# On Some Integral Operators for Certain Classes ofp-Valent Functions

Authors: Lokesh Vijayvargy; Pranay Goswami; Bushra Malik
Journal: International Journal of Mathematics and Mathematical Sciences
 (2011)

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## Abstract

We study some generalized integral operators for the classes ofp-valent
functions with bounded radius and boundary rotation. Our work generalizes many
previously known results. Many of our results are best possible.


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## Body

## 1. Introduction

LetAp denote the class of functions of the form
(1.1)f(z)=zp+∑n=p+1∞anzn,p∈N={1,2,…},
which are analytic in the open unit disc U={z:|z|<1}.Letf and g be analytic functions in U we say that f is subordinate to g, written as (1.2)f≺g;
if there exists a Schwarz function w(z) in U, with w(0)=0 and |w(z)|<1(z∈U), such that
(1.3)f(z)=g(w(z)).
In particular, when g is univalent, then the above subordination is equivalent to (1.4)f(0)=0,f(U)⊆g(U).
For functions f,g∈Ap, given by (1.5)f(z)=zp+∑n=p+1∞anzn,g(z)=zp+∑n=p+1∞bnzn,z∈U,
we define the Hadamard product (or convolution) of f and g by (1.6)(f*g)(z)=zp+∑n=p+1∞anbnzn,z∈U.
Janowski [1] defined the class P[A,B] as follows.Leth be a function, analytic in U, with h(0)=1. Then h is said to belong to the class P[A,B], -1≤B<A≤1, if and only if, for z∈U, (1.7)h(z)=1+Aw(z)1+Bw(z),wherew(z)isaSchwarzfunction.
Or equivalently, we can say that h∈P[A,B], -1≤B<A≤1, if and only if, (1.8)h(z)≺1+Az1+Bz,z∈U.
Geometrically, h(z) is in the class P[A,B], if and only if, h(0)=1 and the image of h(U) lies inside the open disc centered on the real axis with diameter end points, (1.9)D1=h(-1)=1-A1-B,D2=h(1)=1+A1+B,0<D1<1<D2.
Clearly P[A,B]⊂P((1-A)/(1-B)).In the recent paper, Noor [2] introduced the class Pk(α). We define it as follows. Let Pk(α),0≤α<p, be the class of functions p(z) with p(0)=1 and satisfying the property (1.10)p(z)=12∫02π1+(1-2α)ze-it1-ze-itdμ(t),
where μ(t) is a real-valued function of bounded variation on [0,2π] and ∫02πdμ(t)=2 and ∫02π|dμ(t)|≤k.The classesVk(α) and Rk(α) are related to the class Pk(α) and can be defined as (1.11)f∈Vk(α),iff(zf′(z))′pf′(z)∈Pk(α),z∈U,f∈Rk(α),iffzf′(z)pf(z)∈Pk(α),z∈U.
We define a class Pk[A,B] as follows.LetPk[A,B], k≥2, -1≤B<A≤1, denote the class of p-valent analytic functions h(z) that are represented by (1.12)h(z)=(k4+12)h1(z)-(k4-12)h2(z),z∈U,
where h1,h2∈P[A,B]. For A=1-2α(0≤α<p)and B=-1, it reduces to the class Pk(α) and P2(α)=P(α) is the class of p-valent analytic functions h(z)with Reh(z)>α, z∈U. Taking A=1, B=-1, and p=1, we have Pk[1,-1]=Pk(see [3]), and P2[1,-1]=P is the class of functions with positive real part.Definition 1.1.
A functionf, analytic in U, and given by (1.1) is said to be in the class Rk[A,B]; -1≤B<A≤1, k≥2, if and only if,
(1.13)zf′(z)pf(z)∈Pk[A,B],z∈U.
For p=1, Rk[A,B] is introduced and studied by Noor [4]. We note that
(1.14)Rk[A,B]⊂Rk(1-A1-B)⊂Rk,
where Rk is the class of functions with bounded radius rotation (see [5]). For k=2, we have
(1.15)R2[A,B]≡Sp*[A,B]⊂Sp*(1-A1-B)⊂Sp*,
where Sp* is the class of p-valent starlike functions. Similarly, we can define the class Vk[A,B] as follows.Definition 1.2.
A functionf, analytic in U, and given by (1.1) is said to be in the class Vk[A,B]; -1≤B<A≤1, k≥2, if and only if,
(1.16)(zf′(z))′pf′(z)∈Pk[A,B],z∈U.
It is clear that
(1.17)f∈Vk[A,B],iffzf′(z)p∈Rk[A,B],z∈U.
For p=1, Vk[A,B] is the class introduced and studied by Noor [4]. It is easy to see that,
(1.18)Vk[A,B]⊂Vk(1-A1-B)⊂Vk,
where Vk is the class of functions with bounded boundary rotation see [5]. Also
(1.19)V2[A,B]≡Cp[A,B]⊂Cp(1-A1-B)⊂Cp,
where Cp is the class of p-valent convex functions.Very recently, Frasin [6], introduced the following general integral operators for p-valent functions, (1.20)Fp(z)=∫0zptp-1(f1(t)tp)α1⋯(fn(t)tp)αndt,(1.21)Gp(z)=∫0zptp-1(f1′(t)ptp-1)α1⋯(fn′(t)ptp-1)αndt,whereαi∈C,z∈U.
Clearly, we may see that for p=1, these operators become the general integral operators (1.22)F1(z)=Fn(z),G1(z)=Fα1,α2,…,αn(z),
introduced and studied by Breaz and Breaz [7] and Breaz et al. [8], (see also [9, 10]).Forp=n=1, α1=α∈[0,1] in (1.20), we obtain the integral operator ∫0z(f(t)/t)αdt studied in [11] and for p=n=1, α1=δ∈ℂ, |δ|<1/4in (1.21), we obtain the integral operator ∫0z(f′(t))δdt, studied in [12].


## 2. Main Results

Lemma 2.1.
Letβ>0, β+γ>0, α∈[α0,1),with α0=max{(β-γ-1)/2β,-γ/β}. If
(2.1)p(z)+zp′(z)βp(z)+γ≺1+(1-2α)z1-z,
then
(2.2)p(z)≺Q(z)≺1+(1-2α)z1-z,
where Q(z)=1/βG(z)-γ/β,
(2.3)G(z)=∫01(1-z1-tz)2β(1-α)tβ+γ-1dt=2F1(2β(1-α),1,β+γ+1;z1-z),(2.4)ρ=ρ(α,β,γ)=β+γβ2F1(2β(1-α),1,β+γ+1;1/2)-γβ,2F1 denotes the Gauss hypergeometric function. From (2.2), we can deduce the sharp result p∈P(ρ), where ρ is defined in (2.4). This result is a special case of one given in [11].Proof.
To prove this Lemma we use Theorem 3.2j of [11, page 97]. Take h(z)=(1+(1-2α)z)/(1-z), 0≤α<1 and
(2.5)H(z)=βh(z)+γ=a+bz1-z,
where a=β+γ and b=β(1-2α)+γ.
SinceH is convex to apply Theorem 3.2j of [11, page 97] we only need to determine condition ReH(z)>0.
The range of|z|≤1 under H(z) is a half plane. In order to satisfy the required condition this half plane needs to lie in the right half plane. This requirement will be satisfied if ReH(-1)=ReH(i) and ReH(0)>ReH(-1)≥0.Or we can write it as(2.6)β(1-α)>0,βα+γ≥0.
When β>0, β+γ>0, these conditions imply that α∈[-γ/β,1), and if β+γ>1, then α∈[(β-γ-1)/2β,1). Hence all the conditions of Theorem 3.2j of [11, page 97] are satisfied for α∈[α0,1), with α0=max{(β-γ-1)/2β,-γ/β}, thus we have the required result.To show that the solutionQ(z) can be represented in terms of hypergeometric functions we take A=1-2α, B=-1, n=1 in Theorem 3.3d of [11, page 109].Lemma 2.2.
Letf∈Vk(α), 0≤α<p, k≥2. Then f∈Rk(ρ) in U, where
(2.7)ρ=ρ(α,p)=12F1(2p(1-α),1,p+1;1/2).
This result is sharp.Proof.
Let fork≥2, z∈U, we have
(2.8)zf′(z)pf(z)=h(z)=(k4+12)h1(z)-(k4-12)h2(z),
where h, hi are analytic in U with h(0)=1, hi(0)=1, i=1,2.
We define(2.9)ϕp(z)=z+∑n=1∞1p(1+(n-1))zn,z∈U.
By using (2.8), with convolution technique, see [13], we have
(2.10)ϕp(z)z*h(z)=(k4+12)(ϕp(z)z*h1(z))-(k4-12)(ϕp(z)z*h2(z)).
This implies that,
(2.11)h(z)+zh′(z)ph(z)=(k4+12)(h1(z)+zh1′(z)ph1(z))-(k4-12)(h2(z)+zh2′(z)ph2(z)).
Logarithmic differentiation of (2.8) yields,
(2.12)(zf′(z))′pf′(z)=h(z)+zh′(z)ph(z)=(k4+12)(h1(z)+zh1′(z)ph1(z))-(k4-12)(h2(z)+zh2′(z)ph2(z)).
Since (zf′(z))′/pf′(z)∈Pk(α), 0≤α<p, thus
(2.13)hi(z)+zhi′(z)phi(z)∈P(α),i=1,2.
By using Lemma 2.1 (for β=pand γ=0), we deduce that hi∈P(ρ), where ρ is given in (2.7). This estimate is best possible because of the best dominant property of function Q(z), where
(2.14)Q(z)=12F1(2p(1-α),1,p+1;z/(1-z)),z∈U.Forp=1, we have the sharp result proved in [14].We begin with the following theorem.Theorem 2.3.
(i) Letαi>0, fi∈Rk[A,B] for all i=1,2,…,n, and, ∑i=1nαi=1. Then the integral operator Fp∈Vk[A,B] in U, where -1≤B<A≤1, k≥2.
(ii) Letfi∈Rk(α), αi>0for all i=1,2,…,n with α=(1-A)/(1-B), k≥2. If ∑i=1nαi=1, then the integral operator Fp defined by (1.20) also belongs to the class Rk(ρ) in U, where ρ=ρ(α,p)is defined by (2.7). This result is sharp.Proof (i).
From (1.20), we can see that Fp∈Ap in U, and
(2.15)Fp′(z)=pzp-1[(f1(z)zp)α1⋯(fn(z)zp)αn].
Differentiating logarithmically and multiplying by z, we obtain,
(2.16)zFp′′(z)Fp′(z)=(p-1)+∑i=1nαi(zfi′(z)fi′(z)-p),z∈U.
Thus, we have
(2.17)1+zFp′′(z)Fp′(z)=∑i=1nαi(zfi′(z)fi′(z)),
or
(2.18)(zFp′(z))′pFp′(z)=∑i=1nαi(zfi′(z)pfi′(z))=(k4+12)(∑i=1nαipi(z))-(k4-12)(∑i=1nαihi(z)),
where hi,pi∈P[A,B], for all i=1,2,…,n.
SinceP[A,B] is a convex set, see [15], it follows that,(2.19)(zFp′(z))′pFp′(z)=(k4+12)H1(z)-(k4-12)H2(z),
where H1,H2∈P[A,B] and therefore,
(2.20)(zFp′(z))′pFp′(z)∈Pk[A,B],z∈U.
This proves the result.Substitutingp=1, in Theorem 2.3(i), we have the following corollary.Corollary 2.4.
Letαi>0, fi∈Rk[A,B] for all i=1,2,…,n, -1≤B<A≤1, k≥2. Then the integral operator Fp∈Vk[A,B] in U.Remark 2.5.
Lettingα1=α, α2=β, and n=2 in Corollary 2.4, we obtain a result due to Noor [4].Forn=1, α1=α=1, and f1=f in Theorem 2.3(i), we have the following.Corollary 2.6.
Letf∈Rk[A,B] in U, -1≤B<A≤1, k≥2. Then the integral operator ∫0zp(f(t)/t)dt∈Vk[A,B], z∈U.Proof (ii).
TakingA=1-2α, B=-1, with α=(1-A)/(1-B), we have for all i=1,2,…,n,
(2.21)fi∈Rk[1-2α,-1]=Rk(α),
using part (i) of Theorem 2.3, we have
(2.22)Fp∈Vk[1-2α,-1]=Vk(α)inU.
Now using Lemma 2.2 for Fp∈Vk(α), α=(1-A)/(1-B) implies that
(2.23)Fp∈Rk(ρ),whereρ=ρ(α,p)isdefinedin(2.7).
The sharpness of the result is clear from the function Q(z) defined by (2.14).Forp=1, we have the following corollary.Corollary 2.7.
Letαi>0, fi∈Rk(α) for all i=1,2,…,n, with α=(1-A)/(1-B) and A=1-2α, B=-1. Then the integral operator Fp defined by (1.20) also belongs to the class Rk(ρ) in U, where
(2.24)ρ=ρ(α)={2α-12-22(1-α),ifα≠1212ln⁡2,ifα=12.Remark 2.8.
Lettingα1=μ, α2=η, and n=2 in Corollary 2.7, we have the sharp result proved in [14].ForA=1, B=-1, and p=1, we have (2.25)fi∈Rk(0)impliesthatFp∈Vk(12)inU.Theorem 2.9.
(i) Letαi>0,fi∈Vk[A,B] for all i=1,2,…,n. If ∑i=1nαi=1, then the integral operator Gpdefined by (1.21), also belongs to the class Vk[A,B] in U, where -1≤B<A≤1, k≥2.
(ii) Let forαi>0, ∑i=1nαi=1 and fi∈Vk(α), for all i=1,2,…,n with 0≤α<p, α=(1-A)/(1-B), k≥2. Then the integral operator Gp∈Rk(ρ) in U, where ρ=ρ(α,p)is defined by (2.7). This result is sharp.Proof (i).
From definition (1.20), we have
(2.26)1+zGp′′(z)Gp′(z)=p+∑i=1nαi(zfi′′(z)fi′(z)-p+1)=∑i=1nαi(zfi′(z))′fi′(z),
or
(2.27)(zGp′(z))′pGp′(z)=∑i=1nαi(zfi′(z))′pfi′(z)=(k4+12)(∑i=1nαipi(z))-(k4-12)(∑i=1nαihi(z)),
where hi,pi∈P[A,B], for all i=1,2,…,n.
SinceP[A,B] is a convex set, see [15], it follows that,(2.28)(zGp′(z))′pGp′(z)=(k4+12)H1(z)-(k4-12)H2(z),z∈U,
where H1,H2∈P[A,B] and therefore,
(2.29)(zGp′(z))′pGp′(z)∈Pk[A,B]inU.
This implies that Gp∈Vk[A,B].Lettingp=1 in Theorem 2.9(i), we have the following corollary.Corollary 2.10.
Letαi>0, fi∈Vk[A,B] for all i=1,2,…,n and -1≤B<A≤1, k≥2. If ∑i=1nαi=1, then Gp∈Vk[A,B] in U.Proof (ii).
TakingA=1-2α, B=-1, we have for all i=1,2,…,n(2.30)fi∈Vk[1-2α,-1]=Vk(α),whereα=1-A1-B.
Now using part (i) of Theorem 2.9, we have
(2.31)Gp∈Vk[1-2α,-1]=Vk(α)inU.
Now using Lemma 2.2, for α=(1-A)/(1-B), we have
(2.32)Gp∈Vk(α)impliesthatGp∈Rk(ρ),inU,whereρ=ρ(α,p)isdefinedin(2.7).
The sharpness of the result is clear from the function Q(z) defined by (2.14).Forp=1, we have the following corollary.Corollary 2.11.
(i) Letαi>0, fi∈Vk(α), i=1,2,…,n, with α=(1-A)/(1-B) andA=1-2α, B=-1. Then Gp∈Rk(ρ) in U, where ρ=ρ(α) and defined in (2.24).
Also forA=1, B=-1, we have.
(ii) Iffi∈Vk(0) for all i=1,2,…,n, then Gp∈Rk(1/2) in U.



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