ+1,ν+1f(z)Δz,pλ,μ,νf(z)+β(zpΔz,pλ,μ,νf(z))δ+γ}(Δz,pλ,μ,νf(z)zp)δαδ(p-μ)zq(z)+(α+γ)q(z)+β. then (Δz,pλ,μ,νf(z)zp)δq(z) and q(z) is the best dominant.

Proof.

Define the function p(z) by p(z)=(Δz,pλ,μ,νf(z)zp)δ(z𝕌). Then p(z) is analytic in 𝕌 with p(0)=1. By a simple computation, we find from (2.16) that zp(z)p(z)=δz(Δz,pλ,μ,νf(z))Δz,pλ,μ,νf(z)-pδ. By using the identity (1.9) in (2.17), we obtain Δz,pλ+1,μ+1,ν+1f(z)Δz,pλ,μ,νf(z)=1δ(p-μ)zp(z)p(z)+1. Applying (2.16) and (2.18), we have {αΔz,pλ+1,μ+1,ν+1f(z)Δz,pλ,μ,νf(z)+β(zpΔz,pλ,μ,νf(z))δ+γ}(Δz,pλ,μ,νf(z)zp)δ                ={α(1δ(p-μ)zp(z)p(z)+1)+βp(z)+γ}p(z)                =αδ(p-μ)zp(z)+(α+γ)p(z)+β. In view of (2.19), the subordination (2.14) becomes δ(p-μ)(α+γ)p(z)+αzp(z)δ(p-μ)(α+γ)q(z)+αzq(z) and this can be written as (2.1), where θ(w)=δ(p-μ)(α+γ)w,ϕ(w)=α. Since α0, it follows from (2.21) that θ(w) and ϕ(w) are analytic in with ϕ(w)0. Let the functions Q(z) and h(z) be defined by Q(z)=zq(z)ϕ(q(z))=αzq(z),h(z)=θ(q(z))+Q(z)=δ(p-μ)(α+γ)q(z)+αzq(z). Then, by virtue of (2.13), we see that Q(z) is starlike and Re{zh(z)Q(z)}=Re{δ(p-μ)(α+γ)α+(1+zq′′(z)q(z))}>0. Hence, by using Lemma 2.1, we conclude that p(z)q(z), which proves Theorem 2.4.

If we put λ=μ=0 in Theorem 2.4, then we have the following.

Corollary 2.5.

Let α,β,γ,δ and α,δ0, and let 1+pδ(α+γ)/α>0. Suppose that q(z)𝒜0 is univalent in 𝕌 and satisfies Re(1+zq′′(z)q(z))>{-pδ(α+γ)αifδ(α+γ)α0,0ifδ(α+γ)α0. If f(z)𝒜(p) and {αzf(z)f(z)+β(zpf(z))δ+γ}(f(z)zp)δαpδzq(z)+(α+γ)q(z)+β, then (f(z)/zp)δq(z) and q(z) is the best dominant.

By putting δ=α in Corollary 2.5, we obtain the following.

Corollary 2.6.

Let α,β,γ and α0, and let 1+p(α+γ)>0. Suppose that q(z)𝒜0 is univalent in 𝕌 and satisfies Re(1+zq′′(z)q(z))>{-p(α+γ)ifα+γ0,0ifα+γ0. If f(z)𝒜(p) and {αzf(z)f(z)+β(zpf(z))α+γ}(f(z)zp)αzq(z)p+(α+γ)q(z)+β, then (f(z)/zp)αq(z) and q(z) is the best dominant.

By using Lemma 2.1, we obtain the following.

Theorem 2.7.

Let α,β,γ and β0, and let 0λ<1, μ,ν, μ0, μ-ν-p<1, and 1+γ/β>0. Suppose that q(z)𝒜0 is univalent in 𝕌 and satisfies Re(1+zq′′(z)q(z))>{-γβif  γβ0,0ifγβ0. If f(z)𝒜(p) and {αβ[(p-μ-1)Δz,pλ+2,μ+2,ν+2f(z)Δz,pλ+1,μ+1,ν+1f(z)-(p-μ)Δz,pλ+1,μ+1,ν+1f(z)Δz,pλ,μ,νf(z)+1]+γ}·(Δz,pλ+1,μ+1,ν+1f(z)Δz,pλ,μ,νf(z))α    βzq(z)+γq(z), then (Δz,pλ+1,μ+1,ν+1f(z)Δz,pλ,μ,νf(z))αq(z) and q(z) is the best dominant.

Proof.

Define the function p(z) by p(z)=(Δz,pλ+1,μ+1,ν+1f(z)Δz,pλ,μ,νf(z))α(z𝕌). Then p(z) is analytic in 𝕌 with p(0)=1. A simple computation using (1.9) and (2.31) gives 1αzp(z)p(z)=(p-μ-1)Δz,pλ+2,μ+2,ν+2f(z)Δz,pλ+1,μ+1,ν+1f(z)-(p-μ)Δz,pλ+1,μ+1,ν+1f(z)Δz,pλ,μ,νf(z)+1. By using (2.29), (2.31), and (2.32), we get {αβ[(p-μ-1)Δz,pλ+2,μ+2,ν+2f(z)Δz,pλ+1,μ+1,ν+1f(z)-(p-μ)Δz,pλ+1,μ+1,ν+1f(z)Δz,pλ,μ,νf(z)+1]+γ}·(Δz,pλ+1,μ+1,ν+1f(z)Δz,pλ,μ,νf(z))α=    βzp(z)+γp(z). And this can be written as (2.1) when θ(w)=γw and ϕ(w)=β. Note that ϕ(w)0 and θ(w) and ϕ(w) are analytic in . Let the functions Q(z) and h(z) be defined by Q(z)=zq(z)ϕ(q(z))=βzq(z),h(z)=θ(q(z))+Q(z)=γq(z)+βzq(z). Then, by virtue of (2.28), we see that Q(z) is starlike and Re{zh(z)Q(z)}=Re{γβ+(1+zq′′(z)q(z))}>0. Hence, by applying Lemma 2.1, we observe that p(z)q(z), which evidently proves Theorem 2.7.

Finally, we prove

Theorem 2.8.

Let α,β,γ,δ and α,δ0, and let 0λ<1, μ,ν, μp, p+1-μ+ν>0 and 1-δ(p-μ)(α+γ)/α>0. Suppose that q(z)𝒜0 be univalent in 𝕌 and satisfies Re(1+zq′′(z)q(z))>{δ(p-μ)(α+γ)αifδ(p-μ)(α+γ)α0,0ifδ(p-μ)(α+γ)α0. If f(z)𝒜(p) and {αΔz,pλ+1,μ+1,ν+1f(z)Δz,pλ,μ,νf(z)+β(Δz,pλ,μ,νf(z)zp)δ+γ}(zpΔz,pλ,μ,νf(z))δβ+(α+γ)q(z)-αδ(p-μ)zq(z), then (zpΔz,pλ,μ,νf(z))δq(z) and q(z) is the best dominant.

Proof.

If we define the function p(z) by p(z)=(zpΔz,pλ,μ,νf(z))δ(z𝕌), then p(z) is analytic in 𝕌 with p(0)=1. Hence, by using the same techniques as detailed in the proof of Theorem 2.2, we obtain the desired result.

By taking λ=μ=0 in Theorem 2.8 and after a suitable change in the parameters, we have the following.

Corollary 2.9.

Let α{0} and pα<1/2. Suppose that q(z)𝒜0 is univalent in 𝕌 and satisfies Re(1+zq′′(z)q(z))>{2pαifα>0,0ifα<0. If f(z)𝒜(p) and α(1+zf(z)f(z))(zpf(z))α2αq(z)-1pzq(z), then (zp/f(z))αq(z) and q(z) is the best dominant.

Acknowledgment

This work was supported by Daegu National University of Education Research Grant in 2008.

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