# Some Properties of Fractional Calculus and Linear Operators Associated with Certain Subclass of Multivalent Functions

Authors: Sh. Khosravianarab; S. R. Kulkarni; O. P. Ahuja
Journal: International Journal of Mathematics and Mathematical Sciences
 (2009)

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## Abstract

We investigate several distortion inequalities involving fractional calculus,
Ruscheweyh derivatives, and some well-known integral operators. In special cases, the results
presented in this paper provide new approaches to several previously known results.


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## Body

## 1. Introduction

Let𝒜(p) denote the class of functions f 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 disc 𝒰={z∈ℂ:|z|<1}. Also, let 𝒦(p) denote the subclass of 𝒜(p) consisting of all functions f of the form
(1.2)f(z)=zp-∑k=1∞ap+kzp+k(p∈ℕ={1,2,3,…},ap+k≥0),
For functions f,g∈𝒜(p), given by
(1.3)f(z):=zp+∑k=1∞ap+kzp+k,g(z):=zp+∑k=1∞bp+kzp+k,
the Hadamard product (or convolution) of f and g is defined by
(1.4)(f*g)(z):=zp+∑k=1∞ap+kbp+kzp+k.
The Ruscheweyh derivative of f of order δ+p-1 is defined by
(1.5)Dδ+p-1f(z):=zp(1-z)δ+p*f(z)=zp+∑k=1∞(δ+p)k(1)kap+kzp+k,
where f is given by (1.1) and δ>-p. The Ruscheweyh derivative Dδ+p-1 has been studied by several authors; for example, see [1, 2].Forβ<1, γ≥0, p∈ℕ, and δ>-p, let 𝒯δ,p,γ(β) consist of functions f∈𝒦(p) so that
(1.6)ℜ{eiη((1-γ)Dδ+p-1f(z)zp+γ(Dδ+p-1f(z))′pzp-1-β)}>0,z∈𝒰,
for some η∈ℝ. In [3], the authors obtained four containment results for the class 𝒯δ,p,γ(β). We denote 𝒯0,1,γ(β)=𝒯γ(β). The class 𝒯γ(β) was studied by Swaminathan [4–6], Barnard et al. [7], Kim and  Rønning [8], and others.In the present paper, we investigate several distortion inequalities involving fractional calculus, Ruscheweyh derivative, and some well-known integral operators defined on the class𝒯δ,p,γ(β). In special cases, the results presented here provide new approaches to some previously known results.Remark 1.1.
Throughout this section, we assume thatδ+p≥1.


## 2. Definitions and Lemmas

For the functionf given by (1.2), we define
(2.1)qδ,p,γ(z)=(1-γ)Dδ+p-1f(z)zp+γ(Dδ+p-1f(z))′pzp-1.
It is easy to verify that
(2.2)qδ,p,γ(z)=1-∑k=1∞(δ+p)k(p+kγ)p(1)kap+kzk,ap+k≥0.Lemma 2.1.
Let the functionf be given by (1.2). Then f∈𝒯δ,p,γ(β) if and only if
(2.3)∑k=1∞(δ+p)k(p+kγ)p(1)kap+k≤12(|1+eiη(1-β)|-|1-eiη(1-β)|)
for some η∈ℝ.Proof.
Using the fact thatRe(ω)≥0 if and only if |1+ω|≥|1-ω|, it suffices to show that
(2.4)|1+eiη(qδ,p,γ(z)-β)|-|1-eiη(qδ,p,γ(z)-β)|≥0,
where qδ,p,γ(z) is defined by (2.1). Letting
(2.5)𝒜k=(δ+p)k(p+kγ)p(1)k,
and assuming (2.3), we obtain
(2.6)|1+eiη(qδ,p,γ(z)-β)|-|1-eiη(qδ,p,γ(z)-β)|=|1+eiη(1-β)-eiη∑k=1∞Akap+kzk|-|1-eiη(1-β)+eiη∑k=1∞Akap+kzk|≥|1+eiη(1-β)|-|eiη||∑k=1∞Akap+kzk|-|1+eiη(1-β)|-|eiη||∑k=1∞Akap+kzk|≥|1+eiη(1-β)|-|1-eiη(1-β)|-2∑k=1∞Akap+k≥0,
where z∈∂𝒰={z;z∈ℂand|z|=1}. By (2.3), the desired inequality (2.4) follows at once. Conversely, if f∈𝒯δ,p,γ(β), then
(2.7)Re{eiη(qδ,p,γ(z)-β)}>0,
or, equivalently (2.4). This yields
(2.8)|1+eiη(1-∑k=1∞Akap+kzk-β)|≥|1-eiη(1-∑k=1∞Akap+kzk-β)|,
which implies that
(2.9)|1+eiη(1-β)-eiη∑k=1∞Akap+kzk|≥|1-eiη(1-β)+eiη∑k=1∞Akap+kzk|.
Squaring the above inequality, choosing the value of z on the half line z=re-iθ(0≤r<1), and letting r→1- through this line, we obtain
(2.10)|1+eiη(1-β)|2+|eiη|2|∑k=1∞Akap+kzk|2-2|1+eiη(1-β)||∑k=1∞Akap+kzk|≥|1-eiη(1-β)|2+|eiη|2|∑k=1∞Akap+kzk|2+2|1-eiη(1-β)||∑k=1∞Akap+kzk|.
Hence we get
(2.11)2|∑k=1∞Akap+kzk|(|1-eiη(1-β)|+|1+eiη(1-β)|)≤|1+eiη(1-β)|2-|1-eiη(1-β)|2=(|1+eiη(1-β)|-|1-eiη(1-β)|)(|1+eiη(1-β)|+|1-eiη(1-β)|),
which reduces to
(2.12)2∑k=1∞Akap+k≤|1+eiη(1-β)|-|1-eiη(1-β)|.
So the desired inequality (2.3) follows upon using (2.5).Settingη=δ=0 and p=1 in Lemma 2.1, we get the following result.Corollary 2.2 ([5, Theorem  2.4]).
Letf(z) be of the form (1.2). Then necessary and sufficient condition for f to be in 𝒯γ(β) is
(2.13)∑k=2∞[1+γ(k-1)]ak≤1-β,ak≥0.Throughout this paper, we define(2.14)Eη,β:=|1+eiη(1-β)|-|1-eiη(1-β)|.As an immediate consequence of Lemma2.1, we have the following corollary.Corollary 2.3.
Let the functionf be defined by (1.2). If f∈𝒯δ,p,γ(β), then
(2.15)ap+k≤p(1)kEη,β2(δ+p)k(p+kγ)
for some η∈ℝ.Letα1,α2,…,αp and β1,β2,…,βq (p,q∈ℕ∪{0}, p≤q+1) be complex numbers such that βk≠0,-1,-2,…for k∈{1,2,…,q}. The generalized hypergeometric function pFq is given by
(2.16)pFq(α1,…,αp;β1,…,βq;z):=∑n=0∞(α1)n(α2)n⋯(αp)n(β1)n(β2)n⋯(βq)nznn!,z∈𝒰,
where (x)n denotes the Pochhammer symbol defined by
(2.17)(x)n=Γ(x+n)Γ(x)=x(x+1)(x+2)⋯(x+n-1),forn∈ℕ,(x)0=1.
The operator pFq has recently been studied by several authors; for example, [3, 5]. For p=q+1=2, the above series give rise to the Gaussian hypergeometric series F(a,b;c;z).In [9], Hohlov introduced the convolution operator Ha,b;c by
(2.18)Ha,b;c(f)(z):=zF(a,b;c;z)*f(z),f∈𝔸(1).
Motivated by the operator Ha,b;c, the authors in [3] defined the convolution operators Gfp and Ha,b;cp,d as follows:
(2.19)Gfp(a,b;z):=G(z):=(∑k=1∞(1+a)(1+b)(k+a)(k+b)zk+p-1)*f(z),(a>-1,b>-1),(2.20)Ha,b;cp,d(f)(z)=H(z)=zp3F2(a,b,1+pd;pd,c;z)*f(z),
where f∈𝒜(p), p∈ℕ, and d≥0. For p=1, the operator Gf1(a,b;z) was introduced in [7].In Section3, we will make use of the following well-known fractional calculus operators Dz-μ, Dzμ, and Dzn+μ. For an analytic function f defined in a simply connected region of the complex z-plane containing the origin, these operators are defined as follows (See [1, 10]):
(2.21)Dz-μf(z)=1Γ(μ)∫0zf(t)(z-t)1-μdt(μ>0),
where multiplicity of (z-t)μ-1 is removed by requiring log(z-t) to be real when z-t>0;
(2.22)Dzμf(z)=1Γ(1-μ)∫0zf(t)(z-t)μdt(0≤μ<1),
where the multiplicity of (z-t)-μ is removed, as in the definition of Dz-μf(z);
(2.23)Dzn+μf(z):=dndznDzμf(z)(0≤μ<1;n∈ℕ0=ℕ∪{0}).
By virtue of (2.21), (2.22), (2.23) and in terms of Gamma function, it is wellknown (see for details [11]) that(2.24)Dz-μzk=Γ(k+1)Γ(k+μ+1)zk+μ(k∈ℕ,μ>0),Dzμzk=Γ(k+1)Γ(k-μ+1)zk-μ(k∈ℕ,0≤μ<1),Dzq+μzk=dqdzqDzμzk=Γ(k+1)Γ(k-q-μ+1)zk-(q+μ),
where q∈ℕ0, k∈ℕ, 0≤μ<1, and q≤k for μ=0.In Section4, we will investigate the integral operator Jδ,p defined by
(2.25)(Jδ,pf)(z)=δ+pzp∫0ztδ-1f(t)dt,
where f∈𝒜(p), δ>-p, and p∈ℕ. For p=1 and δ=0, the operator was first defined by Bernardi [12]. Later on several authors studied the operator Jδ,p; for example, see [1, 5].


## 3. Distortion Inequalities of Convolution Operators

Theorem 3.1.
Let the functionf defined by (1.2) be in the class 𝒯δ,p,γ(β). Then
(3.1)|qδ,p,γ(z)|≥1-12Eη,β|z|,|qδ,p,γ(z)|≤1+12Eη,β|z|
for some η∈ℝ. Here, qδ,p,γ(z) and Eη,β are defined, respectively, by (2.1) and (2.14).Proof.
From (2.2), we have
(3.2)|qδ,p,γ(z)|≥1-|∑k=1∞(δ+p)k(p+kγ)p(1)kap+kzk|≥1-∑k=1∞(δ+p)k(p+kγ)p(1)kap+k|z|k≥1-|z|∑k=1∞(δ+p)k(p+kγ)p(1)kap+k.
Making use of Lemma 2.1, we get
(3.3)|qδ,p,γ(z)|≥1-12Eη,β|z|
for some η∈ℝ. Similarly,
(3.4)|qδ,p,γ(z)|≤1+|z|∑k=1∞(δ+p)k(1+(kγ/p))(1)kap+k≤1+12Eη,β|z|
for some η∈ℝ. This completes the proof.We next obtain distortion inequalities for the fractional operatersDzμ and Dz-μ.Theorem 3.2.
Supposeμ≤(1+k+p)/(k+b+2) and p<b+1. If f∈𝒯δ,p,γ(β), then for some η∈ℝ, one has
(3.5)|DzμG(z)|≥Γ(p+1)Γ(p-μ+1)|z|p-μ(1-p(1+a)(1+b)(1+p)2(2+a)(2+b)(1-μ+p)(δ+p)(γ+p)|z|Eη,β),|DzμG(z)|≤Γ(p+1)Γ(p-μ+1)|z|p-μ(1+p(1+a)(1+b)(1+p)2(2+a)(2+b)(1-μ+p)(δ+p)(γ+p)|z|Eη,β),
where 0≤μ<1, z∈𝒰, p∈ℕ, and the operator G(z):=Gfp(a,b;z) was defined by (2.19).Proof.
By using (2.19), we deduce that
(3.6)DzμG(z)=Γ(p+1)Γ(p-μ+1)zp-μ-∑k=1∞(1+a)(1+b)Γ(k+p+1)(k+a+1)(k+b+1)Γ(k+p-μ+1)ap+kzp+k-μ.
Then
(3.7)Γ(p-μ+1)Γ(p+1)zμ-pDzμG(z)=1-∑k=1∞θ(k)ap+kzk,
where
(3.8)θ(k)=(1+a)(1+b)Γ(k+p+1)Γ(p-μ+1)(k+a+1)(k+b+1)Γ(k+p-μ+1)Γ(p+1)(k,p∈ℕ,0≤μ<1).
Since θ(k) is a decreasing function of k, when μ≤(1+k+p)/(k+b+2), then
(3.9)0<θ(k)≤θ(1)=(1+a)(1+b)(1+p)(2+a)(2+b)(1-μ+p).
Also, according to Lemma 2.1 and δ+p≥1, we have
(3.10)1p(δ+p)(p+γ)∑k=1∞ap+k=Γ(δ+p+1)(1+(γ/p))Γ(δ+p)∑k=1∞ap+k≤∑k=1∞(δ+p)k(1+(kγ/p))(1)kap+k≤12Eη,β
for some η∈ℝ. Then
(3.11)∑k=1∞ap+k≤p2(δ+p)(p+γ)Eη,β
for some η∈ℝ. From (3.7) and (3.9), we obtain
(3.12)Γ(p-μ+1)Γ(p+1)zμ-pDzμG(z)≥1-θ(1)∑k=1∞ap+kzk.
In view of (3.11), we conclude that
(3.13)|Γ(p-μ+1)Γ(p+1)zμ-pDzμG(z)|≥1-p(1+a)(1+b)(1+p)2(2+a)(2+b)(1-μ+p)(δ+p)(γ+p)|z|Eη,β
for some η∈ℝ, and
(3.14)|Γ(p-μ+1)Γ(p+1)zμ-pDzμG(z)|≤1+p(1+a)(1+b)(1+p)2(2+a)(2+b)(1-μ+p)(δ+p)(γ+p)|z|Eη,β
for some η∈ℝ, which yield (3.5).By lettingp=1 and b=a-1>0 in Theorem 3.2, we deduce the following consequence.Corollary 3.3.
Iff∈𝒯δ,1,b(β), then for 0≤μ<1, z∈𝒰, δ≥0 and some η∈ℝ(3.15)|DzμG(z)|≥1Γ(2-μ)|z|1-μ(1-1(3+b)(2-μ)(1+δ)|z|Eη,β),|DzμG(z)|≤1Γ(2-μ)|z|1-μ(1+1(3+b)(2-μ)(1+δ)|z|Eη,β).Theorem 3.4.
Letμ>0, z∈𝒰 and p∈ℕ. If f∈𝒯δ,p,γ(β), then
(3.16)|Dz-μG(z)|≥Γ(p+1)Γ(p+μ+1)|z|p+μ(1-p(1+a)(1+b)(1+p)2(2+a)(2+b)(1-μ+p)(δ+p)(γ+p)|z|Eη,β),(3.17)|Dz-μG(z)|≤Γ(p+1)Γ(p+μ+1)|z|p+μ(1+p(1+a)(1+b)(1+p)2(2+a)(2+b)(1-μ+p)(δ+p)(γ+p)|z|Eη,β)
for some η∈ℝ. The operator G(z):=Gfp(a,b;z) was defined by (2.19).Proof.
In view of (2.19) and (2.22), we have
(3.18)Γ(p+μ+1)Γ(p+1)z-(μ+p)Dz-μG(z)=1-∑k=1∞τ(k)ap+kzk,
where
(3.19)τ(k)=(1+a)(1+b)Γ(k+p+1)Γ(p+μ+1)(k+a+1)(k+b+1)Γ(k+p+μ+1)Γ(p+1).
Since τ is a decreasing function of k, it follows that
(3.20)0<τ(k)≤τ(1)=(1+a)(1+b)(1+p)(2+a)(2+b)(1+μ+p).
By using (3.11), (3.18), and (3.20), we get
(3.21)|Γ(p+μ+1)Γ(p+1)z-(μ+p)Dz-μG(z)|≥1-p(1+a)(1+b)(1+p)2(2+a)(2+b)(1+μ+p)(δ+p)(γ+p)|z|Eη,β,|Γ(p+μ+1)Γ(p+1)z-(μ+p)Dz-μG(z)|≤1+p(1+a)(1+b)(1+p)2(2+a)(2+b)(1+μ+p)(δ+p)(γ+p)|z|Eη,β
for some η∈ℝ. The last two inequalities yield (3.16) and (3.17), respectively.Lettingδ=0, p=1, and a=b+1=μ+2 in Theorem 3.4, we get the following result.Corollary 3.5.
LetGf1(μ+2,μ+1;z) be defined by (2.19). If f∈𝒯γ(β), then
(3.22)|Dz-μGf1(μ+2,μ+1;z)|≥1Γ(μ+2)|z|μ+1(1-1(μ+4)(γ+1)|z|Eη,β),|Dz-μGf1(μ+2,μ+1;z)|≤1Γ(μ+2)|z|μ+1(1+1(μ+4)(γ+1)|z|Eη,β)
for some η∈ℝ, μ>0, z∈𝒰, and p∈ℕ.We next prove the distortion theorems involving fractional calculus and generalized convolution operator defined by (2.20).Theorem 3.6.
Suppose0≤μ<1, z∈𝒰, p∈ℕ, and η∈ℝ. Also, let a≤1, b≤p-μ+1, and c≥p+1. If f∈𝒯δ,p,γ(β), then
(3.23)|DzμH(f)(z)|≥Γ(p+1)Γ(p-μ+1)|z|p-μ(1-ab(p+1)2c(1-μ+p)(δ+p)|z|Eη,β),|DzμH(f)(z)|≤Γ(p+1)Γ(p-μ+1)|z|p-μ(1+ab(p+1)2c(1-μ+p)(δ+p)|z|Eη,β)
for some η∈ℝ. Here, the operator H(f)(z) is defined by (2.20).Proof.
By making use of (2.20), we have
(3.24)DzμH(f)(z)=Dzμ{zp-∑k=1∞(a)k(b)k(1+(p/γ))k(p/γ)k(c)k(1)kap+kzp+k}=Γ(p+1)Γ(p-μ+1)zp-μ-∑k=1∞(a)k(b)k(1+(p/γ))kΓ(p+k+1)(p/γ)k(c)k(1)kΓ(p+k-μ+1)ap+kzp+k-μ.
It is easy to verify that
(3.25)1+kγp=(1+(p/γ))k(p/γ)k.
This implies that
(3.26)Γ(p-μ+1)Γ(p+1)zμ-pDzμH(f)(z)=1-∑k=1∞λ(k)(1+kγp)ap+kzk,
where
(3.27)λ(k)=(a)k(b)kΓ(p+k+1)Γ(p-μ+1)(c)k(1)kΓ(p+k-μ+1)Γ(p+1).
Since λ is a decreasing function of k, when a≤1, c≥p+1 and b≤p-μ+1, we get
(3.28)0<λ(k)≤λ(1)=ab(p+1)c(p-μ+1).
From Lemma 2.1 and δ+p≥1, we obtain
(3.29)(δ+p)∑k=1∞(1+kγp)ap+k≤∑k=1∞(δ+p)k(1+(kγ/p))(1)kap+k≤12Eη,β
for some η∈ℝ. It follows from (3.26) and (3.28) that
(3.30)|Γ(p-μ+1)Γ(p+1)zμ-pDzμH(f)(z)|≥1-ab(p+1)2c(p-μ+1)(δ+p)|z|Eη,β,|Γ(p-μ+1)Γ(p+1)zμ-pDzμH(f)(z)|≤1+ab(p+1)2c(p-μ+1)(δ+p)|z|Eη,β
for some η∈ℝ, which yield (3.23).We state an obvious variant of Theorem3.6 as follows.Corollary 3.7.
Let the functionf defined by (1.2) be in the class 𝒯δ,p,γ(β). Also let a≤1, b≤p-μ+1, c≥p+1, and δ+p>1. Then
(3.31)|DzμH(f)(z)|≥Γ(p+1)Γ(p-μ+1)|z|p-μ(1-12|z|Eη,β),|DzμH(f)(z)|≤Γ(p+1)Γ(p-μ+1)|z|p-μ(1+12|z|Eη,β)
for some η∈ℝ, 0≤μ<1, z∈𝒰, and p∈ℕ.The proof of Theorem3.8 is much akin to that of Theorem 3.6, and so it is omitted here.Theorem 3.8.
Leta≤1, b≤p+μ+1 and c≥p+1. Also, let μ>0, p∈ℕ, and z∈𝒰. If f∈𝒯δ,p,γ(β), then
(3.32)|Dz-μH(f)(z)|≥Γ(p+1)Γ(p+μ+1)|z|p+μ(1-ab(p+1)2c(1+μ+p)(δ+p)|z|Eη,β),|Dz-μH(f)(z)|≤Γ(p+1)Γ(p+μ+1)|z|p+μ(1+ab(p+1)2c(1+μ+p)(δ+p)|z|Eη,β)
for some η∈ℝ.Next we prove the following.Theorem 3.9.
Leta≤1, b≤p-μ+1, and c≥p+1. Also, let 0≤μ<1, p∈ℕ, and z∈𝒰. If f∈𝒯δ,p,γ(β), then
(3.33)|Dzμ(Dδ+p-1H(f)(z))|≥Γ(p+1)Γ(p-μ+1)|z|p-μ(1-ab(p+1)2c(1-μ+p)|z|Eη,β),|Dzμ(Dδ+p-1H(f)(z))|≤Γ(p+1)Γ(p-μ+1)|z|p-μ(1+ab(p+1)2c(1-μ+p)|z|Eη,β)
for some η∈ℝ.Proof.
We have(3.34)Dδ+p-1H(f)(z)=zp-∑k=1∞Akzp+k,
where
(3.35)Ak=(δ+p)k(a)k(b)k(1+(p/γ))k(1)k(p/γ)k(c)k(1)kap+k.
Therefore
(3.36)Dzμ(Dδ+p-1H(f)(z))=Γ(p+1)Γ(p-μ+1)zp-μ-∑k=1∞AkΓ(p+k+1)Γ(p+k-μ+1)zp+k-μ.
So, from (3.25), we have
(3.37)Γ(p-μ+1)Γ(p+1)zμ-pDzμ(Dδ+p-1H(f)(z))=1-∑k=1∞(δ+p)k(p+kγ)p(1)kλ(k)ap+kzk,
where λ(k) is defined by (3.27). Since λ is a decreasing function of k, when a≤1,b≤p-μ+1 and c≥p+1, then
(3.38)0<λ(k)≤λ(1)=ab(p+1)c(p-μ+1).
From (3.37), (3.38), and Lemma 2.1, we find that
(3.39)|Γ(p-μ+1)Γ(p+1)zμ-pDzμ(Dδ+p-1H(f)(z))|≥1-ab(p+1)2c(1-μ+p)|z|Eη,β,|Γ(p-μ+1)Γ(p+1)zμ-pDzμ(Dδ+p-1H(f)(z))|≤1+ab(p+1)2c(1-μ+p)|z|Eη,β
for some η∈ℝ. The above inequalities lead us to the desired inequalities (3.33).The proof of the following theorem is similar to Theorem3.9, and so it is omitted here.Theorem 3.10.
Leta≤1, b≤p+μ+1, and c≥p+1. Also, let μ>0, z∈𝒰, and p∈ℕ. If f∈𝒯δ,p,γ(β), then
(3.40)|Dz-μ(Dδ+p-1H(f)(z))|≥Γ(p+1)Γ(p+μ+1)|z|p+μ(1-ab(p+1)2c(1+μ+p)|z|Eη,β),|Dz-μ(Dδ+p-1H(f)(z))|≤Γ(p+1)Γ(p+μ+1)|z|p+μ(1+ab(p+1)2c(1+μ+p)|z|Eη,β)
for some η∈ℝ.Upon settingδ=0 and p=1 in Theorems 3.6, 3.8, 3.9, and 3.10, we arrive at the following result.Corollary 3.11.
Leta≤1, c≥2, μ>0, z∈𝒰, and p∈ℕ. If f∈𝒯δ,p,γ(β), then
(3.41)|DzμH(f)(z)|≥1Γ(2-μ)|z|1-μ(1-abc(2-μ)|z|Eη,β),|DzμH(f)(z)|≤1Γ(2-μ)|z|1-μ(1+abc(2-μ)|z|Eη,β)
for some η∈ℝ and b≤2-μ. Furthermore
(3.42)|Dz-μH(f)(z)|≥1Γ(2+μ)|z|1+μ(1-abc(2+μ)|z|Eη,β),|Dz-μH(f)(z)|≤1Γ(2+μ)|z|1+μ(1+abc(2+μ)|z|Eη,β)
for some η∈ℝ and b≤2+μ.Remark 3.12.
Under the hypothesis of Corollary3.11, DzμH(f)(z) and Dz-μH(f)(z) are included in disks with its center at origin and radii r and R, respectively, given by
(3.43)r=1Γ(2-μ)(1+abc(2-μ)Eη,β)(0≤μ<1,b≤2-μ,forsomeη∈ℛ),R=1Γ(2+μ)(1+abc(2+μ)Eη,β)(μ>0,b≤2+μ,forsomeη∈ℛ).


## 4. Distortion Inequalities of Integral Operator

In this section, we obtain the distortion theorems involving the integral operatorJδ,p of functions in the class 𝒯δ,p,γ(β) and fractional calculus operator.Theorem 4.1.
Let0≤μ<1/(δ+p+1), z∈𝒰, and p∈ℕ. If f∈𝒯δ,p,γ(β), then
(4.1)|Dzμ(z-δ(Jδ,pf)(z))|≥1Γ(2-μ)|z|-μ(1-p2(δ+p+1)(p+γ)|z|Eη,β),|Dzμ(z-δ(Jδ,pf)(z))|≤1Γ(2-μ)|z|-μ(1+p2(δ+p+1)(p+γ)|z|Eη,β)
for some η∈ℝ. The operator (Jδ,pf)(z) is defined by (2.21).Proof.
Using the definition (2.25), for function f∈𝒦(p) of the form (1.2), we have
(4.2)(Jδ,pf)(z)=zδ-∑k=1∞δ+pp+k+δap+k.
So
(4.3)Dzμ(z-δ(Jδ,pf)(z))=1Γ(2-μ)z-μ-∑k=1∞(δ+p)Γ(k+1)(δ+p+k)Γ(k-μ+1)ap+kzk-μ.
Therefore, we obtain
(4.4)Γ(2-μ)zμDzμ(z-δ(Jδ,pf)(z))=1-∑k=1∞ϕ(k)ap+kzk,
where
(4.5)ϕ(k)=(δ+p)Γ(k+1)Γ(2-μ)(δ+p+k)Γ(k-μ+1).
Since ϕ(k) is a decreasing function of k, when μ<1/(δ+p+1), then
(4.6)0<ϕ(k)≤ϕ(1)=δ+pδ+p+1.
By using (3.11), (4.4), and (4.6), we get
(4.7)|Γ(2-μ)zμDzμ(z-δ(Jδ,pf)(z))|≥1-p2(δ+p+1)(p+γ)|z|Eη,β,|Γ(2-μ)zμDzμ(z-δ(Jδ,pf)(z))|≤1+p2(δ+p+1)(p+γ)|z|Eη,β
for some η∈ℝ, which prove the inequalities (4.1).The proof of the following theorem is similar to Theorem4.1, and so it is omitted here.Theorem 4.2.
Letμ>0, p∈ℕ, and z∈𝒰. If f∈𝒯δ,p,γ(β), then
(4.8)|Dz-μ(z-δ(Jδ,pf)(z))|≥1Γ(μ+1)|z|μ(1-p2(δ+p+1)(μ+1)(p+γ)|z|Eη,β),|Dz-μ(z-δ(Jδ,pf)(z))|≤1Γ(μ+1)|z|μ(1+p2(δ+p+1)(μ+1)(p+γ)|z|Eη,β)
for some η∈ℝ.



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*Source: 450632-2009-08-20.xml*