# Properties of Certain Classes of Holomorphic Functions Related to Strongly Janowski Type Function

Authors: Bushra Kanwal; Khalida Inayat Noor; Saqib Hussain
Journal: Journal of Mathematics
 (2021)

---

## Abstract

Most subclasses of univalent functions are characterized with functions that map open unit disc∇ onto the right-half plane. This concept was later modified in the literature with those mappings that conformally map ∇ onto a circular domain. Many researchers were inspired with this modification, and as such, several articles were written in this direction. On this note, we further modify this idea by relating certain subclasses of univalent functions with those that map ∇ onto a sector in the circular domain. As a result, conditions for univalence, radius results, growth rate, and several inclusion relations are obtained for these novel classes. Overall, many consequences of findings show the validity of our investigation.


---

## Body

## 1. Introduction

LetA be the class of normalized analytic functions in the open unit disc ∇=z∈ℂ:z<1 of the form(1)gz=z+∑n=2∞bnzn,where g0=0 and g′0=1.Letfz=z+∑n=2∞anzn and gz=z+∑n=2∞bnzn be two analytic functions in A. Then, fz is said to be subordinate to gz, denoted by fz≺gz, if there exist a Schwarz function υz in A, under the conditions υ0=0 and υz<1, such that fz=gυz.LetS denote the subclass of A of univalent functions in ∇ and C,S∗, and K represent the usual subclasses of S that are convex, star-like, and close to convex in ∇, respectively. A number of classes related with strongly star-like and strongly convex functions have been studied; for details, see [1–6]. The Janwoski-type function has been defined and studied in [7]. Here, in this study, we will define strongly Janowski-type function and will discuss some novel classes in relation with this function.The strongly Janowski-type function is defined as(2)φαa,b;z=1+az1+bzα,where −1≤b<a≤1 and 0<α≤1. It is easy to see that this function is univalent and convex in the unit disc ∇.Definition 1.
Let(3)ζz=1+∑n=1∞dnzn,be analytic in ∇ such that ζ0=1. Then, ζz∈Pαa,b if and only if(4)ζz≺φαa,b;z.Definition 2.
Letgz=z+∑n=2∞bnzn∈A. Then, gz∈Sα∗a,b if and only if(5)zg′zgz≺φαa,b;z,and the function gz∈Cαa,b if and only if zg′z∈Sα∗a,b.
It is obvious that, forα=1, we get the classes Pa,b,S∗a,b, and Ca,b, introduced by Janowski in [7].Definition 3.
Letζz=1+∑n=2∞dnzn. Then, ζz∈Pm,αa,b, m≥2, if and only if(6)ζz=m4+12ζ1z−m4−12ζ2z,where ζ1z,ζ2z∈Pαa,b, −1≤b<a≤1 and 0<α≤1. Or equivalently(7)ζz=12∫−ππ1+aze−ιt1+bze−ιtαdμt,where ∫−ππdμt=2 and ∫−ππdμt≤m.
It is easy to note thatPm,1a,b is the well-known class Pma,b defined in [8]. For α=1/2, we have the class Pm,1/2a,b.Definition 4.
Letgz be an analytic function of form (1). Then, gz∈Vm,αa,b if and only if(8)zg′z′g′z∈Pm,αa,b,where m≥2,−1≤b<a≤1, and 0<α≤1.Forα=1, we get the class Vma,b introduced and studied by Noor [8]. Moreover, for a=1−2β and b=−1, we will get the class Vmβ defined by Padmanabhan and Parvatham in [9]. Also, if we take α=1,a=1, and b=−1, then we get the class Vm introduced by Paatero in [10].Definition 5.
Letgz=z+∑n=2∞bnzn∈A. Then, gz∈Tm,αa,b;β if and only if there exists a function ψz∈Vm,αa,b such that(9)g′zψ′z≺1+z1−zβor(10)g′zψ′z∈Pβ1,−1,where m≥2, −1≤b<a≤1, 0<α≤1, and 0<β≤1.Forα=1andβ=1, we get the class Tma,b studied by Noor [11]. For m=2,a=1,b=−1,andα=β=1, we get the well-known class of close-to-convex functions.Definition 6.
Letgz∈A be of form (1). Then, gz∈Qm,αγa,b;s,β if and only if there exists a function g1z∈Tm,αa,b;β such that(11)zg′z+sgz=s+1zg1′zγ,where 0<γ≤1,ℜs>0,m≥2,−1≤b<a≤1,0<α≤1, and 0<β≤1.Remark 1.
(i)
Fors=0 and γ=1, we obtain(12)zg′z=zg1′z,which impliesgz∈Tm,αa,b;β=Qm,α1a,b;0,β.(ii)
Ass⟶∞, we have the class Qm,αγa,b;∞,β, and gz∈Qm,αγa,b;∞,β implies(13)gz=zg1′zγ,whereg1z∈Tm,αa,b;β.(iii)
It can easily be seen thatgz∈Qm,αγa,b;s,β can be represented as the following integral:(14)gz=s+1zs∫0zts−1tg1′tγdt,Some recent work on the classes related to our newly defined classes can be seen in [12, 13].Throughout the present investigation,0<α≤1,−1≤b<a≤1, 0<β≤1, and 0<γ≤1, unless otherwise stated.


## 2. Preliminaries

Lemma 1.
(see [14]). Let ζz=1+∑n=1∞dnzn and Hz=1+∑n=1∞hnzn be two analytic function in ∇ such that ζz≺Hz. If Hz is a convex univalent function in ∇, then dn≤h1,∀n≥1.Lemma 2.
Letζz=1+∑n=1∞dnzn∈Pαa,b. Then, dn≤αa−b,∀n≥1.Proof.
Letζz∈Pαa,b. Then, by definition, ζz≺φαa,b;z. Now, with simple steps, we can show that ℜφα′a,b;z>0. Therefore, by Noshiro Warschawski theorem given in [15], φαa,b;z is univalent in ∇. Moreover,(15)ℜzφα′a,b;z′φα′a,b;z=ℜ1+αa−bz−abz21+az1+bz≥1−αa−br−abr21+ar1+br,wherez=r<1.
Letqr=1−αa−br−abr2. Then, qr is decreasing on 0,1, and therefore, ℜzφα′a,b;z′/φα′a,b;z>0. Hence, φαa,b;z is a convex function. Now, by using (15), we get the required result.


### 2.1. Special Cases

Some special cases of Lemma2 are stated as follows:(i)
Fora=c,b=0,0<c≤1,andα=1/2, we have ζz≺1+cz and dn≤1/2c(ii)
Fora=1andb=−1, we get ζz≺1+z/1−zα and dn≤2α(iii)
Forα=1/2 and ζz≺1+az/1+bz, we have dn≤a−b/2Lemma 3.
Pαa,b⊂Pϱ; that is, if ζz∈Pαa,b, then ζz∈Pϱ, where ϱ=1−a/1−bα.Proof.
Letζz∈Pαa,b. Then, ζz≺1+az/1+bzα. This implies there exists a Schwarz function υz such that υ0=0 and υz<1:(16)ζz=1+aυz1+bυzα.
Now,(17)ℜζz=ℜ1+aυz1+bυzα≥1−aυz1−bυzα>1−a1−bα.
Hence, we get the required result.Lemma 4.
(see [16]). Let pz be convex in ∇ with p0=1. Suppose also that φz is analytic in ∇ with ℜφz≥0, z∈∇. If hz is analytic in ∇ with h0=1, then(18)hz+φzzh′z≺pz⇒hz≺pz,in∇.Lemma 5.
Letζz=1+∑n=1∞dnzn∈Pm,αa,b; then,(i)
Pm,αa,b⊂Pmϱ, where ϱ=1−a/1−bα(ii)
dn≤mα/2a−b(iii)
1/2π∫o2πζreιθ2dθ≤4+m2α2a−b2−4r2/41−r2Proof.
The first result can be easily proved by using Lemma3.
(ii) Letζz∈Pm,αa,b. Then, there exists two functions ζ1z,ζ2z∈Pαa,b such that(19)ζz=m4+12ζ1z−m4−12ζ2z,wherem≥2.
Now, letζ1z=1+∑n=1∞bnzn and ζ2z=1+∑n=1∞cnzn; then,(20)1+∑n=1∞dnzn=m4+121+∑n=1∞bnzn−m4−121+∑n=1∞cnzn,which can be written as(21)∑n=1∞dnzn=∑n=1∞m4+12bn−m4−12cnzn.
Comparing thenth terms of the equality and then by taking modulus on both sides, we obtain(22)dn≤m4+12bn+m4−12cn.
Using Lemma2, we will get the required result.
(iii) Using Parseval’s identity, we have(23)12π∫o2πζreιθ2dθ=∑n=0∞dn2r2n≤1+m2α2a−b2r241−r2,which gives the required result.Corollary 1.
(i)
Forα=1, we have 1/2π∫o2πζreιθ2dθ≤4+m2a−b2−4r2/41−r2(ii)
Form=2,a=1, and b=−1, we have 1/2π∫o2πζreιθ2dθ≤1−r2+4α2r2/1−r2(iii)
Forα=1,m=2,a=1, and b=−1, we have 1/2π∫o2πζreιθ2≤1+3r2/1−r2, which is the result for the well-known class P of Caratheodory functionsLemma 6.
(see [17]). Let gz∈A be of form (1). Then, gz∈Vmϱ, where m≥2 and 0≤ϱ<1 if and only if(i)
There exists someg1z∈Vm such that(24)g′z=g1z1−ϱ,∀z∈∇.(ii)
There exists two star-like functionss1z and s2z such that(25)g′z=s1z/zm/4+1/21−ϱs2z/zm/4−1/21−ϱ.Lemma 7.
Letgz be of form (1). If gz∈Vm,αa,b, then gz∈Vmϱ, where ϱ=1−a/1−bα.Proof.
Letgz∈Vm,αa,b. Then, there exist ζ1z,ζ2z∈Pαa,b, such that(26)zg′z′g′z=m4+12ζ1z−m4−12ζ2z.
By using Lemma3, Pαa,b⊂Pϱ, where ϱ=1−a/1−bα. Therefore, ζiz∈Pϱ, for i=1,2. Hence, gz∈Vϱ.


## 2.1. Special Cases

Some special cases of Lemma2 are stated as follows:(i)
Fora=c,b=0,0<c≤1,andα=1/2, we have ζz≺1+cz and dn≤1/2c(ii)
Fora=1andb=−1, we get ζz≺1+z/1−zα and dn≤2α(iii)
Forα=1/2 and ζz≺1+az/1+bz, we have dn≤a−b/2Lemma 3.
Pαa,b⊂Pϱ; that is, if ζz∈Pαa,b, then ζz∈Pϱ, where ϱ=1−a/1−bα.Proof.
Letζz∈Pαa,b. Then, ζz≺1+az/1+bzα. This implies there exists a Schwarz function υz such that υ0=0 and υz<1:(16)ζz=1+aυz1+bυzα.
Now,(17)ℜζz=ℜ1+aυz1+bυzα≥1−aυz1−bυzα>1−a1−bα.
Hence, we get the required result.Lemma 4.
(see [16]). Let pz be convex in ∇ with p0=1. Suppose also that φz is analytic in ∇ with ℜφz≥0, z∈∇. If hz is analytic in ∇ with h0=1, then(18)hz+φzzh′z≺pz⇒hz≺pz,in∇.Lemma 5.
Letζz=1+∑n=1∞dnzn∈Pm,αa,b; then,(i)
Pm,αa,b⊂Pmϱ, where ϱ=1−a/1−bα(ii)
dn≤mα/2a−b(iii)
1/2π∫o2πζreιθ2dθ≤4+m2α2a−b2−4r2/41−r2Proof.
The first result can be easily proved by using Lemma3.
(ii) Letζz∈Pm,αa,b. Then, there exists two functions ζ1z,ζ2z∈Pαa,b such that(19)ζz=m4+12ζ1z−m4−12ζ2z,wherem≥2.
Now, letζ1z=1+∑n=1∞bnzn and ζ2z=1+∑n=1∞cnzn; then,(20)1+∑n=1∞dnzn=m4+121+∑n=1∞bnzn−m4−121+∑n=1∞cnzn,which can be written as(21)∑n=1∞dnzn=∑n=1∞m4+12bn−m4−12cnzn.
Comparing thenth terms of the equality and then by taking modulus on both sides, we obtain(22)dn≤m4+12bn+m4−12cn.
Using Lemma2, we will get the required result.
(iii) Using Parseval’s identity, we have(23)12π∫o2πζreιθ2dθ=∑n=0∞dn2r2n≤1+m2α2a−b2r241−r2,which gives the required result.Corollary 1.
(i)
Forα=1, we have 1/2π∫o2πζreιθ2dθ≤4+m2a−b2−4r2/41−r2(ii)
Form=2,a=1, and b=−1, we have 1/2π∫o2πζreιθ2dθ≤1−r2+4α2r2/1−r2(iii)
Forα=1,m=2,a=1, and b=−1, we have 1/2π∫o2πζreιθ2≤1+3r2/1−r2, which is the result for the well-known class P of Caratheodory functionsLemma 6.
(see [17]). Let gz∈A be of form (1). Then, gz∈Vmϱ, where m≥2 and 0≤ϱ<1 if and only if(i)
There exists someg1z∈Vm such that(24)g′z=g1z1−ϱ,∀z∈∇.(ii)
There exists two star-like functionss1z and s2z such that(25)g′z=s1z/zm/4+1/21−ϱs2z/zm/4−1/21−ϱ.Lemma 7.
Letgz be of form (1). If gz∈Vm,αa,b, then gz∈Vmϱ, where ϱ=1−a/1−bα.Proof.
Letgz∈Vm,αa,b. Then, there exist ζ1z,ζ2z∈Pαa,b, such that(26)zg′z′g′z=m4+12ζ1z−m4−12ζ2z.
By using Lemma3, Pαa,b⊂Pϱ, where ϱ=1−a/1−bα. Therefore, ζiz∈Pϱ, for i=1,2. Hence, gz∈Vϱ.


## 3. Main Results

In the following theorems, we demonstrate and analyze some novel features of newly defined classes using strongly Janowski-type functions.Theorem 1.
Letgz be of form (1) in the class Vm,αa,b. Then, there exist two analytic functions s1z,s2z∈Sα∗a,b, such that(27)zg′z=s1zm/4+1/2s2zm/4−1/2.Proof.
Letgz∈Vm,αa,b; then, there exist ζ1z,ζ2z∈Pαa,b, such that(28)zg′z′g′z=m4+12ζ1z−m4−12ζ2z.
Then, by using the definition ofCαa,b, we can say that there exist two functions γ1z,γ2z∈Cαa,b, such that(29)zγ1′z′γ1′z=ζ1z,zγ2′z′γ2′z=ζ2z.
Substituting these values in equation (28) and after some simple steps and an integration, we obtain(30)g′z=γ1′zm/4+1/2γ2′zm/4−1/2.
Sinceγ1z,γ2z∈Cαa,b, by the Alexander relation, there exist s1z,s2z∈Sα∗a,b, such that s1z=zγ1′z and s2z=zγ2′z. By putting these values in equation (30), we get the required result.Theorem 2.
Letgz∈Vm,αa,b; then, g∈V2,1ϱ=Cϱ, for z<ro=m−m2−4/2, where ϱ=1−a/1−bα.Proof.
Letgz∈Vm,αa,b; then, by Theorem 1, there exists two functions g1z,g2z∈Cαa,b such that(31)zg′z=zg1′zm/4+1/2zg2′zm/4−1/2.
Logarithmic differentiation of equation (31) yields(32)zg′z′g′z=m4+12zg1′z′g1′z−m4−12zg2′z′g2′z.
Sinceg1,g2∈Cαa,b⊂Cϱ, where ϱ=1−a/1−bα, therefore, there exists p1z,p2z∈Pϱ such that zg1′z′/g1′z=p1z and zg2′z′/g2′z=p2z. We have the well-known distortion result, for class Pϱ,(33)1−1−2ϱr1+r≤Repiz≤piz≤1+1−2ϱr1−r.
So, by means of equation (32), we obtain(34)Rezg′z′g′z−ϱ≥m4+121−1−2ϱr1+r−m4−121+1−2ϱr1−r−ϱ=1−ϱ1−r2r2−mr+1.
Since1−ϱ/1−r2≥0, it follows that ℜzg′z′/g′z−ϱ>0 for z<ro=m−m2−4/2.Theorem 3.
Letg∈Tm,αa,b;β. Then, f∈Cϱ, for z<r1=m1−ϱ+2β−m2−41−ϱ2+4ββ+m1−ϱ/21−ϱ, where ϱ=1−a/1−bα.Proof.
Letg∈Tm,αa,b;β; then, there exists ψz∈Vm,αa,b, such that(35)g′zψ′z∈Pβ1,−1.
On the contrary, we can say that(36)g′z=ψ′zhz,wherehz≺1+z1−zβ.
Then, logarithmic differentiation yields(37)g′z′g′z=ψ′z′ψ′z+h′zhz.
Then, employing equation (34) and the well-known distortion result,(38)zh′zhz<2βr1−r2,where hz∈Pβ1,−1, we obtain(39)ℜzg′z′g′z−ϱ≥1−ϱ1−r2r2−mr+1−2βr1−r2=r21−ϱ−rm−ϱ+2β+1−ϱ1−r2.
Now, let(40)τr=r21−ϱ−rm−ϱ+2β+1−ϱ.
Then,(41)τ0=1−ϱ>0,since0≤ϱ≤1,τ1=1−ϱ1−m−2β<0,sincem≥2and0<β≤1.
Therefore,τr has a root in [0, 1]. Let r1 be the root; then,(42)r1=m1−ϱ+2β−m2−41−ϱ2+4ββ+m1−ϱ21−ϱ.
Hence,g∈Cϱ, for z<r1.Corollary 2.
Forβ=1,α=1,a=1,b=−1, and m=2, we get the radius of convexity r1=2−3 for the class of close-to-convex functions.Corollary 3.
By substitutinga=1andb=−1, we get ϱ=0; also, we have(43)r1=m+2β−m2−4+4ββ+m2,which is the radius of convexity for the class Tm,α1,−1,β.Theorem 4.
Letgz=z+∑n=2∞bnzn∈Vm,αa,b; then, for z=reιϑ,0≤ϑ1≤ϑ2≤2π,(44)∫ϑ1ϑ2ℜzg′z′g′zdϑ>−m2−11−ϱπ,where ϱ=1−a/1−bα.Proof.
Letgz∈Vm,αa,b. Then, by using Lemma 7, we can say that gz∈Vmϱ, where ϱ=1−a/1−bα. Therefore, by using Lemma 6, there exist some g1z∈Vm such that(45)g′z=g1′z1−ϱ.
By using Theorem 2.2 of [17], we have(46)∫ϑ1ϑ2ℜzg1′z′g1′zdϑ>−m2−1π.
Combining equations (45) and (46), we get the required result.Remark 2.
Goodman [18] defined the class of close-to-convex functions of order β as follows.
A normalized analytic functiongz defined in (1) is said to be close to convex of order β if, for β≥0, z=reiϑ, ϑ1<ϑ2, and g′z≠0,(47)∫ϑ1ϑ2ℜzg′z′g′zdϑ>−βπ.
Then, by the criteria defined by Kaplan in [19], we deduce that gz is univalent if β=1.
Hence, it is easy to note that the functiongz∈Vm,αa,b is close to convex of order m/2−11−ϱ. Moreover, the function gz∈Vm,αa,b is univalent for m/2−11−ϱ≤1 or m≤22−ϱ/1−ϱ.
Some noteworthy cases for the univalency of subclasses ofVm,αa,b are stated below.(i)
Fora=1andϱ=0, we have m≤4. So, in this case, gz∈Vm,α1,b is univalent for 2≤m≤4.(ii)
Fora=0,b=−1,andα=1, we have ϱ=1/2 and m≤6. So, gz∈Vm,10,−1 is univalent, for 2≤m≤6.(iii)
Forα=1/2,a=0,b=−1,andϱ=1/2, gz∈Vm,1/20,−1 is univalent for 2≤m≤42−2/2−1.(iv)
Fora=1/2,b=−1/2,andα=1, we have ϱ=1/4 and gz∈Vm,11/2,−1/2 is univalent for 2≤m≤5.Hence, for different values of parameters, we get different classes of analytic functions of bounded boundary rotations and modified limits for the univalency of functions. Since the classVm defined by Paatero in [10] is univalent for 2≤m≤4, here we have discussed some classes of bounded boundary rotations, for which this limit has been improved.Theorem 5.
Letgz=z+∑n=2∞bnzn∈Vm,αa,b; then, bn=O1nσ, where O(1) is a constant. σ=m/2+11−ϱ−2, ϱ=1−a/1−bα, and m≥21+ϱ/1−ϱ.The proof is straightforward by using Lemma6 and Theorem 2.4 of [20].Some noteworthy cases of Theorem5 are stated as follows.Remark 3.
(i)
Fora=1andϱ=0, we have bn=O1nm/2−1. So, m≥2. For m=2, we get a constant.(ii)
Fora=0,b=−1,andϱ=1/2α, we have bn=O1nm/2+11−1/2α−2 and m≥22α+1/2α−1. Here, for α=1, we get bn=O1nm/4−3/2 and m≥6. For m=6, we get a constant. Similarly, if we take α=1/2, we have bn=O1nm/2+11−1/2−2 and m≥21+2/1−2.(iii)
Fora=1/2,b=−1/2,α=1,andϱ=1/3, we have bn=O1nm−4/3 and m≥4. For m=4, we get a constant.(iv)
Fora=1/4,b=−1/4,α=1,andϱ=3/5, we have bn=O1nm−8/5, and we get a constant for m=8.Theorem 6.
Letgz∈Qm,αγa,b;0,β; then, for z=reιϑ and 0≤ϑ1<ϑ2≤2π, we have(48)∫ϑ1ϑ2ℜzg′z′g′zdϑ>−γm2−11−ϱ+βπ,where ϱ=1−a/1−bα.Proof.
Letg∈Qm,αγa,b;0,β; then, there exists g1z∈Tm,αa,b;β such that(49)g′z=g1′zγ.
Logarithmic differentiation of (49) and some manipulations yield(50)zg′z′g′z=γzg1′z′g1′z+1−γ.
For0≤ϑ1<ϑ2≤2π, we have(51)∫ϑ1ϑ2ℜzg′z′g′zdϑ=γ∫ϑ1ϑ2ℜzg1′z′g1′zdϑ+1−γϑ2−ϑ1.
Now, sinceg1z∈Tm,αa,b;β. thus, there exists ψz∈Vm,αa,b such that(52)g1′z=ψ′zhz,where hz∈Pβ1,−1. Logarithmic differentiation of equation (52) and simple computation yield(53)∫ϑ1ϑ2ℜzg1′z′g1′zdϑ=∫ϑ1ϑ2ℜzψ′z′ψ′zdϑ+∫ϑ1ϑ2ℜzh′zhzdϑ.
Sinceψ∈Vm,αa,b, so by Theorem 4, we have(54)∫ϑ1ϑ2ℜzψ′z′ψ′zdϑ>−m2−11−ϱπ.
Forhz∈Pβ1,−1, we have(55)∫ϑ1ϑ2ℜzh′zhzdϑ>−βπ.
By combining equations (51) to (55), we will get the required result.From Theorem6, we can conclude the following important results.Remark 4.
(i)
The necessary condition forg∈Qm,αγa,b;0,β is that the outward drawn normal on the image domain turns back at most γm/2−11−ϱ+βπ.(ii)
The functiongz∈Qm,αγa,b;0,β is univalent if γm/2−11−ϱ+βπ<1, that is, m<21−γβ−1−ϱ/γ1−ϱ.(iii)
The functions in the classQm,αγa,b;0,β are close to convex of order γm/2−11−ϱ+β, that is, Qm,αγa,b;0,β⊂Kγm/2−11−ϱ+β.(iv)
Forγ=1, we get Qm,α1a,b;0,β=Tm,αa,b;β; therefore, g∈Tm,αa,b;β implies(56)∫ϑ1ϑ2ℜzg′z′g′zdϑ>−m2−11−ϱ+βπ,where 0≤ϑ1<ϑ2≤2π and ϱ=1−a/1−bα.Theorem 7.
Letg∈Qm,α1a,b;s,β and gz=z+∑n=2∞bnzn; then, we have(57)bn=O111−δ+δn⋅n1−ϱm/2+1,where ϱ=1−a/1−bα and δ=1/s+1.Proof.
Letg∈Qm,α1a,b;s,β; then,(58)zg′z+sgz=s+1zg1′z,where g1∈Tm,αa,b;β. So, there exists some ψz∈Vm,αa,b such that(59)g1′zψ′z∈Pβ1,−1.
Equation (58) can be written as(60)zg′zψ′z+sgzψ′z=s+1zg1′zψ′z,so we have(61)1s+1g′zψ′z+ss+1gzzψ′z=g1′zψ′z.
Now, let1/s+1=δ. Then, s/s+1=1−δ, so by using (59), we have(62)δg′zψ′z+1−δgzzψ′z∈Pβ1,−1.
Lethz∈Pβ1,−1 such that(63)δzg′z+1−δgz=zψ′zhz.
Sincegz=z+∑n=2∞bnzn, it follows that zg′z=z+∑n=2∞nbnzn. So, by using the Cauchy theorem, we have(64)1−δ+δnbn≤12πrn+1∫02πzψ′zhzdϑ.
By using Lemma6, equation (64) can be written as(65)1−δ+δnbn≤12πrn+1∫02πs1zm/4+1/2s2zm/4−1/21−ϱhzdϑ,where s1,s2∈S∗ and ϱ=1−a/1−bα. Using the distortion result for s2 given in [15], we have(66)1−δ+δnbn≤1rn+14rm/4−1/21−ρ12π∫02πs1zm/4+1/21−ρhzdϑ.
Then, by applying noted Hölder inequality, we have(67)1−δ+δnbn≤1rn+14rm/4−1/21−ϱ12π∫02πs1z2m/4+1/21−ϱdϑ1/212π∫02πhz2dϑ1/2.
Sincehz∈Pβ1,−1, by using Lemma 5, we obtain(68)1−δ+δnbn≤1rn+14rm/4−1/21−ϱ1−r2+4β2r21−r21/212π∫02πs1z2m/4+1/21−ϱdϑ1/2=r⋅22m/4−1/21−ϱ−1/22β11−r1/212π∫02π11−r2m+2/21−ϱdϑ1/2=O111−rm/2+11−ϱ,where O(1) represents the constant term. By taking r=1−1/n, we have the required result.Clearly, we see the following.Remark 5.
(i)
Forα=1, we get bn=O11/1−δ+δn⋅nm/2+1a−b/1−b(ii)
Forδ=1, g∈Tm,αa,b;β and bn=O1nm/2+11−ϱ−1Theorem 8.
Let0<γ1<γ2<1. Then,(69)Qm,αγ1a,b;s,β⊂Qm,αγ2a,b;s,β.Proof.
Letg∈Qm,αγ1a,b;s,β. Then, there exists g1∈Tm,αa,b;β such that(70)zg′z+sgz=s+1zg1′zγ1=s+1zg2′zγ2,where g1′zγ1=g2′zγ2, so g2′z=g1′zγ1/γ2. Using the fact that g1z∈Tm,αa,b;β, this implies there exists ψ1z∈Vm,αa,b;β such that(71)g1′zψ1′z=hz,where hz≺1+z/1−zβ. Now, consider ψ2′z=ψ1′zγ1/γ2. Then, by logarithmic differentiation and simple calculations, we obtain(72)zψ2′z′ψ2′z=1−γ1γ2+γ1γ2zψ1′z′ψ1′z.
Sinceψ1z∈Vm,αa,b implies zψ1′z′/ψ1′z∈Pm,αa,b, it can easily be shown that Pm,αa,b is a convex set and γ1/γ2<1; therefore, ψ2z∈Vm,αa,b. Thus,(73)g2′zψ2′z=g1′zψ1′zγ1/γ2=hzγ1/γ2,where hz∈Pβ1,−1. Sinceγ1/γ2<1, therefore, hzγ1/γ2∈Pβ1,−1. Hence, g2z∈Tm,αa,b;β. Thus, g∈Qm,αγ2a,b;s,β which gives the required inclusion.Theorem 9.
Q2,1γ1,−1;s,β⊂Q2,1γ1,−1;∞,β.Proof.
Letg∈Q2,1γ1,−1;s,β; then, by Definition 5,(74)zg′z+sgz=s+1zg1′zγ,where g1z∈T2,11,−1;β; hence, there exists ψz∈V2,11,−1=C such that(75)g1′z=ψ′zpz,where pz≺1+z/1−zβ. So, equation (74) can be written as(76)zg′z+sgz=s+1zpzγψ′zγ=s+1zp1zψ1′z,where p1z=pzγ and ψ1′z=ψ′zγ. Since pz∈Pβ1,−1 and ψz∈C, therefore, p1z∈Pβ1,−1 and ψ1z∈C. Now, we define Gz such that(77)zG′z+sGz=s+1ψ1z,where ψ1∈C; then,(78)Gz=s+1zs∫0zts−1ψ1tdt.
Gz defined by equation (78) is convex [21]. Now, let(79)gz=hzzG′z.
Then, by combining equations (76)–(79) and some simple computation, we obtain(80)hz+zh′zs+hoz=p1z,where hoz=zG′z′/G′z and ℜhoz>0 since G∈C. Now, we write h1z=1/s+hoz, s≥0; it is obvious that ℜh1z>0, so equation (80) can be written as(81)hz+h1zzh′z=p1z≺1+z1−zβ.
Therefore, by using Lemma4, it follows that hz≺1+z/1−zβ, and hence, g∈Q2,1γ1,−1;∞,β.


## 4. Conclusion

In this study, taking into account the subordination technique, we have introduced certain new classes of analytic functions associated with strong Janowski-type functions. Furthermore, we have obtained several results, and their well-known special cases are apprehended in form of Remarks2–5. We hope that this new methodology will stimulate futuristic research in the fascinating field of geometric function theory and differential subordination.



---

*Source: 1806174-2021-11-08.xml*