# Radius and Differential Subordination Results for Starlikeness Associated with Limaçon Class

Authors: Afis Saliu; Kanwal Jabeen; Isra Al-shbeil; Semiu Oladipupo Oladejo; Adriana Cătaş
Journal: Journal of Function Spaces
 (2022)

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## Abstract

In this study we investigate the sharp radius of starlikeness of subclasses of Ma and Minda class for the ratio of analytic functions which are related to limaçon functions. This survey is connected also to the first-order differential subordinations. In this context, we get the condition onβ for which certain differential subordinations associated with limaçon functions imply Ma and Minda starlike functions. Simple corollaries are provided for certain examples of our results. Finally, we present several geometries related to our study.


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## Body

## 1. Introduction and Preliminaries

LetA be the class of normalized analytic functions fz having the seres form
(1)fz=z+∑n=2∞anznz∈U≔z∈ℂ:z<1.LetPα denotes the class of all functions pz such that Repz>α,z∈U,0≤α<1. The case P0=P is the usual class of Carathéodory functions. Let S denotes the subclasses of A consisting of functions which are univalent in U. We say fz∈A is subordinate to gz∈A (written as f≺g or fz≺gz) if there exists a Schwarz function wz such that fz=gwz for all z∈U. The class S is one of the most vital categories of geometric function theory due to its wide applications in sciences and engineering. For example, univalent functions are extensively used in ODEs and PDEs and operators’ theory. Also, they are important in image processing techniques. Among the earlier subclasses of S that had recieved tremendous attention are the classes C and S∗ of convex and starlike functions, respectively.In 1992, Ma and Minda [1] gave a unified characterization of the subclasses of S, which consist of functions that map U onto starlike domains. For this purpose, they considered analytic functions Vz with ReVz>0 in U and normalized by V0=1 and V′z>0. Thus, the Ma and Minda class of starlike functions denoted by S∗V was defined by the subordination
(2)zf′zfz≺Vz,f∈A,z∈U.In particular, forVz=1+Az/1+Bz,−1≤B<A≤1, the class S∗V reduces to the class S∗A,B of Janowski starlike functions [2]. The class S∗1−2α,−1=S∗α=f∈A:Rezf′z/fz>α,z∈U is a well-known class of starlike functions of order α. For α=0, S∗0≡S∗ of starlike functions. More recently, many researchers have been investigating subclasses of S∗ having nice geometries in the right-half plane. In this direction, let V:U⟶ℂ be defined by
(3)V1z=cosz,V2z=1+sinz,V3z=1+sinh−1z,V4z=ez,V5z=z+1+z2,V6z=1+2π2log1−z1+z2,V7z=1+43z+23z2,V8z=1+z.Then, forVz=Vkz,k=1,2,⋯8, the class S∗V reduces to Scos∗,Ssin∗, Ssinh−1∗,Se∗, Scr∗, Sp∗, Sc∗, and Sl∗. The geometric properties of these classes have been demonstrated in [3–10] and the references therein.In 2020, Masih and Kanas [11] explored another novel subclass of S∗V with Vz=1+sz2,0<s≤1/2. This class was denoted by SL∗ and functions in it map U onto a region bounded by limaçon. Saliu et al. [12] furthered the investigation of this class and obtained the bounds of the Hankel determinants, sharp radius, and differential implications associated with it. Also, Kanaga and Ravichandran [13] examined Lsz=1+sz2,0<s≤1/2 and found the smallest disc DaRa and the largest disc Dara centered at a,0 such that
(4)Dara⊂LU⊂DaRa.This concept was then applied to find the radius of limaçon starlikeness forS∗α. For more findings associated with Lsz, we refer to [14–16].LetP and Q be two subclasses of A defined on U. The Q radius of P denoted by RQP≤1 is the largest radius such that f∈P implies the function fr, defined by frz=r−1frz∈Q, for all 0<r≤RQP. Radius result with ratio of analytic functions f,g∈A satisfying
(5)Refzgz>0orfzgz−1<1,z∈Uwas studied by MacGregor [17], and the ones satisfying
(6)f′zg′z−1<1,z∈Uwas examined by Ratti [18]. Presently, the radius of S∗V of various choices of V for ratio of analytic functions has attracted the interest of researchers. To this end, Ali et al. [19] considered the functions f∈A whose ratio fz/gz,gz/zpz, and pz are each subordinate to 1+z or ez, for some analytic functions gz and pz. They obtained various radii of starlikeness for these classes. In clases where these are subordinate to z+1+z2 or ez, Yadav et al. [20] also obtained various radii of starlikeness. Zhang et al. [21] found the radius of starlikeness connected with subclasses of S∗V for the ratio of analytic functions fz and gz satisfying
(7)Refzgz>0,Re1+z2nzgz>0,z∈U,n∈ℕ.The theory of first-order differential subordinations arose from the work of Goluzin and Robertson in 1935 and 1947, respectively. Later, Miller and Mocanu [22, 23] developed and generalized this idea. Using this theory, many results related with the Ma and Minda class have emerged in different directions and perspectives in the literature. For more information in this direction, we refer to the recent work of Cho et al. [24] and Kumar and Gangania [25] with the references therein.LetPLs be the class of analytic functions pz satisfying the subordination:
(8)pz≺1+sz2,z∈U,0<s≤12.Motivated with these aforecited works, we initiated the following classes of analytic functions:(9)H1=f∈A:fg∈PLs,forsomeg∈Awithgzp∈PLsandp∈PLs,H2=f∈A:fzgz−1<1,forsomeg∈Awithgzp∈PLsandp∈PLs,H3=f∈A:fzp∈PLsforsomep∈PLs.Then, we investigate the sharp radius of starlikeness for various subclasses of Ma and Minda class. Moreover, the first-order differential subordination implications are also studied. Some special cases of our findings are given as a simple corollaries. Finally, we illustrate the geometries of some of our findings.The following lemmas are required for our investigations.Lemma 1.
Letp∈PLs. Then,
(10)zp′zpz≤2sr1−sr,z≤r.
This bound is sharp forpz=1+sz2 with z=−r.Proof.
From the property of subordination, we havepz=1+swz2, where wz is a Schwarz function. Therefore, a simple computation and Schwarz lemma ([26], p. 166) give
(11)zp′zpz=2rsw′z1+swz≤2sr1−wz21−swz1−r2.
Letx=wz with 0≤x≤r. Then,
(12)zp′zpz≤2sr1−r2ϕx,where ϕx=1−x2/1−sx. It is easy to see that ϕx is continous on 0,r. Then, by the elementary theorem of Real analysis, we have that
(13)infx∈0,rϕx,supx∈0,rϕx⊂Rϕ,where Rϕ denotes the range of ϕx. Therefore,
(14)zp′zpz≤2sr1−r2supx∈0,rϕx=2sr1−sr.Lemma 2 (see [23], Theorem 3.4h, p. 132).
Lettz be univalent in U and let ψ and φ be analytic in a domain D containing tU with ψw≠0, where w∈tU. Set Qz=zt′z·ψtz,hz=φtz+Qz, and suppose that either
(i)
hz is convex or Qz is starlike univalent in U(ii)
Rezh′z/Qz=Reφ′tz/ψtz+zQ′z/Qz>0
Ifpz is analytic in U with p0=t0,pU⊂D and
(15)φpz+zp′zψpz≺φtz+zt′zψtz=hz,then pz≺tz, and tz is the best dominant in the sense that p≺q⇒t≺q for all q.


## 2. The ClassH1

Define the functionf1,g1:U⟶ℂ by
(16)f1z=z1+sz6g1z=z1+sz4,0<s≤12.It is easy to see thatf1∈H1 and, thus, H1≠∅. Also, the class contains nonunivalent functions, since
(17)f′z=1+7s1+sz5varnishes at z=−1/7s. Hence, the radius of univalence for H1 is 1/7s for 1/7<s≤1/√2. It is observed that for α=0 in Corollary 4 (i), this radius coincides with the radius of starlikeness for the class.Letf∈H1; then, there exists g∈A such that
(18)fzgz=p1z,zpzgz=p2z,for p,p1,p2∈PLs. Therefore, a computation yields
(19)zf′zfz−1≤zp1′zp1z+zp2′zp2z+zp′zpz.Then, from Lemma1, it follows that
(20)zf′zfz−1≤6sr1−sr,z≤r.


## 3. The ClassH2

Let the functionf2:U⟶ℂ be defined by
(21)f2z=z1+z1+sz4,0<s≤12.Then, the functionsf2z and g1z satisfy
(22)f2zg1z−1<1g1zp∈PLs,for some p=1+sz2. Thus, f2∈H2 and H2≠∅. Since
(23)f2′z=6sz2+5s+2z+11+sz=0forz=−25s+2+25s2−4s+4,the class H2 contains some nonunivalent functions. Hence, the radius of univalence for this class is 2/5s+2+25s2−4s+4.Letf∈H2. Then, there exists an analytic function gz such that
(24)fzgz−1<1,gzp∈PLs,for some p∈PLs. Since the disc w−1<1 implies Re1/w>1/2, then Regz/fz>1/2. Let p1,p2:U⟶ℂ be defined such that
(25)gzfz=p1zgzzpz=p2z,p1,p2∈PLs.Then,(26)fz=zpzp2zp1z,and a computation gives
(27)zf′zfz−1≤zp2′zp2z+zp′zpz+zp1′zp1z.It is known from Lemma 2 of [27] that for p∈Pα, we have
(28)zp′zpz≤21−αr1−r1+1−2αrz≤r.Using this fact for the caseα=1/2 and Lemma 1 in (27), we arrive at
(29)zf′zfz−1≤1+4sr−5sr21−r1−sr.It also follows from (29) that
(30)Rezf′zfz≥6sr2−5s+2r+11−r1−sr>0,provided r≤RS∗H2=2/5s+2+25s2−4s+4. For f2z, we have
(31)zf2′zf2z=6sz2+5s+2z+11+z1+sz=0,at z=−RS∗H2. This shows that the radius is sharp. Hence, the radius of starlikeness for the class H2 is the same as the radius of univalence for the class.


## 4. The ClassH3

Letf3z=g1z. Then, f3∈H3. It is obvious that H3 contains nonunivalent functions, since
(32)f3′z=1+5sz1+sz3varnishes at z=−1/5s for 1/5<s≤1/2. Hence, the radius of univalence for H3 is 1/5s for 1/5<s≤1/2. We notice that this radius coincides with the radius of starlikess of the class for the choice of α=0 in Corollary 4 (iii).Letp1,p:U⟶ℂ be defined such that for f∈H3,
(33)zfzpz=p1z,p1,p∈PLs.Then, an obvious calculation yields(34)zf′zfz−1≤zp′zpz+zp1′zp1z.In view of Lemma1, we arrive at
(35)zf′zfz−1≤4sr1−sr.


## 5. Radius of Starlikeness

In this section, we obtain the radii of starlikeness of the classesH1, H2, and H3 for different Ma and Minda starlike classes of functions.Silverman ([28], pp. 50-51) showed that
(36)w∈ℂ:w−c<d⊂w∈ℂ:w−a<bholds if and only if a−c≤b−d. Using this fact, we obtain the radii of Janowski starlikeness for H1 and H3.Theorem 3.
The following sharp results hold for Janowski starlike classS∗A,B:(i)
RS∗A,BH1=A−B/A−7B+6s,A−B/A−7B+6<s≤1/2(ii)
RS∗A,BH3=A−B/A−5B+4s,A−B/A−5B+4<s≤1/2Proof.
(i)
Letf∈H1. Then, f satisfies (20). To prove our result, it suffices to show that
(37)w∈ℂ:w−1<6sr1−sr⊂w∈ℂ:w−1−AB1−B2<A−B1−B2.Then, from (36), it follows that
(38)1−AB1−B2−1≤A−B1−B2−6sr1−sr,where we have that
(39)r≤RS∗A,B=A−BA−7B+6s,s>A−BA−7B+6.Forf1z in (16), we have
(40)zf1′zf1z=1+7sz1+sz.Atz=−RS∗A,B,
(41)zf1′zf1z−1−AB1−B2=1−A1−B−1−AB1−B2=A−B1−B2(ii)
Letf∈H3. Then, f satisfies (35). We need to show that
(42)w:w−1<4sr1−sr⊂w:w−1−AB1−B2<A−B1−B2.
Then, from (36), it follows that
(43)1−AB1−B2−1≤A−B1−B2−4sr1−sr,which gives
(44)r≤RS∗A,B=A−BA−5B+4s,s>A−BA−5B+4.
Forf3z in Section 4 at z=−RS∗A,B, we have
(45)zf1′zf1z=1+5sz1+sz=1−A1−B,which implies
(46)zf3′zf3z−1−AB1−B2=1−A1−B−1−AB1−B2=A−B1−B2Corollary 4.
The following sharp results hold forS∗α:(i)
RS∗αH1=1−α/7−αs,1−α/7−α<s≤1/2(ii)
RS∗αH2=21−α/5−αs+2−α+5−α2s2+41−s+α14s+α−2α1−αs(iii)
RS∗αH2=1−α/5−αs,1−α/5−α<s≤1/2Proof.
The proof of (i) and (iii) are direct from Theorem3 for A=1−2α.
0<α≤1 and B=−1. For (ii), f∈S∗α provided (30) satisfies
(47)6sr2−5s+2r+11−r1−sr>α.
That is,Tr≔6−αsr2−5−αs+2−αr+1−α≥0. Then, T0=1−α>0 and T1=s−2<0. Therefore, there exists RS∗αH2∈0,1 such that TRS∗αH2=0, where we obtain RS∗αH2 given in the theorem. For sharpness, consider the function f2z in (21). Then, for z0=−RS∗αH2, we have that
(48)zf2′z0f2z0=6sz02+5s+2z0+11+sz01+z0=α.The classSp∗ is the class of parabolic starlike functions introduced by Rønning [8]. This class consists of functions f∈A satisfying the inequality
(49)Rezf′zfz>zf′zfz−1,z∈U.For1/2<a<3/2, Shanmugam and Ravichandran [29] proved the following inclusion relation for the class Sp∗.
(50)w∈ℂ:w−a<a−1/2⊂w∈ℂ:Rew>w−a≔Δp.Theorem 5.
The following sharp results hold forSp∗:(i)
RSp∗H1=1/13s,1/13<s≤1/2(ii)
RSp∗H2=2/31+2s+9+10s+81s2(iii)
RSp∗H3=1/9s,1/9<s≤1/2Proof.
(i)
Fora=1, relation (50) implies that disc (20) lies in Δp provided
(51)6rs1−sr≤12,or if r≤RSp∗H1=1/13s,1/13<s≤1/2. The function f1z shows that the radius of parabolic starlike functions is sharp. For this function, we have
(52)zf1′zf1z=1+6sz1+sz.Atz=−RSp∗H1, a calculation shows that
(53)Rezf1′zf1z=12=zf1′zf1z−1(ii)
Since the center of disc (29) is 1, then it stays inside Δp if
(54)1+4sr−5sr21−r1−sr≤12,or equivalently r≤RSp∗H2=2/31+2s+9+10s+81s2. For f2z at z=−RSp∗H2, a computation gives
(55)zf2′zf2z=6sz2+2+5sz+11+z1+sz,zf2′zf2z−1=12=Rezf2′zf2z(iii)
Since the center of disc (35) is a=1, then inclusion (50) holds provided
(56)4sr1−sr<12,which implies r≤RSp∗H3=1/7s,1/7<s≤1/2. For the function f3z at z=−RSp∗H3, we have
(57)Rezf3′zf3z=12=zf3′zf3z−1The classSe∗ of the exponential starlike functions fz satisfies the inequality
(58)logzf′zfz<1.This class was initiated by Mendiratta et al. [6] in 2015. They also proved that for e−1<a≤e+e−1/2,
(59)w∈ℂ:w−a<a−1/e⊂w∈ℂ:logw<1≔Δe.Theorem 6.
The following sharp results hold forSe∗:(i)
RSe∗H1=e−1/7e−1s,e−1/7e−1<s≤1/2(ii)
RSe∗H2=2e−1/5e−1s+2e−1+5e−12s2−4e2−14e+2s+2e−12(iii)
RSe∗H3=e−1/5e−1s,e−1/5e−1<s≤1/2Proof.
(i)
Fora=1, relation (59) implies that disc (20) lies inside Δe whenever
(60)6rs1−sr≤1−1e.Equivalently, ifr≤RSe∗H2=e−1/7e−1s,e−1/7e−1<s≤1/2. The function f1z shows that this radius is best possible since
(61)zf1′zf1z=1+6rs1−sr,and at z=−RSe∗H2, we have
(62)logzf1′zf1z=loge−1=1(ii)
Sincee−1<1=a<e−1+e/2, inclusion (59) implies that disc (29) is in Δe provided
(63)1+4sr−5sr21−r1−sr≤1−1e,which at the same time means
(64)r≤RSe∗H2=2e−15e−1s+2e−1+5e−12s2−4e2−14e+2s+2e−12.Atz=−RSe∗H2 for f2z, we arrive at
(65)logzf2′zf2z=loge−1=1(iii)
Fora=1, disc (35) lies completely in Δe if
(66)4rs1−sr≤1−1e,or 5e−1sr+1−e≤0. Thus, the Se∗ of the function f∈H3 is RSe∗H3=e−1/5e−1s,e−1/5e−1<s≤1/2. The sharpness is seen from the function f3z at z=−RSe∗H3, i.e.,
(67)logzf3′zf3z=loge−1=1In 2016, Sharma et al. [9] introduced the class Sc∗ of functions that map the open unit disc onto a Cardiod domain. A function f∈Sc∗ if it satisfies the subordination:
(68)zf′zfz≺1+43z+23z2.The following inclusion relation was also established in [9]:
(69)w∈ℂ:w−a<3a−1/3⊂Δc,provided 1/3<a<5/3, where
(70)Δc=x+iy:9x2+9y2−18x+52−169x2+9y2−6x+1=0.Theorem 7.
The following sharp results hold for the classSc∗:(i)
RSc∗H1=1/10s,1/10<s≤1/2(ii)
RSc∗H2=4/5+14s+196s2+4s+25(iii)
RSc∗H3=1/7s,1/7<s≤1/2Proof.
(i)
Since the center of (20) is 1, then by relation (69), we have that f∈H1 belongs to Sc∗ provided
(71)6sr1−sr≤1−13,or r≤RSc∗H1=1/10s,1/10<s≤1/2. This radius cannot be improved since the function f1z assumes the extremum, i.e., at z=−RSc∗H1,
(72)zf1′zf1z=13=V7−1∈∂V7U(ii)
From (69), it follows that disc (29) stays inside Δc whenever
(73)1+4sr−5sr21−r1−sr≤23.This givesr≤RSc∗H2=4/5+14s+196s2+4s+25. To prove that this radius is sharp, we consider the function f2z such that at z=−RSc∗H2, we arrive at
(74)zf2′zf2z=13=V7−1∈∂V7U(iii)
Letf∈H3. Then, we see that f∈Sc∗ if
(75)4sr1−sr≤23,which gives r≤RSc∗H3=1/7s,1/7<s≤1/2. For the function f3z, we see that
(76)zf3′zf3z=13=V7−1∈∂V7U,which shows that the radius is sharpIn 2019, Cho et al. [4] considered and investigated the class Ssin∗=f∈A:zf′z/fz≺1+sinz. They also proved the following:
(77)w∈ℂ:w−a<sin1−a−1⊂Δs,if 1−sin1<a<1+sin1, where Δs is the image of U under the function V2z.Theorem 8.
The following sharp results hold forSsin∗:(i)
RSsin∗H1=sin1/6+sin1s,sin1/6+sin1<s≤1/2(ii)
RSsin∗H2=2sin1/4+sin1s+1+sin1+4+sin12s2+8−10sin1−2sin21s+1+sin12(iii)
RSsin∗H3=sin1/4+sin1s,sin1/4+sin1<s≤1/2Proof.
(i)
It is easy to see from (77) that disc (20) stays inside the region Δsin provided
(78)6sr1−sr≤sin1,which implies that r≤RSsin∗H1=sin1/6+sin1s,sin1/6+sin1<s≤1/2. To show that this radius is best possible, we consider the function f1z so that
(79)zf1′zf1z=1+6sz1+sz.Atz=−RSsin∗H1, we arrive at
(80)zf1′zf1z=1−sin1=V2−1∈∂V2U(ii)
Since1−sin1<1=a<1+sin1, then disc (29) satisfies relation (77) provided
(81)1+4sr−5sr21−r1−sr≤sin1.This givesr≤RSsin∗H2=2sin1/4+sin1s+1+sin1+4+sin12s2+8−10sin1−2sin21s+1+sin12.For the functionf2z, the result is sharp. Indeed, at z=−RSsin∗H2, we have
(82)zf2′zf2z=1−sin1=V2−1∈∂V2U(iii)
Proceeding as in the above cases and using inclusion (77), we find that f∈Ssin∗ if
(83)4sr1−sr≤sin1,which holds for r≤RSsin∗H3=sin1/4+sin1s,sin1/4+sin1<s≤1/2. The sharpness of the radius is assumed by the function f3zKumar and Arora [5] investigated the subclass Ssinh−1∗ of Ma and Minda class of functions that map U onto a petal domain. They also proved the inclusion relation
(84)w∈ℂ:w−a<a−1−sinh−11⊂Δsinh−1,where Δsinh−1 is the image of U under the function V3z. In view of the procedure of Theorem 8, (20), (29), and (35), we have the following theorem.Theorem 9.
The following sharp results hold forSsinh−1∗:(i)
RSsinh−1∗H1=sinh−11/6+sinh−11s,sinh−11/6+sinh−11<s≤1/2(ii)
RSsinh−1∗H2=2sinh−11/4+sinh−11s+1+sinh−11+4+sinh−112s2+8−10sinh−11−2sinh−112s+1+sinh−112(iii)
RSsinh−1∗H3=sinh−11/4+sinh−11s,sinh−11/4+sinh−11<s≤1/2In 2021, Bano and Mohsan [3] introduced the subclass Scos∗ of analytic functions characterized by the subordination
(85)zf′zfz≺cosz,z∈U.They also proved that(86)w∈ℂ:w−a<a−cos1⊂Δcos,where Δcos is the image domain of U under the mapping V1z=cosz.Theorem 10.
The following sharp results hold forScos∗:(i)
RScos∗H1=1−cos1/7−cos1s,1−cos1/7−cos1<s≤1/2(ii)
RScos∗H2=21−cos1/5−cos1s+2−cos1+5−cos12s2−2cos21−14cos1+4s+2−cos12(iii)
RScos∗H3=1−cos1/5−cos1s,1−cos1/5−cos1<s≤1/2Proof.
(i)
Since disc (20) has a center 1, then relation (86) implies that this disc lies inside Δcos whenever
(87)6rs1−sr≤1−cos1,or r≤RScos∗H1=1−cos1/7−cos1s,1−cos1/7−cos1<s≤1/2. The sharpness of this radius is achieved by the function f1z at z=−RScos∗H1(ii)
Fora=1, disc (29) is inside the domain Δcos provided
(88)1+4sr−5sr21−r1−sr≤1−cos1,which means that r≤RScos∗∗H2=21−cos1/5−cos1s+2−cos1+5−cos12s2−2cos21−14cos1+4s+2−cos12.The functionf2z shows that the radius cannot be improved since at z=−RScos∗∗H2, we have
(89)zf2′zf2z=cos1=V1−1∈∂V1U(iii)
It is observed that disc (35) is in Δcos if
(90)4sr1−sr≤1−cos1,which equivalently implies r≤RScos∗H3=1−cos1/5−cos1s,1−cos1/5−cos1<s≤1/2. For the function f3z, we see that at z=−RScos∗H3,
(91)zf3′zf3z=cos1=V1−1∈∂V1U


## 6. Differential Subordination Implications

In this section, we present some first-order differential subordinations associated withLsz.Theorem 11.
Letpz be analytic in U with p0=1, and suppose
(92)1+βzp′z≺Lsz,z∈U.
Then, we have the following subordination results:(i)
pz≺1+Az/1+Bz, for β≥maxβ1,β2 and −1<B<A≤1 with β1=s4−s1−B/2A−B,β2=s4+s1+B/2A−B and
(93)maxβ1,β2=β2,ifB>−s4,β1,ifB<−s4(ii)
pz≺V8zforβ≥s4+s/22−1(iii)
pz≺V4zforβ≥es4−s/2e−1(iv)
pz≺V7z for β≥3s4−s/4(v)
pz≺V2zforβ≥s4+s/2sin1(vi)
pz≺V3zforβ≥s4+s/2sinh−11(vii)
pz≺V5zforβ≥s4−s/22−2
These bounds are sharp.Proof.
Consider the analytic functiontβ:U¯⟶ℂ such that
(94)tβz=1+szsz+42βis a solution of the differential equation
(95)1+βztβ′z=1+sz2.
Letφw=1 and ψw=β. Then, the function Q:U¯⟶ℂ is defined by Qz=ztβ′zψtβz=szsz+2 such that
(96)RezQ′zQz=21−Re12+sz≥21−s2−s>0.
This means thatQz is a starlike univalent function in U. It is easy to see that the function hz=φtβz+Qz=1+sz2 satisfies
(97)Rezh′zQz=RezQ′zQz>0.
Therefore, by Lemma2, we have that
(98)1+βzp′z≺1+βztβ′z⇒pz≺tβz.
Each subordinations of Theorem11 is equivalent to
(99)pz≺ℙz,for each superordinate function ℙz in the theorem, which holds if tβz≺ℙz,z∈U. Then,
(100)ℙ−1<tβ−1<tβ1<ℙ1.
This yields the necessary condition for whichpz≺ℙz,z∈U. Looking at the geometry of each of the functions ℙz, it is observed that this condition is also sufficient.
(i)
Letℙz=1+Az/1+Bz. Then,
(101)tβ−1≥1−A1−B,wheneverβ≥s4−s1−B2A−B≔β1,tβ1≤1+A1+B,wheneverβ>s4+s1+B2A−B≔β2.We notice that(102)maxβ1,β2=β=β2,ifB>−s4,β1,ifB<−s4.Thus,(103)tβz≺1+Az1+Bifβ≥maxβ1,β2(ii)
ℙz=1+z. Then, the inequalities tβ−1≥0 and tβ1≤2 reduce to β≥s4−s/2=β1 and β≥s4+s/22−1=β2. We note that β1−β2<0. Thus,
(104)tβz≺1+zifβ≥maxβ1,β2=β2(iii)
Letℙz=ez. Then, from (100), we have
(105)tβ−1≥e−1,wheneverβ≥es4−s2e−1≔β1,tβ1≤e,wheneverβ>es4+s2e−1≔β2.We notice thatβ1−β2>0. Therefore, the subordination tβz≺ez holds if β≥maxβ1,β2=β1(iv)
Letℙz=1+4/3z+2/3z2. Then, the inequalities tβ−1≥1/3 and tβ1≤3 give β≥β1 and β≥β2, where β1=3s4−s/4 and β1=s4+s/4. A calculation shows that β1−β2=s2−s>0. Therefore, the subordination tβz≺1+4/3z+2/3z2 holds if β≥maxβ1,β2=β1(v)
Letℙz=1+sinz. Then, the inequalities tβ−1≥1−sin1 and tβ1≤1+sin1 give β≥β1 and β≥β2, where β1=s4−s/2sin1 and β2=s4+s/2sin1. A calculation shows that β1−β2=s2−s>0. Therefore, the subordination tβz≺1+sinz holds provided β≥maxβ1,β2=β2(vi)
Letℙz=1+sinh−1z. Then, following the same line of proof of v, we show that the subordination tβz≺1+sinh−1z holds provided β≥s4+s/2sinh−11(vii)
Letℙz=z+1+z2. Then, the inequalities tβ−1≥2−1 and tβ1≤1+2 yield β≥β1=s4−s/22−2 and β≥β2=s4+s/22. A calculation shows that β1−β2>0. Therefore, the subordination tβz≺z+1+z2 holds provided β≥s4−s/22−2As a direct consequence of Theorem11, we have the following results.Corollary 12.
Letf∈A and suppose it satisfies the following subordination
(106)zf′zfzzf′z′f′z−zf′zfz≺2z+1/2z2β,then
(i)
f∈S∗A,B provided β≥maxβ1,β2 with
(107)maxβ1,β2=β=β2,ifB>−28,β1,ifB<−28,where
(108)β1=42−11−BA−B,β2=42−11+B4A−B,−1<B<A≤1(ii)
f∈Sl∗ provided β≥52+9/4≈4.0178(iii)
f∈Se∗ provided β≥e28−2/8e−1≈1.8418(iv)
f∈Sc∗ provided β≥122−3/8≈1.7463(v)
f∈Ssin∗ provided β≥28+2/8sin1≈1.9778(vi)
f∈Ssinh−1∗ provided β≥28+2/8loge1+2≈1.882(vii)
f∈Scr∗ provided β≥72+6/8≈1.9874Theorem 13.
Letpz be analytic in U with p0=1, and suppose
(109)1+βzp′zpz≺Lsz,z∈U.
Then, the following subordination results are true:(i)
pz≺1+Az/1+Bz, for βmaxβ1,β2 and −1<B<A≤1 with
(110)β1=s4−sloge1−B2−loge1−A2,β2=s4+sloge1+A2−loge1+B2(ii)
pz≺V8z for β≥s4+s/loge2(iii)
pz≺V4z for β≥s4+s/2(iv)
pz≺V7z for β≥s4+s/loge9(v)
pz≺V2z for β≥s4+s/2logesin1(vi)
pz≺V3z for β≥s4+s/2loge1+sinh−11(vii)
pz≺V5z for β≥s4+s/2loge2+1
These bounds onβ cannot be improved.Proof.
Let the analytic functionqβ:U¯⟶ℂ be defined by
(111)qβz=exp4sz+s2z22β.
Then,qβz is a solution of the differential equation
(112)1+βzqβ′zqβz=1+sz2.
Letφw=1 and ψw=β/w. Then, the function Q:U¯⟶ℂ is defined by Qz=zqβ′zψqβz=szsz+2. As it was demonstrated in Theorem 11, we note that Qz is starlike in U and that hz=φqβz+Qz=1+sz2 satisfies
(113)Rezh′zQz=RezQ′zQz>0.
Therefore, by Lemma2, we have that
(114)1+βzp′zpz≺1+βzqβ′zqβz⇒pz≺qβz.
Each subordinations of Theorem11 is equivalent to
(115)pz≺ℙz,for each superordinate function ℙz in the theorem. From here, we obtain the required results by following the same line of proof as in Theorem 11.The following results are direct consequences of Theorem13Corollary 14.
Letf∈A and suppose it satisfies the following subordination
(116)zf′z′f′z−zf′zfz≺2z+1/2z2β,then
(i)
f∈S∗A,B provided β≥maxβ1,β2, where for −1<B<A≤1,
(117)β1=28−24loge1−B2−loge1−A2,β2=28+24loge1+A2−loge1+B2(ii)
f∈Sl∗ provided β≥22+8/4loge2≈4.8020(iii)
f∈Se∗ provided β≥e28+2/8≈1.6642(iv)
f∈Sc∗ provided β≥22+8/8loge3≈1.5149(v)
f∈Ssin∗ provided β≥28+2/8loge1+sin1≈2.7256(vi)
f∈Ssinh−1∗ provided β≥28+2/loge1+loge1+2≈2.6332(vii)
f∈Scr∗ provided β≥28+2/8loge1+2≈1.8882Theorem 15.
Letpz be analytic in U with p0=1, and suppose
(118)1+βzp′zp2z≺Lsz,z∈U.
Then, the following subordination results hold:(i)
pz≺1+Az/1+Bz, for β≥maxβ1,β2 and −1≤B<A<1 with β1=s4−s1−A/2A−B,β2=s4+s1+A/2A−B and
(119)maxβ1,β2=β1,ifA<−s4,β2,ifA>−s4(ii)
pz≺V8z for β≥s4+s2+2/2(iii)
pz≺V4z for β≥es4+s/2e−1(iv)
pz≺V7z for β≥3s4+s/4(v)
pz≺V2z for β≥s4+s1+sin1/2sin1(vi)
pz≺V3z for β≥s4+s1+loge1+2/2loge1+2(vii)
pz≺V5z for β≥s4+s2+2/4
These bounds onβ are sharp.Proof.
Consider the analytic functionrβ:U¯⟶ℂ such that
(120)rβz=1−szsz+42β−1.
Then,rβ is a solution of the differential equation
(121)1+βzrβ′zrβ2z=1+sz2,z∈U.
Letφw=1 and ψw=β/w2. Then, the function Q:U¯⟶ℂ is defined by Qz=zrβ′zψrβz=szsz+2. A computation shows that Qz is starlike in U. Also, we have that hz=φrβz+Qz=1+sz2 satisfies
(122)Rezh′zQz=RezQ′zQz>0.
Therefore, by Lemma2, we have that
(123)1+βzp′zp2z≺1+βzrβ′zrβ2z⇒pz≺rβz.
Hence, we obtain the desired results by following the same line of proof of Theorem11From Theorem15, we have the following consequences.Corollary 16.
Letf∈A and suppose it satisfies the following subordination
(124)zf′zfz−1zf′z′f′z−zf′zfz≺2z+1/2z2β,then
(i)
f∈S∗A,B provided β≥maxβ1,β2 with
(125)maxβ1,β2=β=β1,ifA<−28,β2,ifA>−28,where
(126)β1=42−11−A4A−B,β2=42−11+A4A−B,−1≤B<A<1(ii)
f∈Sl∗ provided β≥92+10/4≈5.6820(iii)
f∈Se∗ provided β≥e28+2e/8e−1≈2.6328(iv)
f∈Sc∗ provided β≥122+3/8≈2.4963(v)
f∈Ssin∗ provided β≥28+21+sin1/8sin1≈3.6421(vi)
f∈Ssinh−1∗ provided β≥28+21+log1+2/loge1+2≈3.5526(vii)
f∈Scr∗ provided β≥92+10/8≈2.8410The following geometries illustrate the sharpness of the Corollaries12, 14 and 16 from (ii) to (vii) as we can see in Figure 1.Figure 1


## 7. Conclusion

We introduced the classesH1, H2, and H3 and then obtained the sharp radii for which normalized analytic functions fz in these classes belonged to Ma and Minda starlike class. Moreover, applying the well-known theory of differential subordination developed by Miller and Mocanu, we found the restriction on β for which the differential subordination
(127)1+βzp′zpkz≺Lsz,k=0,1,2,z∈Uimplies certain Ma and Minda subclasses. We gave some special cases of our findings. Lastly, geometries of parts of our investigations were also illustrated.



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*Source: 8264693-2022-08-31.xml*