# On Differential Subordinations of Multivalent Functions Involving a Certain Fractional Derivative Operator

Authors: Jae Ho Choi
Journal: International Journal of Mathematics and Mathematical Sciences
 (2010)

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## Abstract

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.


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## Body

## 1. Introduction and Definitions

Let𝒜(p) denote the class of functions f(z) of the form(1.1)f(z)=zp+∑k=1∞ap+kzp+k(p∈ℕ:={1,2,3,…}),
which are analytic in the open unit disk 𝕌={z:z∈ℂ,|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 f≺g or f(z)≺g(z), if there exists the Schwarz function w, analytic in 𝕌 such that w(0)=0, |w(z)|<1(z∈𝕌), and f(z)=g(w(z))(z∈𝕌). In particular, if the function g is univalent, then the above subordination is equivalent to f(0)=g(0) and f(𝕌)⊂g(𝕌).Leta, b, and c be complex numbers with c≠0,-1,-2,…. Then the Gaussian hypergeometric function 2F1(a,b;c;z) is defined by(1.2)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(1.3)(η)k=Γ(η+k)Γ(η)={1(k=0),η(η+1)⋯(η+k-1)(k∈ℕ).
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 [3]; see also [4]).Definition 1.1.
Let0≤λ<1 and μ,ν∈ℝ. Then the generalized fractional derivative operator 𝒥0,zλ,μ,ν of a function f(z) is defined by
(1.4)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
(1.5)f(z)=O(|z|ϵ)(z→0)
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 Definition1.1, the fractional derivative operator 𝒥0,zλ+m,μ+m,ν+m of a function f(z) is defined by
(1.6)𝒥0,zλ+m,μ+m,ν+mf(z)=dmdzm𝒥0,zλ,μ,νf(z)(z∈𝕌;m∈ℕ0:={0}∪ℕ).With the aid of the above definitions, we define a modification of the fractional derivative operatorΔz,pλ,μ,ν by(1.7)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:(1.8)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(z∈𝕌;0≤λ<1;μ-ν-p<1;f∈𝒜(p)).
It is easily verified from (1.8) that(1.9)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 operatorΔz,pλ,μ,ν.


## 2. Main Results

In order to establish our results, we require the following lemma due to Miller and Mocanu [7].Lemma 2.1.
Letq(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 w∈q(𝕌). Set Q(z)=zq′(z)ϕ(q(z)), h(z)=θ(q(z))+Q(z) and suppose that(1)
Q(z) is starlike (univalent) in 𝕌,(2)
Re{zh′(z)/Q(z)}=Re{θ′(q(z))/ϕ(q(z))+zQ′(z)/Q(z)}>0(z∈𝕌).If p(z) is analytic in 𝕌, with p(0)=q(0), p(𝕌)⊂𝒟, and
(2.1)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 followingTheorem 2.2.
Letα,β,γ∈ℝ and β≠0, and let 0≤λ<1,  μ,ν∈ℝ, μ≠p-1, μ-ν-p<1, and γ(p-μ-1)/β<2. Suppose that q(z)∈𝒜0 is univalent in 𝕌 and satisfies
(2.2)Re(1+zq′′(z)q′(z))>{γ(p-μ-1)-ββifγ(p-μ-1)β≥1,0ifγ(p-μ-1)β≤1.
If f(z)∈𝒜(p) and
(2.3)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
(2.4)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 functionp(z) by
(2.5)p(z)=Δz,pλ,μ,νf(z)Δz,pλ+1,μ+1,ν+1f(z)(z∈𝕌).
Then p(z) is analytic in 𝕌 with p(0)=1. A simple computation using (2.5) gives
(2.6)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
(2.7)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
(2.8){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
(2.9)[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
(2.10)θ(w)=[γ(p-μ-1)-β]w,ϕ(w)=-β.
Since β≠0, we find from (2.10) that θ(w) and ϕ(w) are analytic in ℂ with ϕ(w)≠0. Let the functions Q(z) and h(z) be defined by
(2.11)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
(2.12)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 putλ=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.
Letα,β,γ,δ∈ℝ and α,δ≠0, and let 0≤λ<1,  μ,ν∈ℝ, μ≠p, μ-ν-p<1, and 1+δ(p-μ)(α+γ)/α>0. Suppose that q(z)∈𝒜0 is univalent in 𝕌 and satisfies
(2.13)Re(1+zq′′(z)q′(z))>{δ(μ-p)(α+γ)αifδ(p-μ)(α+γ)α≤0,0ifδ(p-μ)(α+γ)α≥0.
If f(z)∈𝒜(p) and
(2.14){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≺αδ(p-μ)zq′(z)+(α+γ)q(z)+β.
then
(2.15)(Δz,pλ,μ,νf(z)zp)δ≺q(z)
and q(z) is the best dominant.Proof.
Define the functionp(z) by
(2.16)p(z)=(Δz,pλ,μ,νf(z)zp)δ(z∈𝕌).
Then p(z) is analytic in 𝕌 with p(0)=1. By a simple computation, we find from (2.16) that
(2.17)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
(2.18)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
(2.19){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
(2.20)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
(2.21)θ(w)=δ(p-μ)(α+γ)w,ϕ(w)=α.
Since α≠0, it follows from (2.21) that θ(w) and ϕ(w) are analytic in ℂ with ϕ(w)≠0. Let the functions Q(z) and h(z) be defined by
(2.22)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
(2.23)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 putλ=μ=0 in Theorem 2.4, then we have the following.Corollary 2.5.
Letα,β,γ,δ∈ℝ and α,δ≠0, and let 1+pδ(α+γ)/α>0. Suppose that q(z)∈𝒜0 is univalent in 𝕌 and satisfies
(2.24)Re(1+zq′′(z)q′(z))>{-pδ(α+γ)αifδ(α+γ)α≤0,0ifδ(α+γ)α≥0.
If f(z)∈𝒜(p) and
(2.25){αzf′(z)f(z)+β(zpf(z))δ+γ}(f(z)zp)δ≺αpδzq′(z)+(α+γ)q(z)+β,
then (f(z)/zp)δ≺q(z) and q(z) is the best dominant.By puttingδ=α in Corollary 2.5, we obtain the following.Corollary 2.6.
Letα,β,γ∈ℝ and α≠0, and let 1+p(α+γ)>0. Suppose that q(z)∈𝒜0 is univalent in 𝕌 and satisfies
(2.26)Re(1+zq′′(z)q′(z))>{-p(α+γ)ifα+γ≤0,0ifα+γ≥0.
If f(z)∈𝒜(p) and
(2.27){αzf′(z)f(z)+β(zpf(z))α+γ}(f(z)zp)α≺zq′(z)p+(α+γ)q(z)+β,
then (f(z)/zp)α≺q(z) and q(z) is the best dominant.By using Lemma2.1, we obtain the following.Theorem 2.7.
Letα,β,γ∈ℝ and β≠0, and let 0≤λ<1,  μ,ν∈ℝ, μ≠0, μ-ν-p<1, and 1+γ/β>0. Suppose that q(z)∈𝒜0 is univalent in 𝕌 and satisfies
(2.28)Re(1+zq′′(z)q′(z))>{-γβifγβ≤0,0ifγβ≥0.
If f(z)∈𝒜(p) and
(2.29){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
(2.30)(Δz,pλ+1,μ+1,ν+1f(z)Δz,pλ,μ,νf(z))α≺q(z)
and q(z) is the best dominant.Proof.
Define the functionp(z) by
(2.31)p(z)=(Δz,pλ+1,μ+1,ν+1f(z)Δz,pλ,μ,νf(z))α(z∈𝕌).
Then p(z) is analytic in 𝕌 with p(0)=1. A simple computation using (1.9) and (2.31) gives
(2.32)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
(2.33){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 ℂ. Let the functions Q(z) and h(z) be defined by
(2.34)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
(2.35)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 proveTheorem 2.8.
Letα,β,γ,δ∈ℝ and α,δ≠0, and let 0≤λ<1, μ,ν∈ℝ, μ≠p, 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
(2.36)Re(1+zq′′(z)q′(z))>{δ(p-μ)(α+γ)αifδ(p-μ)(α+γ)α≥0,0ifδ(p-μ)(α+γ)α≤0.
If f(z)∈𝒜(p) and
(2.37){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))δ≺β+(α+γ)q(z)-αδ(p-μ)zq′(z),
then
(2.38)(zpΔz,pλ,μ,νf(z))δ≺q(z)
and q(z) is the best dominant.Proof.
If we define the functionp(z) by
(2.39)p(z)=(zpΔz,pλ,μ,νf(z))δ(z∈𝕌),
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 takingλ=μ=0 in Theorem 2.8 and after a suitable change in the parameters, we have the following.Corollary 2.9.
Letα∈ℝ∖{0} and pα<1/2. Suppose that q(z)∈𝒜0 is univalent in 𝕌 and satisfies
(2.40)Re(1+zq′′(z)q′(z))>{2pαifα>0,0ifα<0.
If f(z)∈𝒜(p) and
(2.41)α(1+zf′(z)f(z))(zpf(z))α≺2αq(z)-1pzq′(z),
then (zp/f(z))α≺q(z) and q(z) is the best dominant.



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