# Properties of Functions with Symmetric Points Involving Subordination

Authors: Malik Ali Raza; Syed Zakar Hussain Bukhari; Imtiaz Ahmed; Muhammad Ashfaq; Maryam Nazir
Journal: Journal of Function Spaces
 (2022)

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## Abstract

We study a new subclass of functions with symmetric points and derive an equivalent formulation of these functions in term of subordination. Moreover, we find coefficient estimates and discuss characterizations for functions belonging to this new class. We also obtain distortion and growth results. We relate our results with the existing literature of the subject.


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## Body

## 1. Introduction and Definitions

LetHΔ represent analytic functions f in the disc Δ≔z:∣z∣<1 and A⊂HΔ be defined as:
(1)A≔f∈HΔ:fz=z+a2z2+⋯z∈Δ.LetQ denote “Carathéodory functions” h such that h0=1,Rehz>0 and hz=1+h1z+h2z2+⋯,z∈Δ. The Möbius function k0z=1+z/1−z∈Q or its rotation acts as an extremal function for the class Q and maps Δ onto Rek0z>0. Recall that Qε⊂Q,0≤ε<1 consists of functions h∈Q such that Rehz>ε in Δ. For f,g∈H, we say that the function f is subordinate to g and write f≺g, if for
(2)Φ∈HΔ,withΦ0=0and∣Φz∣<1,fz=gΦz.For a univalent functiong, f≺g if and only if f0=g0 and fΔ⊂gΔ. For reference, see [1]. Applying subordination, Janowski [2] defined the class QA,A1 for −1≤A1<A≤1. A function h∈QA,A1, if
(3)hz=1+AΦz1+A1Φz≺1+Az1+A1zz∈Δ.Geometrically, the imagehΔ lies inside the disk centered on Imz=0, and diameter ends at h−1 and h1. Clearly, QA,A1⊂Q1−A/1−A1. The class QA,A1 is related with the class Q as: h∈Q iff, we write
(4)A+1hz−A−1A1+1hz−A1−1∈QA,A1.AlsoΛk, k≥0 is given by Λk=Φ=u+iv:u>ku−12+v2, represents various plane curves for the specific values of k. Let S be the class of complex-valued injective functions and S∗ represents the class of starlike functions whereas C denotes the class of convex functions. A function f∈S is close-to-convex, if and only if there exists a function g∈C such that
(5)Rezf′zgz∈Q,z∈Δ.We denote the class of close-to-convex functions byK. This class was introduced by Kaplan in [3]. Sakaguchi (see [4]) defined the class SSP as:Definition 1.
Letf∈S. Then, f∈SSP, if
(6)Re2zf′zfz−f−z>0,z∈Δ.
Forf∈A, f∈CSP [5] iff zf′∈SSP, where CSP is the class of convex functions with respect to symmetric points. Various authors studied the class CSP and its subclasses, for detail, see [3, 6–8]. Obviously, it represents the univalent functions. Moreover, it includes the class of convex and odd starlike functions, see [4]. This and other classes are investigated in the literature of the subject; for example, see [9–14].Definition 2.
Letf be analytic in Δ defined by (1). We say that f∈KSPε,0≤ε<1, if for g∈S∗1/2 we have
(7)Rez2f′zgzg−z<−ε,z∈Δ.
For more details, see [15]. We see that KSP0=KSP, where KSP is the class of functions defined in [16]. We study a new class KSPε,η involving g∈S∗1/2.Definition 3.
Letf∈A. Then, f∈KSPε,η,0≤ε<1,0≤η≤1 if for g∈S∗1/2 we have
(8)Rezf′z+ηz2f′′z1−ηGz+ηzG′z<−εz∈Δ,where Gz=gzg−z/z. By a simple calculations, we see that (8) is equivalent to
(9)zf′z+ηz2f′′z1−ηGz+ηzG′z+1<zf′z+ηz2f′′z1−ηGz+ηzG′z−1+2ε,z∈Δ.From [16], we have the following lemma.Lemma 4.
Forg∈S∗1/2 such that
(10)gz=z+∑k=2∞bkzk,z∈Δ,if we put
(11)zGz=gzg−z=z2+∑k=1∞ck+1z2k+1,where
(12)ck=2b2k−1−2b2b2k−2+..⋯+−1k2bk−1bk+1+−1k+1bk2,then G∈S∗.Remark 5.
Sinceg∈S∗1/2, then Lemma 4 proves that G∈S∗. Also from (8), we see that KSPε,η contains close-to-convex functions.


## 2. Main Results

In the following theorem, we have an equivalent formulation of condition (9) in terms of subordination.Theorem 6.
A functionf∈KSPε,η0≤ε<1,0≤η≤1 iff for g∈S∗1/2, we write
(13)−zf′z−ηz2f′′z1−ηGz+ηzG′z≺1+1−2εz1−z,z∈Δ,where Gz=gzg−z/z.Proof.
Letf∈KSPε,η. Then, for g∈S∗1/2, we write
(14)zf′z+ηz2f′′z1−ηGz+ηzG′z+1<zf′z+ηz2f′′z1−ηGz+ηzG′z−1+2ε,z∈Δ,or
(15)Re−zf′z−ηz2f′′z1−ηGz+ηzG′z>ε,z∈Δ,where Gz=gzg−z/z. Using subordination, we write
(16)Hz=−zf′z−ηz2f′′z1−ηGz+ηzG′z≺1+1−2εz1−z,z∈Δ,because k0ε,z=1+1−2εz/1−z∈S and H0=k0ε,0=1, where Gz=gzg−z/z. Conversely, we assume that (13) holds. Then, there exists Φ with Φ0=0 and Φz<1 such that
(17)−zf′z−ηz2f′′z1−ηGz+ηzG′z=1+1−2εΦz1−Φz.
Hence, usingΦz<1, we obtain (9) equivalent to (8), so f∈KSPε,η.Now, we prove sufficient conditions forf∈KSPε,η.Theorem 7.
Letg∈S∗1/2 be a function given by (10) and 0≤ε<1,0≤η≤1. If f defined by (1) satisfies.
(18)2∑k=2∞k+ηkk−1ak+1−2ε+1∑k=2∞2ηk−1+1ck≤21−ε,where the coefficients ck are given by (12), then f∈KSPε,η. In particular, if
(19)∑k=2∞k+ηkk−1ak≤1−ε,then f∈KSPε,η.Proof.
We setFη,z=zf′z+ηz2f′′z for f given by (1), zGz=gzg−z where g is defined by (10) and have
(20)Λ=Fη,z−1−ηGz+ηzG′z−Fη,z−1−2η1−ηGz+ηzG′z=∑k=2∞k+ηkk−1akzk−∑k=2∞2ηk−1+1ckz2k−1−2−2εz+∑k=2∞k+ηkk−1akzk+1−2ε∑k=2∞2ηk−1+1ckz2k−1.
Hence, forz∈Δ, we have the inequality
(21)Λ≤∑k=2∞k+ηkk−1akzk+∑k=2∞2ηk−1+1ckz2k−1−2−2εz+∑k=2∞k+ηkk−1akzk+1−2ε∑k=2∞2ηk−1+1ckz2k−1=−2−2εz+∑k=2∞2k+ηkk−1akzk+1−2ε+1∑k=2∞2ηk−1+1ckz2k−1≤−2−2ε+∑k=2∞2k+ηkk−1ak+1−2ε+1∑k=2∞2ηk−1+1ck≤0.
From these calculations, we see thatΛ<0. Also by (20), we can write
(22)Fη,z−1−ηGz+ηzG′z<Fη,z+1−2η1−ηGz+ηzG′z,which is equivalent to (9) and (8). Thus, f∈KSPε,η, and it completes the proof.The next theorem deals with the coefficient estimatesf∈KSPε,η.Theorem 8.
Let0≤ε<1 and 0≤η≤1. Suppose that f given by (1) and g∈S∗1/2 given by (10) are such that (8) holds. Then, for k=2,3,⋯ we have
(23)2k21+η2k−12−21−ε2≤1−ε∑k=2n22k−1ηk2a2k−1c2k−1+2ε−1+1ηk2c2k−12,where ηk=1+2ηk−1 and ck is defined by (12). In particular, if gz=z, then
(24)k1+η2k−1a2k≤1−ε.Proof.
Iff∈KSPε,η for some g∈S∗1/2, then (9) holds. Using Lemma 4, with α=β=1, we have
(25)zf′z+ηz2f′′z1−ηGz+ηzG′z=1+2ε−1zΦz1+zΦz,z∈Δ,where Φ is an analytic function in Δ, Φz≤1 for z∈Δ, and G is given by (11). Then
(26)ηz2f′′z+zf′z−2ε−11−ηGz+ηzG′zzΦz=1−ηGz+ηzG′z−zf′z−ηz2f′′z.
Now,zΦz=∑k=1∞skzk. We see that Φz≤z for z∈Δ. Thus
(27)2−2εz+∑k=2∞k+ηkk−1akzk−2ε−1∑k=2∞2ηk−1+1ckz2k−1∑k=1∞skzk=∑k=2∞2ηk−1+1ckz2k−1−∑k=2∞k+ηkk−1akzk.
Equating coefficients in (27), for k≥2, we can also write
(28)2−2εz+∑k=1n−12kη2ka2kz2k+∑k=2nηk2k−1a2k−1−2ε−1c2k−1z2k−1zΦz=∑k=2nηkc2k−1−2k−1a2k−1z2k−1−∑k=1k2kη2ka2kz2k+∑k=2k+1∞ckzk,where ηk=1+2ηk−1 and η2k=1+η2k−1.
Then, we square and integrate alongz=r<1. After using the fact Φz≤z<1, we obtain
(29)∑k=2nηkc2k−1−2k−1a2k−12r4k−2+∑k=1k2kη2ka2k2r4k+∑k=2k+1∞ck2r2k<2−2ε2r2+∑k=1n−12kη2ka2k2r4k+∑k=2nηk2k−1a2k−1−2ε−1c2k−12r4k−2.
Lettingr⟶1, we have
(30)∑k=2nηkc2k−1−2k−1a2k−12+∑k=1k2kη2ka2k2<2−2ε2+∑k=1k−12kη2ka2k2+∑k=2nηk2k−1a2k−1−2ε−1c2k−12.
Hence,(31)4k21+η2k−12−41−ε2≤∑k=2nηk2k−1a2k−1−2ε−1c2k−12−ηkc2k−1−2k−1a2k−12≤21−ε∑k=2n2ηk2a2k−1c2k−1+2ε−1+1ηk2c2k−12.
Thus, we have the inequality (23) which finishes the proof.In the following theorem, we prove the growth and distortion theorems forf in the class KSPε,η.Theorem 9.
Iff∈KSPε,η, where 0≤ε<1 and 0≤η≤1, then
(32)χ1r,ε≤f′z≤χ2r,ε,where
(33)χ1r,ε=1−1−2εr1+r1+r2−−11/η1−ε1+1+rarctanhrr1/η1+r,χ2r,ε=1+1−2εr1−r1−r2−−11/η1−ε4r1/ηln1+r1−r−1−ε−11/η1−ε+12r1/η1−r2.
Also, we have(34)χ3r,ε,η≤fz≤χ4r,ε,η+χ5r,ε,η,where
(35)χ3r,ε,η=1−εln1+r1+r2+εarctanr−ηη−1r−1/η+1−−11/ηln1+r,χ4r,ε,η=ε2ln1+r1−r+2−εr1−r−η−11/η1−ε4η−1ln1+r1−rr−1/η+1−ln1+r−11/η1−r,χ5r,ε,η=−1−ε2ηη−1−11/η+1r−1/η+1−−11/η+1ln1−r,and z=r,0≤r<1.Proof.
Iff∈KSPε,η, then for g∈S∗1/2, (8) holds. It follows from Lemma 4 that G in (11) is an odd starlike function. Then
(36)r1+r2≤Gz≤r1−r2,z=r,0≤r<1,(37)1−r21+r22≤G′z≤1+r21−r22,z=r,0≤r<1.
For detail, see [17]. From (8), we obtain a function h with real part greater then ε such that
(38)zf′z+ηz2f′′z1−ηGz+ηzG′z=hz,z∈Δ.
It is known, see [18], that
(39)1−1−2εr1+r≤hz≤1+1−2εr1−r,z=r,0≤r<1.
Thus, from (36), (38), and (39), we obtain (32). From (32) for z=teiθ, we have
(40)fz=∫0zf′sds≤∫0rf′teiθdt≤∫0r1+1−2εt1−t21+t−−11/η1−ε4t1/ηln1+r1−r−1−ε−11/η1−ε+12t1/η1−t2dt=ε2ln1+r1−r+2−εr1−r−η−11/η1−ε4η−1ln1+r1−rr−1/η+1−ln1+r−11/η1−r−1−ε2ηη−1−11/η+1r−1/η+1−−11/η+1ln1−r.
This gives us the right-hand side of the inequality (34). To prove the left-hand side of the inequality (34), we must show that it holds for the nearest point fzo from zero, where zo=r and 0<r<1. Moreover, we have fz≥fzo for z=r. Since f∈K, we know that the function f is univalent in the unit disc Δ. We conclude that the original image of the line segment
(41)fz=∫fCdΦ=∫Cf′zdz≥∫0r1−1−2εt1+t1+t2−−11/η1−ε1+1+rarctanhtt1/η1+tdt=1−εln1+r1+r2+εarctanr−ηη−1r−1/η+1−−11/ηln1+r.
This finishes the proof of the inequality (34).


## 3. Conclusions

In this research, we studied a new subclass of functions with symmetric points and derived an equivalent formulation of these functions in terms of subordination. Moreover, we determined coefficient estimates and discussed characterizations for functions belonging to this new class. We also obtained distortion and growth results. We observed that our findings are related with the existing literature of the subject.



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