# On a New Subclass ofp-Valent Close-to-Convex Mappings Defined by Two-Sided Inequality

Authors: Yi-Hui Xu; Jin-Lin Liu
Journal: Journal of Function Spaces
 (2018)

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## Abstract

A new subclassKp(α1,α2,β) of p-valent close-to-convex mappings defined by two-sided inequality is introduced. Some sufficient conditions for functions to be in Kp(α1,α2,β) are given.


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## Body

## 1. Introduction

LetA(p) be the class of functions of the form(1)fz=zp+∑n=1∞ap+nzp+np∈Nwhich are p-valent analytic in the open unit disk U={z∈C:z<1}. There and in the following, let N,C, and R be the sets of positive integers, complex numbers, and real numbers, respectively. A function f analytic in U is said to be close-to-convex if there is a convex function g such that(2)Ref′zg′z>0for all z∈U. The concept of close-to-convex was introduced by Kaplan [1] in 1952. A number of results for close-to-convex functions in U have been obtained by several authors (see, e.g., [2–14]).A functionf∈A(p) is said to be in the class Kp(α1,α2,β) if it satisfies the following two-sided inequality:(3)-α2π2<arg⁡f′zpzp-1-β<α1π2z∈Ufor 0<α1,α2≤1 and 0≤β<1. Note that if f∈Kp(α1,α2,β), then f is p-valent close-to-convex in U. Furthermore, if f∈Kp(α,α,β)(0<α≤1,0≤β<1), then (3) becomes (4)arg⁡f′zpzp-1-β<απ2z∈U.A function f∈Kp(α,α,β) is called p-valent close-to-convex of order α and type β in U.Given two functionsf and g, which are analytic in U, we say that the function g is subordinate to f and write g≺f or g(z)≺f(z)(z∈U), if there exists a Schwarz function w, analytic in U with w(0)=0 and |w(z)|<1(z∈U) such that g(z)=f(w(z))(z∈U). In particular, if f is univalent in U, we have the following equivalence:(5)gz≺fzz∈U⟺g0=f0,gU⊂fU.Throughout this paper, we let(6)0<α1≤1,0<α2≤1,0<β<1,γ=α1-α2α1+α2,c=eγπi.In order to prove our main results, we need the following lemmas.Lemma 1.
The functiong defined by(7)gz=1+cz1-zα1+α2/2g0=1is analytic and univalent convex in U and(8)gU⊂w∈C:-α2π2<arg⁡w<α1π2.Proof.
In view of (6), it is easy to see that -1<γ<1 and the transformation(9)ζ=e-γπi/2w2/α1+α2maps the convex region (10)G=w∈C:-α2π2<arg⁡w<α1π2conformally onto the right-half ζ-plane -π/2<arg⁡ζ<π/2 so that w=1 corresponding to ζ=e-γπi/2. Since(11)z=ζ-e-γπi/2ζ+eγπi/2maps the right-half ζ-plane Re(ζ)>0 onto U, from (7), (9), and (11) we find that (12)w=eγπi/2ζα1+α2/2=1+cz1-zα1+α2/2=gzmaps U conformally onto G=g(U) with g(0)=1. The proof of Lemma 1 is completed.Lemma 2 (see [15]).
Let the functiong be analytic and univalent in U and let the functions θ and φ be analytic in a domain D containing g(U), with φ(w)≠0(w∈g(U)). Set(13)Qz=zg′zφgz,hz=θgz+Qzand suppose that(i)
Q is univalent starlike in U,(ii)
Re⁡zh′(z)/Q(z)=Re⁡θ′(g(z))/φ(g(z))+zQ′(z)/Q(z)>0(z∈U).
Ifq is analytic in U with q(0)=g(0), q(U)⊂D, and(14)θqz+zq′zφqz≺θgz+zg′zφgz=hzz∈U,then q(z)≺g(z)(z∈U). The function g is the best dominant of (14).Lemma 3 (see [16]).
Letp(z) be analytic function in U of the form (15)pz=1+∑n=m∞cnzn,cm≠0,m≥1,with p(z)≠0 in U. If there exists a point z0∈U such that (16)arg⁡pz<π2for |z|<|z0| and (17)arg⁡pz0=π2,then(18)z0p′z0pz0=ik,where(19)k≥m2a+1a,whenarg⁡pz0=π2k≤-m2a+1a,whenarg⁡pz0=-π2,where p(z0)=±ia and a>0.In this paper we shall derive some criteria for a functionf∈A(p) to be in the class Kp(α1,α2,β).


## 2. Main Results

Our first result is the following theorem.Theorem 4.
Letμ∈C and s∈R. Also let s+1≤2/(α1+α2) and s+1Re⁡μ≥0. If the function q is analytic in U with q(0)=1 and q(z)≠0(z∈U) and satisfies(20)μqzs+1+zq′zqzs≺hzz∈U,where(21)hz=1+cz1-z1/2s+1α1+α2μ+α1+α22z1-z+cz1+czis close-to-convex in U, then(22)-α2π2<arg⁡qz<α1π2z∈U.The bounds α1 and α2 in (22) are sharp for the function g defined by (7).Proof.
We choose(23)gz=1+cz1-zα1+α2/2,θw=μws+1,φw=wsin Lemma 2. By Lemma 1, the function g is analytic and univalent convex in U and(24)arg⁡gz<max⁡α1π2,α2π2≤π2z∈U.Clearly, θ and φ are analytic in a domain D containing g(U) and q(U), with φ(w)≠0 when w∈g(U). For (25)-2α1+α2≤s+1≤2α1+α2,the function Q given by (26)Qz=zg′zφgz=α1+α21+cz21-z1+1/2s+1α1+α21+cz1-1/2s+1α1+α2is univalent starlike in U because(27)Re⁡zQ′zQz=1+1+12s+1α1+α2Re⁡z1-z-1-12s+1α1+α2Re⁡cz1+cz>1-121+12s+1α1+α2-121-12s+1α1+α2=0z∈U.Further, we have (28)θgz+Qz=μ1+cz1-z1/2s+1α1+α2+α1+α21+cz21-z1+1/2s+1α1+α21+cz1-1/2s+1α1+α2=hz,where h is given by (21) and so(29)zh′zQz=μs+1+zQ′zQz.For s+1Re⁡μ≥0, it follows from (27) and (29) that (30)Re⁡zh′zQz>0z∈U.The other conditions of Lemma 2 are also satisfied. Therefore, we conclude that (31)qz≺gz=1+cz1-zα1+α2/2z∈Uand the function g is the best dominant of (20). The proof of the theorem is completed.Theorem 5.
Iff∈A(p) satisfies f′(z)≠0(0<z<1) and(32)-β2π<arg⁡f′zpzp-11+zf′′zf′z-pβ<β1πfor all z∈U, where(33)βj=αj2+1πtan-1⁡α1+α22ptan⁡αjπ2α1+α2j=1,2,then f∈Kp(α1,α2,β). The bounds β1 and β2 in (32) are the largest numbers such that (3) holds true.Proof.
Forf∈A(p) satisfying f′(z)≠0(0<z<1), we define the function q by(34)f′zpzp-1=β+1-βqz.Then q is analytic in U with q(0)=1 and β+(1-β)q(z)≠0 for all z∈U. Taking the logarithmic differentiations in both sides of (34), we have(35)1+zf′′zf′z=p+1-βzq′zβ+1-βqzand(36)11-βf′zpzp-11+zf′′zf′z-pβ=pqz+zq′zfor all z∈U. Putting s=0 and μ=p in Theorem 4 and using (36), we find that if(37)11-βf′zpzp-11+zf′′zf′z-pβ≺hz,where(38)hz=1+cz1-zα1+α2/2p+α1+α22z1-z+cz1+cz,is (close-to-convex) univalent in U, then (39)-α2π2<arg⁡qz<α1π2z∈U,that is, f∈Kp(α1,α2,β).
Forz=eiθ(θ∈R), z≠1, and z≠-1/c, we have(40)z1-z=-12+i2cot⁡θ2,cz1+cz=12+i2tan⁡θ+γπ2and(41)1+cz1-z=1+eiθ+γπ1-eiθ=cos⁡θ+γπ/2sin⁡θ/2eα1πi/α1+α2≠0.
Now we consider the following two cases.
(i) If(42)Kθ=cos⁡θ+γπ2sin⁡θ2=12sin⁡θ+γπ2-sin⁡γπ2>0,then we deduce from (38), (40), and (41) that (43)heiθ=cos⁡θ+γπ/2sin⁡θ/2α1+α2/2·eα1π/2ip+iα1+α24cot⁡θ2+tan⁡θ+γπ2,which yields(44)arg⁡heiθ=α1π2+tan-1⁡α1+α2cos⁡γπ/24pKθ.Let θ1=(π/2)(1-γ). Then(45)0<Kθ≤Kθ1=121-sin⁡γπ2and it follows from (44) and (45) that(46)π>arg⁡heiθ≥arg⁡heiθ1=α1π2+tan-1⁡α1+α2cos⁡γπ/22p1-sin⁡γπ/2=α1π2+tan-1⁡α1+α22ptan⁡α1π2α1+α2=β1π>0.
(ii) IfK(θ)<0, then it follows from (38) to (41) that (47)heiθ=-cos⁡θ+γπ/2sin⁡θ/2α1+α2/2·e-α2π/2ip+iα1+α24cot⁡θ2+tan⁡θ+γπ2,and so(48)arg⁡heiθ=-α2π2+tan-1⁡α1+α2cos⁡γπ/24pKθ.Let θ2=-(π/2)(1+γ). Then(49)0>Kθ≥Kθ2=-121+sin⁡γπ2and from (48) and (49) we have(50)-π<arg⁡heiθ≤arg⁡heiθ2=-α2π2-tan-1⁡α1+α2cos⁡γπ/22p1+sin⁡γπ/2=-α2π2-tan-1⁡α1+α22ptan⁡α2π2α1+α2=-β2π<0.
Noting thath(0)=p>0, we deduce from (46) and (50) that h(U) properly contains the region -β2π<arg⁡{w}<β1π in the complex w-plane. Therefore, if a function f satisfies (32), then the subordination relation (37) holds true. This shows that f∈Kp(α1,α2,β).
Furthermore, for the functiong defined by (7), we have(51)11-βg′zpzp-11+zg′′zg′z-pβ=hz.Hence, by using (46) and (50), we conclude that the bounds β1 and β2 in (32) are the best possible ones. The proof of the theorem is completed.Theorem 6.
Letδ>0. Also let α∈C and 0≤θ=arg⁡{α}<π/2. If f∈A(p) satisfies(52)-π2+θ<arg⁡α1+zf′′zf′z-p+f′zpzp-1δ<π2+tan-1⁡αsin⁡θ2δ+αcos⁡θfor z∈U, then(53)arg⁡f′zzp-1<π2δz∈U.In particular, if δ≥1, then f∈Kp(1/δ,1/δ,0) or f is p-valent close-to-convex of order 1/δ.Proof.
We denote(54)qz=f′zpzp-1δ.Then condition (52) becomes(55)-π2+θ<arg⁡qz+αδzq′zqz<π2+tan-1⁡αsin⁡θ2δ+αcos⁡θz∈U.We want to prove that (56)arg⁡qz<π2forz∈U.If there exists a point z0(|z0)|<1) such that (57)arg⁡qz<π2z<z0,arg⁡qz0=π2,then, from Lemma 3, we have (58)z0q′z0qz0=ik,where q(z0)=±ai, a>0 and (59)k≥a2+12a,whenarg⁡qz0=π2,k≤-a2+12a,whenarg⁡qz0=-π2.For the case arg⁡{q(z0)}=π/2, we have(60)arg⁡qz0+αδz0q′z0qz0=arg⁡ai+ikαδ=π2+arg⁡a+kαδ=π2+tan-1⁡Im⁡a+kα/δRe⁡a+kα/δ=π2+tan-1⁡kα/δsin⁡θa+kα/δcos⁡θ≥π2+tan-1⁡αa2+1/2aδsin⁡θa+αa2+1/2aδcos⁡θ≥π2+Mα,β,where (61)Mα,β=mina>0⁡tan-1⁡αa2+1sin⁡θ2δ+αcos⁡θa2+αcos⁡θ.The function(62)ga=αa2+1sin⁡θ2δ+αcos⁡θa2+αcos⁡θ,a>0,has a negative derivative(63)g′a=-4aδαsin⁡θ2δ+αcos⁡θa2+αcos⁡θ2≤0for0≤θ<π2.Hence (64)Mα,β=lima→∞⁡tan-1⁡ga=tan-1⁡αsin⁡θ2δ+αcos⁡θ.Therefore, (60) becomes (65)arg⁡qz0+αδz0q′z0qz0≥π2+tan-1⁡αsin⁡θ2δ+αcos⁡θ,which contradicts (55). Thus (66)arg⁡qz<π2forz∈U.
For the casearg⁡{q(z0)}=-π/2, we have k<0. Applying the same method as the above, we get (67)arg⁡qz0+αδz0q′z0qz0=arg⁡-ai+ikαδ=-π2+arg⁡a-kαδ≤-π2+θ.This contradicts (55). The proof of the theorem is completed.Applying the same method as the above we can prove the following theorem.Theorem 7.
Letδ>0. Also let α∈C and -π/2<θ=arg⁡{α}≤0. If f∈A(p) satisfies (68)-π2+tan-1⁡αsin⁡θ2δ+αcos⁡θ<arg⁡α1+zf′′zf′z-p+f′zpzp-1δ<π2+θfor z∈U, then (69)arg⁡f′zzp-1<π2δz∈U.In particular, if δ≥1, then f∈Kp(1/δ,1/δ,0) or f is p-valent close-to-convex of order 1/δ.



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