# Some Inclusion Relations Associated with Generalized Fractional Differintegral Operator

Authors: Amit Soni; Shashi Kant
Journal: International Journal of Analysis
 (2013)

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## Abstract

A generalized fractional differintegral operator is used to define some
new subclasses of analytic functions in the open unit disk𝕌. For each of these new function
classes, several inclusion relationships are established.


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## Body

## 1. Introduction and Definitions

Let𝔸 be the class of normalized functions f of the form
(1)f(z)=z+∑n=2∞‍anzn
which are analytic in the open unit disk 𝕌={z∈D,|z|<1}. If f∈𝔸 is given by (1) and g∈𝔸 given by g(z)=z+∑n=2∞bnzn in z∈𝕌, then the Hadamard product (or convolution) of f and g is defined by
(2)(f*g)(z)=z+∑n=2∞‍anbnzn.
Let Pk(α) denote the class of functions h(z) analytic in the unit disk 𝕌, satisfying the properties h(0)=1 and
(3)∫02π|ℝ(h(z)-α1-α)|dθ≤kπ(z=reiθ;0≤α<1;k≥2).
This class Pk(α) has been introduced in [1]. Note that, for α=0, we obtain the class Pk defined and studied in [2], and for k=2, we have the class P(α) of functions with positive real part greater than α. In particular,P(0) is the class P of functions with positive real part. From (3), we can easily deduce that h∈Pk(α) if and only if
(4)h(z)=(k4+12)h1(z)-(k4-12)h2(z),h1,h2∈P(α).
Following the recent investigations [3] (see also [4, 5]), we have the following subclasses:
(5)Rk(α)={f∈A:zf′(z)f(z)∈Pk(α),z∈𝕌},Vk(α)={f∈A:(zf′(z))′f′(z)∈Pk(α),z∈𝕌},Pk′(α)={f∈A:f′(z)∈Pk(α),z∈𝕌},Tk(β,α)={f∈A:g∈R2(α),zf′(z)g(z)∈Pk(β),z∈𝕌},Tk*(β,α)={f∈A:g∈V2(α),(zf′(z))′g′(z)∈Pk(β),z∈𝕌}.
We note that the class R2(α)=S*(α) and V2(α)=𝒦(α) are, respectively, the subclasses of 𝔸, consisting of functions which are starlike of order α and convex of order α in 𝕌. The class T2*(β,α)=C*(β,α) was considered by Noor [6], and T2*(0,0)=C* is the class of quasiconvex univalent functions which was first introduced and studied in [7]. It can be easily seen from the above definition that
(6)f(z)∈Vk(α)⟺zf′(z)∈Rk(α),(7)f(z)∈Tk*(β,α)⟺zf′(z)∈Tk(β,α).
We recall here the following family of generalized fractional integral operators due to Srivastava et al. [8] and generalized fractional derivative operators due to Raina and Nahar [9] (see also [10, 11]).Definition 1.
Letλ>0 and μ,η∈ℝ. Then the fractional integral operator ℐ0,zλ,μ,η is defined by
(8)ℐ0,zλ,μ,ηf(z)=z-λ-μΓ(λ)∫0z‍(z-ζ)λ-1F12(λ+μ,-η;λ;1-ζz)f(ζ)dζ,
where the function f(z) is analytic in a simply connected region of the complex z-plane containing the origin, with the order
(9)f(z)=O(|z|ε)(z→0;ε>max{0,μ-η}-1),
and the multiplicity (z-ζ)λ-1 is removed by requiring log(z-ζ) to be real when z-ζ>0.Definition 2.
Form-1≦λ<m and μ,η∈ℝ and m∈ℕ, the fractional derivative operator 𝒥0,zλ,μ,η is defined by
(10)𝒥0,zλ,μ,ηf(z)=dmdzm(zλ-μΓ(m-λ)dmdzmc×∫0z‍(z-ζ)m-λ-1F21(ζzμ-λ,m-η;m-λ;dmdzmc×∫0z‍(z-ζ)m-λ-1F21c1-ζz)f(ζ)dζzλ-μΓ(m-λ)),
where the multiplicity (z-ζ)m-λ is removed as in the above definition.The operatorsℐ0,zλ,μ,η and 𝒥0,zλ,μ,η include the Riemann-Liouville and Erdelyi-Kober operators of fractional calculus (see, e.g., [10]). Using the hypotheses of Definitions 1 and 2 the generalized fractional differintegral operator
(11)𝒮0,zλ,μ,η:𝔸→𝔸
is defined by (see also [12–15])
(12)𝒮0,zλ,μ,ηf(z)={Γ(2-μ)Γ(2+η-λ)Γ(2+η-μ)zμ𝒥0,zλ,μ,ηf(z),(0≤λ<η+2),Γ(2-μ)Γ(2+η-λ)Γ(2+η-μ)zμℐ0,z-λ,μ,ηf(z),(λ<0),(z∈𝕌,μ,η∈ℝ,min{-μ,η-λ,η-μ}>-2).
It is easily seen from (12) that for a function f of the form (1), we have
(13)𝒮0,zλ,μ,ηf(z)=z+Γ(2-μ)Γ(2+η-λ)Γ(2+η-μ)×∑n=2∞‍Γ(n+1)Γ(n+η-μ+1)Γ(n-μ+1)Γ(n+η-λ+1)anzn=z3F2(1,2,2+η-μ;2-μ,2+η-λ;z)*f(z).We note that the operator𝒮0,zλ,μ,η is a generalization of many other operators, for example,
(14)(i)𝒮0,z0,0,0f(z)=f(z),(ii)𝒮0,z1,1,1f(z)=𝒮0,z1,0,0f(z)=zf′(z),(iii)𝒮0,z2,1,1f(z)=zf′(z)+z2f′′(z),(iv)𝒮0,zλ,λ,ηf(z)=𝒮0,zλ,μ,0f(z),(v)𝒮0,z-λ,0,α-1f(z)=Ωαλf(z)(v)𝒮0,z-λ,0,α-1f(z)=(λ+αα)λzα(v)𝒮0,z-λ,0,α-1f(z)×∫0z‍ζα-1(1-ζz)λ-1f(ζ)dζ,
which is a multiplier transformation operator Ωαλ studied by Jung et al. [16],
(15)(vi)𝒮0,z-1,0,α-1f(z)=ℐαf(z)=α+1zα∫0z‍ζα-1f(ζ)dζ(α>-1)
which is the generalized Bernardi-Libra-Livingston integral operator. It can easily be verified from (12) that
(16)z(𝒮0,zλ,μ,ηf(z))′=(η-λ+1)𝒮0,zλ+1,μ,ηf(z)-(η-λ)𝒮0,zλ,μ,ηf(z).
For the generalized fractional differintegral operator 𝒮0,zλ,μ,η, we now define the following subclasses of 𝔸.Definition 3.
Letf(z)∈𝔸. Then f(z)∈Rλ,μ,η(k,α) if and only if 𝒮0,zλ,μ,ηf(z)∈Rk(α), for z∈𝕌.Definition 4.
Letf(z)∈𝔸. Then f(z)∈Vλ,μ,η(k,α) if and only if 𝒮0,zλ,μ,ηf(z)∈Vk(α), for z∈𝕌.Definition 5.
Letf(z)∈𝔸. Then f(z)∈Tλ,μ,η(k,β,α) if and only if 𝒮0,zλ,μ,ηf(z)∈Tk(β,α), for z∈𝕌.Definition 6.
Letf(z)∈𝔸. Then f(z)∈T*λ,μ,η(k,β,α) if and only if 𝒮0,zλ,μ,ηf(z)∈Tk*(β,α), for z∈𝕌.Definition 7.
Letf(z)∈𝔸. Then f(z)∈Pk′(λ,μ,η,α) if and only if 𝒮0,zλ,μ,ηf(z)∈Pk′(α), for z∈𝕌.In this paper we establish some inclusion relationships and some other interesting properties for these subclasses.


## 2. Main Inclusion Relationships

We recall first the following necessary lemma.Lemma 8 (see [17]).
Letu=u1+iu2 and v=v1+iv2, and let ϕ(u,v) be a complex-valued function satisfying the conditions:(i)
ϕ(u,v) is continuous in D⊂ℂ2,(ii)
(1,0)∈D and ℝϕ(1,0)>0,(iii)
ℝϕ(iu2,v1)≤0 whenever (iu2,v1)∈D and v1≤-(1/2)(1+u22).Ifh(z)=1+∑n=2∞‍cnzn is a function analytic in 𝕌 such that (h(z),zh′(z))∈D and ℝe(ϕ(h(z),zh′(z)))>0, for z∈𝕌, then ℝe(h(z))>0 for z∈𝕌.Our first main inclusion relationship is given by the theorem below.Theorem 9.
Letf∈𝔸,  0≤α<1,  η-λ>1/8, and  min{-μ,η-μ}>-2; then
(17)Rλ+1,μ,η(k,α)⊂Rλ,μ,η(k,α1),
where
(18)α1=2(2αη-2αλ+1)×(4(η-λ+α)2+4(η-λ-α)+9(2(η-λ-α)+1)c+4(η-λ+α)2+4(η-λ-α)+9)-1
and 0≤α<α1<1.Proof.
Letf∈Rλ+1,μ,η(k,α). Then upon setting
(19)z(𝒮0,zλ,μ,ηf(z))′𝒮0,zλ,μ,ηf(z)=p(z)=(k4+12)p1(z)-(k4-12)p2(z)(z∈𝕌),
we see that the function p(z) is analytic in 𝕌, with p(0)=1 in z∈𝕌. Using the identity (16) in (19) and differentiating with respect to z, we get
(20)z(𝒮0,zλ+1,μ,ηf(z))′𝒮0,zλ+1,μ,ηf(z)=(p(z)+zp′(z)p(z)+η-λ)∈Pk(α)(z∈𝕌).
Let
(21)ϕ(z)=∑j=1∞‍η-λ+jη-λ+1zj.
Then, by convolution technique, (see [18]), we have
(22)p(z)*ϕ(z)z=p(z)+zp′(z)p(z)+η-λ=(k4+12)(p1(z)*ϕ(z)z)-(k4-12)(p2(z)*ϕ(z)z)
and this implies that
(23)(pi(z)+zpi′(z)pi(z)+η-λ)∈P(α)(z∈𝕌,i=1,2).
We want to show that pi(z)∈P(α1), where α1 is given by (18), and this will show that p(z)∈Pk(α1) for z∈𝕌. Let
(24)pi(z)=(1-α1)hi(z)+α1(z∈𝕌,i=1,2).
Then in view of (23) and (24), we obtain, for z∈𝕌,  i=1,2(25)ℝ((1-α1)hi(z)+(α1-α)+(1-α1)zhi′(z)(1-α1)hi(z)+α1+η-λ)>0.
We now form a function ϕ(u,v) by choosing u=hi(z) and v=zhi′(z) in (25). Thus
(26)ϕ(u,v)=(1-α1)u+α1-α+(1-α1)v(1-α1)u+α1+η-λ.
We can easily see that the first two conditions of Lemma 8 are easily satisfied as ϕ(u,v) is continuous in D=ℂ-(-(α1+η-λ)/(1-α1))×ℂ,  (1,0)∈D, and ℝ{ϕ(1,0)}>0. Now for v1≤-(1/2)(1+u22) we obtain
(27)ℝ{ϕ(iu2,v1)}=ℝ((1-α1)iu2+α1-α+(1-α1)v1(1-α1)iu2+α1+η-λ)=α1-α+(1-α1)v1{α1+η-λ}(α1+η-λ)2+(1-α1)2u22≤α1-α-12(1-α1)(α1+η-λ)(1+u22)(α1+η-λ)2+(1-α1)2u22=A+Bu222C,
where A=(α1+η-λ){2(α1-α)(α1+η-λ)-(1-α1)},  B=(1-α1){2(α1-α)(1-α1)-(α1+η-λ)}, and C=(α1+η-λ)2+(1-α1)2u22>0.
We note thatℝ{ϕ(iu2,v1}≤0 if and only if A≤0 and B≤0. From A≤0, we obtain α1 as given by (18), and B≤0 gives us 0≤α1<1. This completes the proof.Theorem 10.
Letf∈𝔸,  0≤α<1,  η-λ>1/8, and min{-μ,η-μ}>-2; then
(28)Vλ+1,μ,η(k,α)⊂Vλ,μ,η(k,α1),
where α1 is given by (18).Proof.
To prove the inclusion relationship, we observe (in view of Theorem9) that
(29)f(z)∈Vλ+1,μ,η(k,α)⟺zf′(z)∈Rλ+1,μ,η(k,α)⟹zf′(z)∈Rλ,μ,η(k,α1)⟺f(z)∈Vλ,μ,η(k,α1),
which establishes Theorem 10.Theorem 11.
Letf∈𝔸,   0≤α,  β<1,  η-λ>1/8, and min{-μ,η-μ}>-2; then
(30)Tλ+1,μ,η(k,β,α)⊂Tλ,μ,η(k,β1,α1),
where α1 is given by (18) and β<β1<1 is defined in the proof.Proof.
Letf(z)∈Tλ+1,μ,η(k,β,α). Then there exists g(z)∈Rλ+1,μ,η(2,α) such that
(31)z(𝒮0,zλ+1,μ,ηf(z))′𝒮0,zλ+1,μ,ηg(z)∈Pk(β)(z∈𝕌,0≤β<1).
Since Rλ+1,μ,η(2,α)⊂Rλ,μ,η(2,α1), therefore we have
(32)z(𝒮0,zλ,μ,ηg(z))′𝒮0,zλ,μ,ηg(z)=(1-α1)q(z)+α1=(1-α1)(q1+ιq2)+α1,
where q(z)∈P. Next consider,
(33)z(𝒮0,zλ,μ,ηf(z))′𝒮0,zλ,μ,ηg(z)=(1-β1)h(z)+β1
and proceed on the same technique of Theorem 9 (see also [5]); we obtain f(z)∈Tλ,μ,η(k,β1,α1), where
(34)β1=(2β[((1-α1)q1+α1+η-λ)2+(1-α1)2q22]c+[(1-α1)q1+α1+η-λ][((1-α1)q1+α1+η-λ)2+(1-α1)2q22])×([((1-α1)q1+α1+η-λ)2+(1-α1)2q22]2[((1-α1)q1+α1+η-λ)2+(1-α1)2q22]c+[(1-α1)q1+α1+η-λ][((1-α1)q1+α1+η-λ)2+(1-α1)2q22])-1.In view of relation (7), an application of Theorem 11 is obtained as follows.Theorem 12.
Letf∈𝔸,   0≤α,  β<1,  η-λ>1/8, and min{-μ,η-μ}>-2; then
(35)T*λ+1,μ,η(k,β,α)⊂T*λ,μ,η(k,β1,α1),
where β1 and α1 are as in Theorem 11.Lemma 13 (see [19]).
Letp(z) be analytic in 𝕌 with p(0)=1 and ℝp(z)>0,  z∈𝕌. Then, for s>0 and η1≠-1 (complex),
(36)ℝ(p(z)+szp′(z)p(z)+η1)>0,for|z|<r0,
where r0 is given by r0=|1+η1|/m+(m2-|η12-1|)1/2 and m=2(s+1)2+|η1|2-1, and this result is the best possible.Theorem 14.
Letz∈𝕌,  f∈Rλ,μ,η(k,0). Then f∈Rλ+1,μ,η(k,0) for |z|<r0, where
(37)r0=η-λ+1m+m2-|(η-λ)2-1|,m=7+(η-λ)2.
This radius is exact.Proof.
Let(38)z(𝒮0,zλ,μ,ηf(z))′𝒮0,zλ,μ,ηf(z)=p(z)=(k4+12)p1(z)-(k4-12)p2(z),
where p∈Pk and p1,p2∈P in 𝕌. Using a similar argument as in Theorem 9, we obtain
(39)z(𝒮0,zλ+1,μ,ηf(z))′𝒮0,zλ+1,μ,ηf(z)=p(z)+zp′(z)η-λ+p(z)=(k4+12)(p1(z)+zp1′(z)p1(z)+η-λ)-(k4-12)(p2(z)+zp2′(z)p2(z)+η-λ).
Applying Lemma 13, we get
(40)ℝ(pi(z)+zpi′(z)pi(z)+η-λ)>0for|z|<r0,
where r0 is given by (37). This completes our proof.Lemma 15 (see [20]).
Letp be analytic in 𝔼 with p(0)=1, and δ is a complex number satisfying ℝδ≥0,  (δ≠0); then ℝ[p(z)+δzp′(z)]>β,  (0≤β<1) implies ℝp(z)>β+(1-β)(2γ-1), where γ is given by
(41)γ=∫01(1+tℝδ)-1dt,
which is an increasing function of ℝδ and 1/2≤γ≤1. The estimate is sharp in the sense that bound cannot be improved.Theorem 16.
Letmin{η-λ-1,-μ,η-μ}>-2. Then
(42)Pk′(λ+1,μ,η,α)⊂Pk′(λ,μ,η,α+(1-α)(2γ-1)),
where
(43)γ=∫01(1+t1/(η-λ+1))-1dt,
which is an increasing function of 1/(η-λ+1) and 1/2≤γ<1.Proof.
Letf(z)∈Pk′(λ+1,μ,η,α)(44)(𝒮0,zλ,μ,ηf(z))′=H(z)=(k4+12)h1(z)-(k4-12)h2(z),
where H(z) is analytic and H(0)=1 in 𝕌. Identity (16) gives us
(45)(𝒮0,zλ+1,μ,ηf(z))′=H(z)+zH′(z)η-λ+1=(k4+12)(h1(z)+zh1′(z)η-λ+1)-(k4-12)(h2(z)+zh2′(z)η-λ+1).
This implies that
(46)ℝ(hi(z)+zhi′(z)η-λ+1)>α,i=1,2.
Now using Lemma 15, we get the desired result.Lemma 17 (see [21]).
Letϕ be convex and g starlike in 𝕌. Then, for F analytic in 𝕌 with F(0)=1,  (Φ*Fg)/(Φ*g) is contained in the convex hull of F(𝕌). By convex hull of a set X, one means the intersection of all convex sets that contain X.Theorem 18.
Letϕ be a convex function and f∈Rλ,μ,η(2,α). Then ϕ*f∈Rλ,μ,η(2,α).Proof.
LetG=ϕ*f, and let
(47)ϕ(z)=z+∑n=2∞bnzn,f(z)=z+∑n=2∞‍anzn.
Then
(48)𝒮0,zλ,μ,ηG(z)=𝒮0,zλ,μ,η[z+∑n=2∞‍anbnzn]=z+Γ(2-μ)Γ(2+η-λ)Γ(2-μ+η)×∑n=2∞‍Γ(n+1)Γ(n-μ+η+1)Γ(n-μ+1)Γ(n+η-λ+1)anbnzn=(ϕ*𝒮0,zλ,μ,ηf)(z).
Also, f(z)∈Rλ,μ,η(2,α). Therefore, 𝒮0,zλ,μ,ηf(z)∈R2(α)=S*(α). By logarithmic differentiation of (48), we have
(49)z(𝒮0,zλ,μ,ηG(z))′𝒮0,zλ,μ,ηG(z)=ϕ(z)*F𝒮0,zλ,μ,ηf(z)ϕ(z)*𝒮0,zλ,μ,ηf(z),
where
(50)F(z)=z(𝒮0,zλ,μ,ηf(z))′𝒮0,zλ,μ,ηf(z)
is analytic in 𝕌, and F(0)=1. From Lemma 17, we see that z(𝒮0,zλ,μ,ηG(z))′/𝒮0,zλ,μ,ηG(z) is contained in the convex hull of F(𝕌). Since z(𝒮0,zλ,μ,ηG(z))′/𝒮0,zλ,μ,ηG(z) is analytic in 𝕌 and F(𝕌)⊂Ω={W:(z(𝒮0,zλ,μ,ηW(z))′)/(𝒮0,zλ,μ,ηW(z))∈P2(α)}, then z(𝒮0,zλ,μ,ηG(z))′/𝒮0,zλ,μ,ηG(z) lies in Ω. This implies that G=ϕ*f∈Rλ,μ,η(2,α).This is application of Theorem18.Theorem 19.
The classRλ,μ,η(2,α) is invariant under the following differintegral operators. That is if f∈Rλ,μ,η(2,α), then so does fi, where fi are given to us:(i)
f1(z)=∫0z(f(t)/t)dt,(ii)
f2(z)=(2/z)∫0zf(t)dt,(iii)
f3(z)=∫0z((f(t)-f(xt))/(t-xt))dt,  |x|≤1,  x≠1,(iv)
f4(z)=((1+c)/zc)∫0ztc-1f(t)dt,  ℝe(c)>0.The proof immediately follows from Theorem18. Since we can write fi=f*ϕi with
(51)ϕ1(z)=-log(1-z),ϕ2(z)=-2[z+log(1-z)z],ϕ3(z)=11-xlog(1-xz1-z),ϕ4(z)=∑m=1∞‍1+cm+czmℝec>0,each ϕi is convex for i=1,2,3, and 4.



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