# An Extension to the Owa-Srivastava Fractional Operator with Applications to Parabolic Starlike and Uniformly Convex Functions

Authors: Oqlah Al-Refai; Maslina Darus
Journal: International Journal of Differential Equations
 (2009)

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## Abstract

Let𝒜 be the class of analytic functions in the open unit disk 𝕌. We define Θα,β:𝒜→𝒜 by (Θα,βf)(z):=Γ(2−α)zαDzα(Γ(2−β)zβDzβf(z)),(α,β≠2,3,4…), where Dzγf is the fractional derivative of f of order γ. If α,β∈[0,1], then a function f in 𝒜 is said to be in the class SPα,β if Θα,βf is a parabolic starlike function. In this paper, several properties and characteristics of the class SPα,β are investigated. These include subordination, characterization and inclusions, growth theorems, distortion theorems, and class-preserving operators. Furthermore, sandwich theorem related to the fractional derivative is proved.


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## Body

## 1. Introduction and Definitions

Let𝒜 be the class of functions analytic in the open
unit disk 𝕌:={z:|z|<1} and let 𝒜[a,n] be the subclass of 𝒜 consisting of functions of the form(1.1)g(z)=a+anzn+an+1zn+1+⋯,and 𝒜0 be the class of functions f in 𝒜 of the form(1.2)f(z)=z+∑n=1∞anzn+1.Let 𝒯 be the subclass of 𝒜0 consisting of functions f of the form(1.3)f(z)=z−∑n=1∞anzn+1,(an≥0).A function f in 𝒜0 is said to be uniformly convex in 𝕌 if f is a univalent convex function along with the
property that, for every circular arc γ contained in 𝕌,
with center γ also in 𝕌,
the image curve f(γ) is a convex arc. The class of uniformly convex
functions is denoted by UCV (for details, see [1]). It is well known from
[2, 3] that(1.4)f∈UCV⟺|zf′′(z)f′(z)|<Re{1+zf′′(z)f′(z)},(z∈𝕌).Condition (1.4) implies
that(1.5)1+zf′′(z)f′(z)lies in the interior of the
parabolic region(1.6)ℛ:={w:w=u+iv,v2<2u−1},for every value of z∈𝕌.
A function f in 𝒜0 is said to be in the class of parabolic
starlike functions, denoted by SP (cf. [3]), if(1.7)zf′(z)f(z)∈ℛ,(z∈𝕌).Let the functionφ(a,b;z) be given by(1.8)φ(a,b;z):=∑n=0∞(a)n(b)nzn+1(b≠0,−1,−2,…;z∈𝕌),where (x)n is the Pochhammer symbol defined
by(1.9)(x)n:=Γ(x+n)Γ(x)={1,n=0,x(x+1)(x+2)⋯(x+n−1),n∈ℕ:={1,2,3,…}.Further, let (cf. [4, 5])(1.10)L(a,b)f(z)=φ(a,b;z)*f(z)(f∈𝒜).In terms of Hadamard product
or convolution,
note that L(a,a) is the identity operator and(1.11)L(a,c)=L(a,b)L(b,c)(b,c≠0,−1,−2,…).It is well known that if b>a>0,
then L maps 𝒜 into itself. We also need the following
definitions of a fractional derivative.Definition 1.1  (cf. [5, 6], see also [7, 8]).
Let the
functionf(z) be analytic in a simply connected domain of
the z-plane containing the origin. The fractional
derivative of f of order α is defined by(1.12)Dzαf(z)=1Γ(1−α)ddz∫0zf(ξ)(z−ξ)αdξ,0≤α<1,
where the multiplicity of (z−ξ)−α is removed by requiring log(z−ξ) to be real when z−ξ>0.Using Definition1.1 and its known extensions involving
fractional derivatives and fractional integrals, Owa and Srivastava [5] introduced the operator Ωα:𝒜→𝒜 defined by(1.13)Ωαf(z)=Γ(2−α)zαDzαf(z),α≠2,3,4,…=∑n=0∞Γ(n+2)Γ(2−α)Γ(n+2−α)anzn+1=φ(2,2−α;z)*f(z)=L(2,2−α)f(z).Note that Ω0f(z)=f(z).Corresponding to the operatorΩα defined in (1.13),
Srivastava and Mishra [9] studied the class SPα(0≤α≤1) of functions f∈𝒜0 satisfying the inequality(1.14)|z(Ωαf(z))′Ωαf(z)−1|<Re{z(Ωαf(z))′Ωαf(z)},(z∈𝕌).In  Definition1.2, we
generalize the Owa-Srivastava operator defined in (1.13) as
follows.Definition 1.2.
Letf be in 𝒜.
One defines an operator Θα,β:𝒜→𝒜 by(1.15)(Θα,βf)(z)=Γ(2−α)zαDzα(Γ(2−β)zβDzβf(z)),(α,β≠2,3,4,…),where Dzγf is the fractional derivative of f of order γ.From Definition1.2, we note that(1.16)Θα,βf(z)=Θβ,αf(z),Θ0,0f(z)=f(z),Θα,0f(z)=Θ0,αf(z)=Ωαf(z),Θα,1f(z)=z(Ωαf(z))′,(1.17)Θα,βf(z)=Γ(2−α)zαDzα(Γ(2−β)zβDzβf(z)),(α,β≠2,3,4,…)=∑n=0∞Γ(n+2)Γ(2−α)Γ(n+2−α)Γ(n+2)Γ(2−β)Γ(n+2−β)anzn+1=φ(2,2−β;z)*φ(2,2−α;z)*f(z)=φ(2,2−β;z)*L(2,2−α)f(z)=φ(2,2−β;z)*Ωαf(z)=L(2,2−β)Ωαf(z)=Ωβ(Ωαf(z))=Ωα(Ωβf(z)).In the present paper, we study a class of analytic
functions, related toUCV, SP, and SPα,
using the operator Θα,β defined in Definition 1.2.Definition 1.3.
LetSPα,β,
where α,β∈[0,1] be the class of functions f∈𝒜0 satisfying the inequality(1.18)|z(Θα,βf(z))′Θα,βf(z)−1|<Re{z(Θα,βf(z))′Θα,βf(z)},(z∈𝕌).It
follows that(1.19)SPα,β≡SPβ,α,SPα,0≡SP0,α≡SPα,SP1,0≡SP0,1≡UCV,SP0,0≡SP.Remark 1.4.
f(z)∈SP1,1 if and only if zf′(z) is uniformly convex function.Using the definition ofΘα,β,
we start with proving sandwich theorem related to the fractional derivative.
Then, we investigate several properties and characteristics of the general
class SPα,β using similar techniques to [9]. These include
subordination, inclusions and characterization, growth theorems, and
class-preserving operators (like the Hadamard product and various integral
transforms).


## 2. Sandwich Theorem

In order to
prove our sandwich result, we need first to recall
the principle of subordination between analytic functions, let the functionsf and F
be in 𝒜.
We say that f is subordinate to F or F is superordinate to f in 𝕌,
written as f≺F,
if F is univalent in 𝕌,(2.1)f(0)=F(0),f(𝕌)⊆F(𝕌).Let p,h∈𝒜 and let ϕ(s,t;z):ℂ2×𝕌→ℂ.
If p and ϕ(p(z),zp′(z);z) are univalent and p satisfies the first-order differential
superordination,(2.2)h(z)≺ϕ(p(z),zp′(z);z),then p is a solution of the differential
superordination (2.2). An analytic function q is called a subordination if q≺p for all p satisfying (2.2). A univalent subordinant q˜ that satisfies q≺q˜ for all subordinations q of (2.2) is said to be the best subordinant.
An analytic function q is said to be dominant if p≺q for all p satisfying(2.3)ϕ(p(z),zp′(z);z)≺h(z).A univalent dominant q˜ that satisfies q˜≺q for all dominants q of (2.3) is said to be the best dominant.We also need the following definition and
lemma.Definition 2.1 (see [10, page 817, Definition 2]).
Denoted byQ,
the set of all functions f(z) that are analytic and injective on 𝕌¯−E(f),
where(2.4)E(f)={ζ∈∂𝕌:limz→ζf(z)=∞},and are such that f′(ζ)≠0 for ζ∈∂𝕌−E(f).Lemma 2.2 (see [11]).
Letq1,q2 be two nonzero univalent functions in 𝕌,
and let λ≠0,μ∈ℂ.
Further assume that ℜ[μλ¯qi(z)]≥0 and for (i=1,2), zqi′(z)/qi(z) is starlike univalent in 𝕌. If g∈𝒜0, zkg(k)(z)/g(k−1)(z)∈𝒜[1,1]∩Q  (k∈ℕ, g(k) is the kth derivative of g) and
zg(k+1)(z)/g(k)(z)+(μzk/λ−z)g(k)(z)/g(k−1)(z)+k is univalent in 𝕌,
then(2.5)μq1(z)+λzq1′(z)q1(z)≺λ(zg(k+1)(z)g(k)(z)+(μzk/λ−z)g(k)(z)g(k−1)(z)+k)≺μq2(z)+λzq2′(z)q2(z)implies(2.6)q1(z)≺zkg(k)(z)g(k−1)(z)≺q2(z),and q1,q2 are, respectively, the best subordinant and
the best dominant.As an application of Lemma2.2, we prove the following theorem.Theorem 2.3.
Letq1,q2 be two nonzero univalent functions in 𝕌,
and let k∈ℕ, α≠2,3,4,…, and 0<β<1.
Further, assume that zq1′(z)/q1(z) and zq2′(z)/q2(z) are starlike univalent in 𝕌.
If f∈𝒜0,(2.7)Φ(α,β,k;z)=k+zDzβ+k+1(Γ(2−α)zαDzαf(z))(1−β−k)Dzβ+k(Γ(2−α)zαDzαf(z))isunivalentin𝕌,Ψ(α,β,k;z)=zk+1Dzβ+k(Γ(2−α)zαDzαf(z))(2−β−k)Dzβ+k−1(Γ(2−α)zαDzαf(z))∈𝒜[1,1]∩Q,then(2.8)zq1′(z)q1(z)≺Φ(α,β,k;z)≺zq2′(z)q2(z)implies(2.9)q1(z)≺Ψ(α,β,k;z)≺q2(z),and q1,q2 are, respectively, the best subordinant and
the best dominant.Proof.
Letg(z)=Θα,βf(z) be defined as in Definition 1.2, where f∈𝒜0,0<β<1 and (α≠2,3,4,…).
Then from (1.17), we have for k∈ℕ,(2.10)g(k)(z)=dkdzk(Θα,βf(z))=L(2,2−α)(dkdzk(Ωβf(z)))=L(2,2−α)Ωβ+kf(z)=Θα,β+kf(z).This yields(2.11)zkg(k)(z)g(k−1)(z)=zkΘα,β+kf(z)Θα,β+k−1f(z)=zk+1Dzβ+k(Γ(2−α)zαDzαf(z))(2−β−k)Dzβ+k−1(Γ(2−α)zαDzαf(z)),zg(k+1)(z)g(k)(z)=zΘα,β+k+1f(z)Θα,β+kf(z)=k+zDzβ+k+1(Γ(2−α)zαDzαf(z))(1−β−k)Dzβ+k(Γ(2−α)zαDzαf(z)).By applying Lemma 2.2 for λ=1 and μ=0,
we get the result.Puttingα=0 in Theorem 2.3, we get the following
corollary.Corollary 2.4.
Letq1,q2 be two nonzero univalent functions in 𝕌,
and let k∈ℕ,  0<β<1.
Further, assume that zq1′(z)/q1(z) and zq2′(z)/q2(z) are starlike univalent in 𝕌.
If f∈𝒜0,(2.12)Φ(β,k;z)=k+zDzβ+k+1f(z)(1−β−k)Dzβ+kf(z)isunivalentin𝕌,Ψ(β,k;z)=zk+1Dzβ+kf(z)(2−β−k)Dzβ+k−1f(z)∈𝒜[1,1]∩Q,then(2.13)zq1′(z)q1(z)≺Φ(β,k;z)≺zq2′(z)q2(z)implies(2.14)q1(z)≺Ψ(β,k;z)≺q2(z),and q1,q2 are, respectively, the best subordinant and
the best dominant.In particular, fork=1,
Corollary 2.4 reduces to the following remark.Remark 2.5.
Letq1,q2 be two nonzero univalent functions in 𝕌,
and assume that zq1′(z)/q1(z) and zq2′(z)/q2(z) are starlike univalent in 𝕌.
For 0<β<1,
if f∈𝒜0, 1−(zDzβ+2f(z)/βDzβ+1f(z)) is univalent in 𝕌 and z2Dzβ+1f(z)/(1−β)Dzβf(z)∈𝒜[1,1]∩Q,
then(2.15)zq1′(z)q1(z)≺1−1β(zDzβ+2f(z)Dzβ+1f(z))≺zq2′(z)q2(z)implies(2.16)q1(z)≺11−β(z2Dzβ+1f(z)Dzβf(z))≺q2(z),and q1,q2 are, respectively, the best subordinant and
the best dominant.Remark 2.6.
Takingα=1 in Theorem 2.3 yields that Corollary
2.4 and
Remark 2.5 are also hold true for f(z)=zg′(z),
where g∈𝒜0.


## 3. Some Properties of the ClassSPα,β

We need the
following results in our investigation of the classSPα,β.Lemma 3.1 (see [12]).
LetF and G be univalent convex functions in 𝕌.
Then the Hadamard product F*G is also univalent convex in 𝕌.Lemma 3.2 (see [13]).
LetF and G be univalent convex functions in 𝕌.
Also let f≺F and g≺G.
Then f*g≺F*G.Lemma 3.3 (see [12]).
Let each of the functionsf and g be univalent starlike of order 1/2.
Then, for every function F∈𝒜,(3.1)f(z)*g(z)F(z)f(z)*g(z)∈CH¯{F(𝕌)},(z∈𝕌),where CH¯ denotes the closed convex hull.Theorem 3.4.
If(0≤μ<α≤1) and 0≤β≤1,
then(3.2)SPα,β⊂SPμ,β.Proof.
Letf∈SPα,β.
Then(3.3)Θμ,βf=L(2,2−μ)Ωβf=L(2−α,2−μ)Θα,βf=φ(2−α,2−μ;z)*Θα,βf,z(Θμ,βf)′=L(2,1)L(2−α,2−μ)Θα,βf=φ(2−α,2−μ;z)*{z(Θα,βf)′}.Also it is known that (cf.
[14])(3.4)φ(2−α,2−μ;z)∈S*(12).Since ℛ is a convex region, using Lemma 3.3, we
get(3.5)z(Θμ,βf)′Θμ,βf=φ(2−α,2−μ;z)*(z(Θα,βf)′/Θα,βf)Θα,βfφ(2−α,2−μ;z)*Θα,βf∈ℛ.Thus, f∈SPμ,β.
This completes the proof of Theorem 3.4.Corollary 3.5.
Let0<α<1 and 0≤β≤1.
Then(3.6)SP1,1⊂SPα,1⊂(SP0,1≡UCV)⊂(SPα,0≡SPα)⊂(SP0,0≡SP),SP1,β⊂SPα,β⊂SPβ.In particular, the functions in SPα,β are parabolic starlike and they are uniformly
convex when β=1.Corollary 3.6.
Let0≤λ<α≤1 and 0≤μ<β≤1.
Then SPα,β⊂SPλ,μ.It can be verified that the Riemann mapq of 𝕌 onto the region ℛ,
satisfying q(0)=1 and q′(0)>0,
is given by(3.7)q(z)=1+2π2(log1+z1−z)2=1+8π2∑n=1∞(1n∑k=0n−112k+1)zn=∑n=0∞Bnzn=1+8π2(z+23z2+2345z3+44105z4+⋯),(z∈𝕌).We define the functionH by(3.8)H(z):=1z{L(2,2−α)L(2,2−β)zexp(∫0zq(s)−1sds)},(z∈𝕌).Theorem 3.7.
Letα,β∈[0,1) and let H(z) be defined by (3.8). Then H(z) is a convex univalent function. Furthermore,
if f∈SPα,β,
then(3.9)f(z)z≺H(z).Proof.
We first note that(3.10)H(z)=φ(2,2−α;z)z*φ(2,2−β;z)z*exp(∫0zq(s)−1sds),(z∈𝕌),where each member of the
Hadamard product in (3.10) is known to be a convex univalent function (cf.
[2, 14]). Therefore, by
Lemma 3.1, H(z) is a univalent convex function. Next, if f∈SPα,λ,
then(3.11)z(Θα,βf(z))′Θα,βf(z)≺q(z).Thus, there exists a function ω satisfying the Schwarz Lemma such
that(3.12)Θα,βf(z)z=exp(∫0zq(ω(s))−1sds),(z∈𝕌).Since q(z)−1 is a univalent convex function, a result of
[15] (see also
[16, page 50])
yields(3.13)Θα,βf(z)z≺exp(∫0zq(s)−1sds).It now follows from a known
result of [14, page
508, Theorem 2] that(3.14)f(z)z≺H(z).The proof of Theorem 3.7 is
evidently completed.Remark 3.8.
(i) Lettingα or β equal to zero in Theorem 3.7, we immediately
obtain a subordination result due to Srivastava and Mishra (see [9]).
(ii) Takingα=1,β=0 in Theorem 3.7, we get a result of [2,
page 169, Theorem 3].Theorem 3.9.
Letα,β∈[0,1).
If f∈SPα,β,
then(3.15)H(−r)≤|f(z)z|≤H(r),(|z|=r),(3.16)|arg(f(z)z)|≤maxθ∈[0,2π]{arg(H(reiθ))},(z=reiθ),where H(z) is defined by (3.8). Equality holds true in
(3.15) and (3.16) for some z≠0 if and only if f is a rotation of zH(z).Proof.
Letf∈SPα,β.
Then, by Theorem 3.7 and the Lindelöf principle of subordination, we
get(3.17)inf|z|=rRe{H(z)}≤inf|z|≤rRe{f(z)z}≤sup|z|≤rRe{f(z)z}≤sup|z|≤r|f(z)z|≤sup|z|≤rRe{H(z)}.Since H(z) is a univalent convex function and has real
coefficients, H(𝕌) is a convex region symmetric with respect to
real axis. Hence,(3.18)inf|z|≤rRe{H(z)}=inf−r≤x≤rH(x)=H(−r),sup|z|≤rRe{H(z)}=sup−r≤x≤rH(x)=H(r).Thus, (3.17) gives the assertion
(3.15) of Theorem 3.9. Also, we readily have the assertion
(3.16) of
Theorem 3.9.
The sharpness in (3.15) and
(3.16) is also a consequence of the principle of
subordination. This completes the proof of Theorem 3.9.Corollary 3.10.
Letf∈SPα,β,
where α,β∈[0,1].
Then,(3.19){w:|w|≤H(−1)}⊆f(𝕌).The result is sharp.Remark 3.11.
(i) Lettingα or β equal to zero in Theorem 3.9, we obtain a result
due to Srivastava and Mishra (see [9]).
(ii) Takingα=1,β=0 in Theorem 3.9, we get a result of [2, page 170,
Corollary 3].Next, we investigate characterization forf to be in the class SPα,β∩𝒯.
We need first the following lemma.Lemma 3.12.
IfΘα,βf∈𝒯, where α,β∈[0,1],
then(3.20)∑n=1∞(n+1)((n+1)!)2(2−α)n(2−β)nan≤1.Proof.
Suppose∑n=1∞((n+1)((n+1)!)2/(2−α)n(2−β)n)an>1.
We can write(3.21)∑n=1∞(n+1)((n+1)!)2(2−α)n(2−β)nan=1+ε,(ε>0).Then, there exists an integer N such that(3.22)∑n=1N(n+1)((n+1)!)2(2−α)n(2−β)nan>1+ε2.For (1/(1+ε/2))1/N<z<1,
we have(3.23)(Θα,βf)′(z)=1−∑n=1∞(n+1)((n+1)!)2(2−α)n(2−β)nanzn≤1−∑n=1N(n+1)((n+1)!)2(2−α)n(2−β)nanzn≤1−zN∑n=1N(n+1)((n+1)!)2(2−α)n(2−β)nan<1−zN(1+ε2)<0.Since (Θα,βf)′(0)=1>0,
there exists a real number z0, 0<z0<1,
such that (Θα,βf)′(z0)=0.
Hence, Θα,βf is not univalent.Theorem 3.13.
Letα,β∈[0,1].
Then, a function f∈SPα,β∩𝒯 if and only if(3.24)∑n=1∞(2n+1)((n+1)!)2(2−α)n(2−β)nan<1.Proof.
First, consider(3.25)|z(Θα,βf)′(z)Θα,βf(z)−1|−Re{z(Θα,βf)′(z)Θα,βf(z)−1}≤2|z(Θα,βf)′(z)Θα,βf(z)−1|≤∑n=1∞(2n((n+1)!)2/(2−α)n(2−β)n)|an||z|n+11−∑n=1∞(((n+1)!)2/(2−α)n(2−β)n)|an||z|n+1≤∑n=1∞(2n((n+1)!)2/(2−α)n(2−β)n)an1−∑n=1∞(((n+1)!)2/(2−α)n(2−β)n)an,where 1−∑n=1∞(((n+1)!)2/(2−α)n(2−β)n)|an||z|n+1>0.
Hence, if (3.24) holds, then the above expression is less than 1,
and consequently(3.26)|z(Θα,βf)′(z)Θα,βf(z)−1|<Re{z(Θα,βf)′(z)Θα,βf(z)}.Conversely, if f∈SPα,β∩𝒯 and z is real, we get(3.27)1−∑n=1∞((n+1)((n+1)!)2/(2−α)n(2−β)n)anzn1−∑n=1∞(((n+1)!)2/(2−α)n(2−β)n)anzn>∑n=1∞(n((n+1)!)2/(2−α)n(2−β)n)anzn1−∑n=1∞(((n+1)!)2/(2−α)n(2−β)n)anzn.Let z→1− along the real axis, then we
get(3.28)1−∑n=1∞((2n+1)((n+1)!)2/(2−α)n(2−β)n)an1−∑n=1∞(((n+1)!)2/(2−α)n(2−β)n)an>0.Using Lemma 3.12, we
have(3.29)1−∑n=1∞((n+1)!)2(2−α)n(2−β)nan>1−∑n=1∞(n+1)((n+1)!)2(2−α)n(2−β)nan≥0.Therefore, the denominator in
(3.28) is positive, and hence
(3.24) holds. This completes the proof of Theorem
3.13.Remark 3.14.
Theorem3.13 is sharp for functions
of the form(3.30)Θα,βfn(z)=z−(2−α)n(2−β)n(2n+1)((n+1)!)2zn+1,(n≥1).Corollary 3.15.
Ifα∈[0,1],
then(3.31)f∈SPα∩𝒯⟺∑n=1∞(n+1)!(2n+1)(2−α)nan<1.In particular, f∈SP∩𝒯 if and only if ∑n=1∞(2n+1)an<1 and f∈UCV∩𝒯 if and only if ∑n=1∞(2n+1)(n+1)an<1.Corollary 3.16.
Iff∈SPα,β∩𝒯, where α,β∈[0,1],
then(3.32)an≤(2−α)n(2−β)n(2n+1)((n+1)!)2,(n≥1).Corollary 3.17.
Iff∈SPα,β∩𝒯, where α,β∈[0,1],
then(3.33)|z|−(2−α)(2−β)12|z|2<|f(z)|<|z|+(2−α)(2−β)12|z|2,(z∈𝕌).Proof.
Letf∈SPα,β∩𝒯, where α,β∈[0,1].
Clearly,(3.34)12(2−α)(2−β)∑n=1∞an≤∑n=1∞(2n+1)((n+1)!)2(2−α)n(2−β)nan<1.Therefore,(3.35)|f(z)|≥|z|−|z|2∑n=1∞an>|z|−(2−α)(2−β)12|z|2,|f(z)|≤|z|+|z|2∑n=1∞an<|z|−(2−α)(2−β)12|z|2.Remark 3.18.
Under the hypothesis of Corollary3.17, f(z) lies in a disc centered at the origin with
radius r given by(3.36)r=1+(2−α)(2−β)12.In particular, we have
(i)
iff∈SP∩𝒯,
then f(z) lies in a disc centered at the origin with
radius 4/3;(ii)
iff∈SPα∩𝒯,
then f(z) lies in a disc centered at the origin with
radius (8−α)/6;(iii)
iff∈UCV∩𝒯,
then f(z) lies in a disc centered at the origin with
radius 7/6;(iv)
iff∈SP1,1∩𝒯,
then f(z) lies in a disc centered at the origin with
radius 13/12.Consequently, letΘα1,α2,…,αk be defined by(3.37)Θα1,α2,…,αkf(z)=(Ωα1∘Ωα2∘⋯∘Ωαk)f(z),αj∈[0,1].Then, f belonging to the corresponding class SP1,1,…,1︸k-times∩𝒯 implies f(z) lies in a disc centered at the origin with radius rk given by(3.38)rk=1+13×2k.


## 4. Class-Preserving Operators and Transforms

Theorem 4.1.
LetΘλ,βf be univalent starlike function of order 1/2  (f∈𝒜0) and let g∈SPα,β(λ≤α).
Then,(4.1)Θλ,βf*Θα,βg∈SPα,β.In particular, if Θα,βf is univalent starlike function of order 1/2 and g∈SPα,β,
then(4.2)Θα,βf*Θα,βg∈SPα,β.Proof.
LetΘλ,βf be univalent starlike function of order 1/2(f∈𝒜0) and g∈SPα,β(λ≤α).
By definition,(4.3)f∈S*(12),Θα,βg∈SP⊂S*(12).The commutative and associative
properties of the Hadamard product yield(4.4)z(Θλ,βf*Θα,βg)′=L(2,1)(Θλ,βf*Θα,βg)=Θλ,βf*L(2,1)Θα,βg=Θλ,βf*{z(Θα,βg)′}.Therefore, using Lemma 3.3, we
get(4.5)z(Θλ,βf*Θα,βg)′Θλ,βf*Θα,βg=Θλ,βf*((z(Θα,βg)′/Θα,βg)Θα,βg)Θλ,βf*Θα,βg∈ℛ,(z∈𝕌).This completes the proof of
Theorem 4.1.Takingα=1 and β=0 in Theorem 4.1, then we have the following
corollary.Corollary 4.2.
IfΩλf(z)∈S*(1/2) and g(z)∈UCV,
then Ωλf(z)*zg′(z)∈UCV.
In particular, if Ωλf(z)∈SP and g(z)∈UCV,
then Ωλf(z)*zg′(z)∈UCV.
Moreover, if Ωλf(z)∈UCV and g(z)∈UCV,
then Ωλf(z)*zg′(z)∈UCV.Takingλ=0 or 1, α=1, and β=0 in Theorem 4.1, then we have the following
corollary.Corollary 4.3.
Iff(z)∈S*(1/2) and g(z)∈UCV,
then f(z)*zg′(z)∈UCV.
In particular, if f(z)∈SP and g(z)∈UCV,
then f(z)*zg′(z)∈UCV.
Moreover, if f(z)∈UCV and g(z)∈UCV,
then f(z)*zg′(z)∈UCV.Corollary 4.4 (see [9]).
IfΩλf∈S*(1/2) and g∈SPμ(λ≤μ),
then(4.6)Ωλf*Ωμg∈SPμ.In particular, if Ωλf∈S*(1/2) and g∈SPλ,
then(4.7)Ωλf*Ωλg∈SPλ.Corollary 4.5 (see [3]).
Iff∈S*(1/2) and g∈SP,
then f*g∈SP.
In particular, if f∈SP and g∈SP,
then f*g∈SP.Theorem 4.6.
Letf∈SPβ and g∈SPα,β,
where α,β∈[0,1].
Then, f*g∈SPα,β.Proof.
The proof of Theorem4.6 is similar
to that of Theorem 4.1. Let f∈SPβ and g∈SPα,β.
We first note that(4.8)z(Θα,β(f*g)(z))′=Ωαg(z)*z(Ωβf(z))′,Θα,β(f*g)(z)=Ωαg(z)*Ωβf(z).Therefore, using Lemma 3.3, we
get(4.9)z(Θα,β(f*g)(z))′Θα,β(f*g)(z)=Ωαg(z)*(z(Ωβf(z))′/Ωβf(z))Ωβf(z)Ωαg(z)*Ωβf(z)∈ℛ,(z∈𝕌).Thus, f*g∈SPα,β.
This completes the proof of Theorem 4.6.Corollary 4.7.
The classSPα(0≤α≤1) is closed under convolution, and in particular
the classes SP and UCV are so.Theorem 4.8.
Letf∈SP1,β(0≤β≤1) and g∈SPα,β(0≤α≤1).
Then, f*g∈SP1,β.
In particular, the class SP1,β(0≤β≤1) is closed under convolution.Proof.
The proof of Theorem4.8 is similar
to that of Theorem 4.6. Let f∈SP1,β and g∈SPα,β.
We first note that(4.10)z(Θ1,β(f*g)(z))′=zf′(z)*z(Ωβg(z))′,Θ1,β(f*g)(z)=zf′(z)*Ωβg(z).Therefore, using Lemma 3.3, we
get(4.11)z(Θ1,β(f*g)(z))′Θ1,β(f*g)(z)=zf′(z)*(z(Ωβg(z))′/Ωβg(z))Ωβg(z)zf′(z)*Ωβg(z)∈ℛ,(z∈𝕌).Thus, f*g∈SP1,β.
By taking α=1,
we see that the class SP1,β(0≤β≤1) is closed under convolution. This completes
the proof of Theorem 4.8.Theorem 4.9.
Letfj∈SPα,β(j=1,2,…,n).
Also let(4.12)μj>0,∑j=1nμj=1.Define a function g by(4.13)Θα,βg=∏j=1n(Θα,βfj)μj.Then, g∈SPα,β.Proof.
Letfj∈SPα,β(j=1,2,…,n) and let g be defined by (4.13). Direct calculation
gives(4.14)|z(Θα,βg)′Θα,βg−1|=|∑j=1nμjz(Θα,βfj)′Θα,βfj−1|<∑j=1nμjRe(z(Θα,βfj)′Θα,βfj)=Re(z(Θα,βg)′Θα,βg).Thus, by Definition 1.3, g∈SPα,β.
This completes the proof of Theorem 4.9.Theorem 4.10.
Letf∈SPα,β,
where α,β∈[0,1].
Then, the function F(z) defined by the integral
transform(4.15)F(z):=c+1zc∫0ztc−1f(t)dt,(z∈𝕌;c>−1)is also in the class SPα,β.Proof.
We begin by noting
that(4.16)F(z)=L(c+1,c+2)f(z),z(Θα,βF(z))′=L(2,1)L(2,2−α)L(2,2−β)L(c+1,c+2)f(z)=L(c+1,c+2)(z(Θα,βf(z))′)=φ(c+1,c+2;z)*(z(Θα,βf(z))′).Using a result of Bernardi
[17], it can be
verified that(4.17)φ(c+1,c+2;z)∈S*(12).Also, by hypothesis, Θα,βf(z)∈SP⊂S*(1/2).
Thus, using Lemma 3.3, we get(4.18)z(Θα,βF(z))′Θα,βF(z)=φ(c+1,c+2;z)*(z(Θα,βf(z))′/Θα,βf(z))Θα,βf(z)φ(c+1,c+2;z)*Θα,βf(z)∈ℛ,which completes the proof of
Theorem 4.10.



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*Source: 597292-2009-02-12.xml*