# On Convolution and Convex Combination of Harmonic Mappings

Authors: Ahmad Sulaiman Ahmad El-Faqeer; Zhen Chuan Ng; Shamani Supramaniam
Journal: Journal of Mathematics
 (2021)

---

## Abstract

In this paper, the subclass of harmonic univalent functions by shearing construction is studied and this subclass of harmonic mappings needs a necessary and adequate condition to be convex in the horizontal direction. Furthermore, convolutions of two special subclasses of univalent harmonic mappings are shown to be convex in the horizontal direction. Also, the family of univalent harmonic mappings of the unit disk onto a region convex in the direction of the imaginary axis is introduced. Sufficient conditions for convex combinations of harmonic mappings of this family to be univalently convex in the direction of the imaginary axis are obtained.


---

## Body

## 1. Motivation and Preliminaries

A complex-valued functionf=u+iv defined on the unit disk U=z∈ℂ:z<1 is called harmonic mapping if u and v are real-valued harmonic functions. In addition, since U is a simply connected domain, f has a unique representation f=h+g¯, where h and g are analytic and co-analytic parts of f, respectively. It is known from Lewy [1] that the mapping f=h+g¯ defined on U is locally univalent and sense-preserving if and only if(1)g′z<h′z,∀z∈U.The class of all univalent, harmonic sense-preserving mappingsf=h+g¯ defined in U is denoted by ℋ. Moreover, let SH be the class of all functions f∈ℋ which is normalized by f0=fz0−1=0. Therefore, each member of f∈SH has the following representation:(2)fz=z+∑n=2∞bnzn+∑n=1∞c¯nz¯nc1<1,for all z∈U.The class of such type of functions given in (2) with c1=0=fz¯0 is a subclass of SH and is denoted by SH0. The dilatation of f belonging to SH0 is the function ω:U⟶ℂ given by ω=g′z/h′z and satisfies ωz<1 for all z∈U. Further, denote by KH (or KH0) all f∈SH (or SH0) which are mapping U onto convex regions. For comprehensive and fundamental knowledge on planar harmonic mappings, see Duren [2]. The subclass of SH0 denoted by SCHD0 consists of all univalent harmonic functions which maps onto domain convex in the direction of the real axis.Section2 demonstrates that the convolution of two special subclasses of univalent harmonic mappings is convex in the horizontal direction. The convolution of any two arbitrary harmonic functions(3)fz=hz+g¯z=z+∑n=2∞anzn+∑n=1∞b¯nz¯n,Fz=Hz+G¯z=z+∑n=2∞Anzn+∑n=1∞B¯nz¯nis defined by(4)f∗˜Fz=h∗Hz+g∗G¯z=z+∑n=2∞anAnzn+∑n=1∞b¯nB¯nz¯n.The convolution of harmonic functions is different from the convolution of univalent functions (for more details, one can refer to [3]). Indeed, the convolution of two harmonic functions does not preserve the convexity and the convolution of any two univalent harmonic functions f∈SH0 need not be univalent. These facts generated significant interest in the analysis of harmonic convolutions of univalent harmonic functions, and several articles on this subject recently appeared in the literature [3–11]. Specifically, the collection of the mappings F=H+G¯∈SH0 that map U onto the right half-plane Ψ=w:Rew>−1/2 and have the form H+Gz=z/1−z with G′<H′ for all z∈U. Kumar et al. [12] proposed a class of right half-plane harmonic mappings, Fa=Ha+G¯a,−1<a<1, satisfying(5)Haz+Gaz=z1−z,Ga′z=a−z1−azHa′z.Using the shear construction of Clunie and Sheil-Small [13], it follows that(6)Haz=z/1+a−1/2z21−z2,Gaz=a/1+az−1/2z21−z2.Lettinga=0 in (5), the mapping F0=H0+G0 with F0+G0z=z/1−z and G0′z=−zH0′z, which is called the standard right half-plane mapping, is obtained. In [9], the following result was obtained.Theorem 1 (see [9]).
Letf=h+g¯∈KH0 satisfy the condition h+gz=z/1−z and dilatation ωz=g′z/h′z=eiθznn∈ℤ+;θ∈ℝ. Ifn=1,2, thenF0∗f∈SCHD0.
Recently, Liu and Ponnusamy [8] generalized Theorem 1 in the case n=1 as follows.Theorem 2 (see [8]).
Letf=h+g¯∈KH0satisfy the conditionh+gz=1+az/1−zwith dilatationω1z=z+a/1+az, wherea∈−1,1, andf1=h1+g¯1∈KH0with dilatationω2z=eiθzθ∈ℝ. Then, f∗f1∈SCHD0.
Liu and Ponnusamy [8] also proposed the following problem.Problem 1.
Letf=h+g¯∈KH with h+gz=1+az/1−z and dilatation function ω1z=z+a/1+az, where a∈−1,1, and let fn=hn+g¯n∈KH0 with dilatation ω2z=eiθznθ∈ℝ,n∈ℕ+. Then, f∗fn∈SCHD0.
Ali et al. [14] solved Problem 1 for ω2z=−z. In [8], the authors conjectured that f∗fn is not locally univalent for n≥2. In Section 2, new subclass of harmonic mappings Paϕ is obtained. It is also shown that the necessary and sufficient condition for Paϕ∈SCHD0, if and only if ϕz is convex. The function fn=h+g¯ satisfies the condition h−gz=z/1−z and dilatation ωz=zn;n∈ℕ with the mapping Paℓ belonging to subclass SCHD0 for a∈−1,2−n/2+n and for all n∈ℕ are also explored.
The results proposed by Hengartner and Schober [15] and Pommenrenke [16] are very useful in checking the convexity of an analytic function in the direction of the imaginary axis as well as the convexity in the direction of the real axis, respectively.Lemma 1 (see [15]).
Supposefis a nonconstant analytic function ofU. Then,(7)Re1−z2f′z≥0,if and only if(1)
fis univalent inU.(2)
fis convex in the direction of imaginary axis.(3)
There exist sequenceszn′converging toz=1andzn″converging toz=−1, such that(8)limn⟶∞Refzn′=supz<1Refz,limn⟶∞Refzn″=infz<1Refz.Lemma 2 (see [16]).
Supposefis a nonconstant analytic function ofUsatisfying the conditionf0=0andf′0=0, and suppose(9)φz=z1+zeiθ11+zeiθ2,where θi,i=1,2∈ℝ. If(10)Rezf′zφz>0,then f is convex in the direction of the real axis.
Recall that the regionΩ is said to be convex in the direction θ, 0≤θ<π, if every line parallel to the line through 0 and eiθ has either connected or empty intersection with Ω, and if Ω is convex in every direction, then it is called convex. If θ=0 (or θ=π/2), then Ω is convex in the direction of the real axis “CHD” or (is convex in the direction of the imaginary axis “CID”). Clunie and Sheil [13] proposed an integrated way to construct a univalent harmonic mapping convex in a given direction.Theorem 3 (see [13]).
Let0≤θ<π. A locally univalent harmonic functionf=h+g¯inUis a univalent mapping ofUonto a domain convex in the direction ofθif and only ifh−e2iθgis a univalent mapping ofUonto a domain convex in the direction ofθ.
The study of the geometric properties of the convex combination of the univalent mappings is another important topic in the geometric function theory. Dorff and Rolf [17] developed a necessary condition for convex combination to be univalent maps onto a domain convex in the direction of the imaginary axis.Theorem 4 (see [17]).
Letf1=h1+g¯1andf2=h2+g¯2be two univalent harmonic mappings defined onU, withf1andf2having the same second complex dilatation and satisfying conditions in (4); then,(11)f3=h3+g¯3=th1z+1−th2+tg¯1z+1−tg¯2z=tf1z+1−tf2is univalent and is convex in the direction of the imaginary axis.
It is known that even iff and g are convex analytic functions, the convex combination of f and g may not be a univalent function (see [18], and for more recent studies of linear combinations of harmonic mappings, see [17, 19–23]). Moreover, Kumar et al. [20] studied the convexity of linear combination of harmonic mappings, which are shears of the analytic mapping z/1−z and z1−βz/1−z2, for −1≤β≤1. Beig et al. [23] studied and found necessary conditions for the convex combination of the right half-plane mappings, the vertical strip mapping, their rotations, and some other harmonic mappings to be univalent and convex in a particular direction. In Section 3, the convex combination of mappings of the family of sense-preserving and locally univalent harmonic mappings fα=hα+gα, by shearing the function hα+gα=Kα where(12)Kαz=hαz+gαz=z1−αz21−z2,α∈0,1,gα′hα′<1,are studied. Further, the appropriate conditions for the convex combination in the vertical direction between univalent harmonic mappings to be univalent and convex in the vertical direction are also established.


## 2. Convolutions of Subclasses of Univalent Harmonic Right Half-Plane Mappings

In this section, we consider the harmonic mappingPa=ha+g¯a with ha−gaz=1−az/1−z with the second complex dilatation ωz=a+z/1+az, a∈−1,1. Hence, we can solve for ha and ga as follows:(13)haz=z−1/21−az21−z2=121+az1−z2+1−az1−z,gaz=haz−1−az1−z=121+az1−z2−1−az1−z,and thus(14)Paz=haz+g¯az=121+az1−z2+1−az1−z+1+az1−z2−1−az1−z¯.Takingℓz=z/1−z, Pa given in (14) can be rewritten as(15)Paz=121+azℓ′z+1−aℓz+1+azℓ′z−1−aℓz¯.For an analytic univalent functionϕ:U⟶ℂ normalized by ϕ0=ϕ′0−1=0, define(16)Paϕz=121+azϕ′z+1−aϕz+1+azϕ′z−1−aϕz¯=Haϕz+G¯aϕz,for all z∈U. It is clear that Paℓ=Pa. Therefore,(17)Paℓz=121+azℓ′z+1−aℓz+1+azℓ′z−1−aℓz¯=Haz+G¯az.IfF is an analytic function with F0=0, then(18)Ha∗Fz=121+azF′z+1−aFz,and(19)Ga∗Fz=121+azF′z−1−aFz.Lemma 3.
Letf=h+g¯satisfy the conditionh−gz=z/1−zwith second complex dilatationω=g′/h′such thatω<1andPaℓbe family of harmonic mappings given in (17). Then, ω^, the second complex dilatation off∗Paℓ, is given by(20)ω^z=2a+zω1−ω+1+azω′1−z21+az1−ω+1+azω′1−z.Proof.
Since(21)f∗Paℓ=h+g¯∗Ha+G¯a=h∗Ha+g∗Ga¯,then(22)ω^z=g∗Ga′zh∗Ha′z.
Now, from (18) and (19), it follows that(23)ω^z=1+azg′z−1−agz′1+azh′z+1−ahz′=2ag′z+1+azg″z2h′z+1+azh″.
Sinceg″=ω′h′+ωh″, then(24)ω^z=2aω+1+azω′h′z+1+azωh″z2h′z+1+azh″.
Fromh−gz=z/1−z, it follows that(25)h′z=11−ω1−z2,h″z=21−ω+ω′1−z1−ω21−z3,and thus(26)ω^z=2aω+1+azω′1−ω1−z+1+azω21−ω+ω′1−z21−ω1−z+1+az21−ω+ω′1−z=2a+zω1−ω+1+azω′1−z2+2az1−ω+1+azω′1−z.Lemma 4.
Letf=h+g¯∈SH0withh−gz=z/1−z. IfPaℓ∗fis locally univalent, thenPaℓ∗fis inSCHD0.Proof.
Since(27)Ha−Ga=1+aℓz,h−g=ℓz,then(28)1+ah+g=Ha−Ga∗h+g=Ha∗h+Ha∗g−Ga∗h−Ga∗g,Ha+Ga=Ha+Ga∗h−g=Ha∗h−Ha∗g+Ga∗h−Ga∗g.
Thus,(29)Ha∗h−Ga∗g=121+ah+g+Ha+Ga,Ha∗h−Ga∗g′=121+ah′+g′+Ha′+Ga′.
Letφz=z/1−z2. Consequently, we have(30)RezHa∗h−Ga∗g′φ=Re1+az2φh′+g′+z2φHa′+Ga′=12Re1+azφh′−g′h′+g′h′−g′+RezφHa′−Ga′Ha′+Ga′Ha′−Ga′=1+a2Reh′+g′h′−g′+ReHa′+Ga′Ha′−Ga′=1+a2Rep1z+Rep2z>0,where p1z=1+ωf/1−ωf and p2z=1+ωPaℓ/1−ωPaℓ. Thus, from Lemma 3, we deduce that Ha∗h−Ga∗g is convex in the direction of the real axis. Theorem 3 implies that Paℓ∗f∈SCHD0.Lemma 5.
The mappingPaϕgiven in (16) is locally univalent if and only if the functionϕis convex.Proof.
The mappingPaϕ=Haϕ+Ga¯ϕ is given in (16). Since Paϕ is locally univalent if and only if Ga′ϕ/Ha′ϕ<1,(31)2aϕ′z+1+azϕ″z2ϕ′z+1+azϕ″z<1,and hence(32)2a/1+a+zϕ″z/ϕ′z2/1+a+zϕ″z/ϕ′z<1.
Thus,(33)a−11+a+1+zϕ″zϕ′z<1−a1+a+1+zϕ″zϕ′z.
It can be easily shown that the above inequality is identical to(34)Re1+zϕ″zϕ′z>0.
Hence,Paϕ is locally univalent if ϕz is convex function.Theorem 5.
The mappingPaϕis in the classSCHD0, if and only ifϕis convex.Proof.
The mappingPaϕ=Haϕ+G¯aϕ is given in (16), and Haϕ−Gaϕ=1−aϕ where ϕ is convex. This implies that Haϕ−Gaϕ is convex in the direction of the real axis. Consequently, from Lemma 5 and Theorem 3, we deduce that Paϕ in SCHD0.
In order to prove the next result, we will employ the distribution theory of the polynomial function roots of the disc unit. Given a polynomial function of degreem:(35)pz=amzm+am−1zm−1+⋯+a1z+a0,with complex coefficients, the parallel algorithm for discovering polynomial zeros for (35) inside U is worth exploring. Let(36)p∗z=znp1z¯¯=a¯m+a¯m−1z+⋯+a¯1zm−1+a¯0zm.
The inverse zeros for (35) and (36) can be easily checked with respect to U. We need the following lemma to prove the main theorem.Lemma 6 (Cohn’s rule) (see [24]).
Letpbe a polynomial function as given in (35) of degreemand letp∗be as given in(36). The number of zeros ofpin or on the unit circle is denoted byrands, respectively. Ifa0<am, then(37)p1z=a¯mp1z−a0p∗zzis a polynomial function of degreem−1withr1=r−1ands1=s, wherer1ands1are the number of zeros of the polynomialp1in or on the unite circle, respectively.Theorem 6.
LetPaℓbe mapping given in (17) and letfn=h+g¯withh−gz=z/1−zand dilatationω1z=zn,n∈ℕ. Then, Paℓ∗fn∈SCHD0fora∈−1,2−n/2+nand for all n belonging toℕ.Proof.
According Lemma4, it suffices to show that the mapping Paℓ∗fn has dilatation ω such that ω<1 for all z∈U. Setting ω1z=zn in (20), we obtain(38)ωz=−zna2+n+n+2−n−anz−2azn−2z1+n−2−2az+2−n−anzn+a2+n+nz1+n=−znqzq∗z,where(39)qz=a2+n+n+2−n−anz−2azn−2z1+n=a0+a1z+anzn+an+1zn+1,q∗z=zn+1q1z¯¯=−2−2az+2−n−anzn+a2+n+nz1+n=a¯n+1+a¯n+a¯1zn+a¯0zn+1.
Ifz0≠0, is zero of the polynomial function qz, then 1/z0 will be zero of the polynomial function q∗. Hence, we can rewrite ω as(40)ωz=−znz+A1z+A2z+A3,…,z+An+11+A¯1z1+A¯2z1+A¯3z,…,1+A¯n+1z.
In order to show thatω<1 in U, it suffices to prove that Ai<1 for i=1,2,3,…,n+1, or equivalently, all the zeros Ai for i=1,2,3,…,n+1 of the polynomial function qz lie in or on the unit circle z=1.
Note that(41)a0=a2+n+n<2=an+1,∀a∈−1,2−n2+n,n∈ℕ,so Cohn’s rule can be applied on q. Consider the polynomial function q1z given by(42)q1z=a¯n+1qz−a0q∗zz=1+a−2+n+a2+n2+nzn−1−2+nzn.
It is easy to verify that Cohn’s rule is applicable toq1 also. So, let a polynomial function q2 given by(43)q2z=−2+nq1z−2q∗zz=1+a−2+n+a2+n4n+n2−1+zzn−2.
Then, the zeros of polynomial functionq2 lie on ∂U or in U. We deduce that all zeros of q1 lie inside U or on ∂U, and this implies that all the zeros of q lie inside U or on ∂U.Corollary 1.
LetPaℓbe mapping given in (17) and letf2=h+g¯withh−gz=z/1−zand dilatationω1z=−zn,n∈ℕ. Then, Paℓ∗fn∈SCHD0fora∈−1,2−n/2+n.
The following examples illustrate Theorems5 and 6.Example 1.
In Theorem6, if n=2, h−gz=z/1−z, and ωf2z=z2, then(44)f2z=hz+g¯z=182z3−2z1−z2+log1+z1−z+182z2z−11−z2+log1+z1−z¯.
It follows from (17) that(45)Paℓz=ha+ga¯=121+az1−z2+1−az1−z+1+az1−z2−1−az1−z¯,and Paℓz∗f2z=Hz+G¯z. Further,(46)Hz=ha∗hz=1161+az1−z2+1−az1−z∗2z3−2z1−z2+log1+z1−z=11681+az1−z31+z+1−a2z3−2z1−z2+log1+z1−z,Gz=ga∗gz=1161+az1−z2−1−az1−z∗2z2z−11−z2+log1+z1−z=11681+az31−z31+z−1−a2z2z−11−z2+log1+z1−z.
Figures1–4, respectively, display representations of concentrated circles within U under the harmonic mapping Paℓz and concentrated circles under the convolution map Paℓz∗f2z. We take various values of a=−0.001,−0.5∈−1,0.Figure 1
Image ofP−0.001ℓz.Figure 2
Image ofP−0.5ℓz.Figure 3
Image ofP−0.001ℓz∗f2z.Figure 4
Image ofP−0.5ℓz∗f2z.Example 2.
Consider an analytic univalent functionϕz=z+z2/4, where ϕ0=0 and ϕ′0=1. Since ϕ′z=1+z/2 and ϕ″z=1/2,(47)Re1+zϕ″zϕ′z=Re1+1/2z1+1/2z≥0.
This implies that the functionϕz maps U into convex region. In view of Theorem 1, the mapping Paϕz is in SCHD0 for −1<a<1. Now(48)Paϕz=121−az24+z+a+1z22+z+12a+1z22+z−1−az24+z¯,z∈U.
Figures5–7 display images of U under Paz+1/4z2 for various values of a∈−1,1, respectively. The image of U under Paz+1/4z2 is convex in the direction of the real axis.Figure 5
Image ofP−0.09z+z2/4.Figure 6
Image ofP0z+z2/4.Figure 7
Image ofP0.09z+z2/4.


## 3. Convex Combination of a Family of Univalent Harmonic Mappings

Letfαz=hαz+g¯αz, where(49)Kαz=hαz+gαz=z1−αz21−z2,α∈0,1,gα′hα′<1,be normalized, sense-preserving, and locally univalent harmonic mapping defined on U. First of all, we have to show that fα belongs to class SH and maps onto region convex in the direction of the imaginary axis. Let(50)Ψz=1−z2Kα′z=1+αz4+1−3αz21−z2,for all z∈U, 0≤α≤1, and we will prove that(51)ReΨz=Re1+αz4+1−3αz21−z2>0.Since1−z22≠0, for all z in U, 0≤α≤1 and(52)1−z¯21+αz4+1−3αz2=1−z¯2+z2−z4+αz4−3z2+3z4−z4z2,it follows that(53)ReΨz=Re1−z¯2+z2−z4+αz4−3z2+3z4−z4z21−z22.Note thatΨ0=1 and for each γ∈ℝ,(54)ReΨeiγ=Re1−e−iγ2+eiγ2−eiγ4+αeiγ4−3eiγ2+3eiγ4−eiγ4eiγ21−eiγ22=Re1−e−iγ2+eiγ2−1+αeiγ4−3eiγ2+3−eiγ21−eiγ22=Re−4icosγsinγ+αeiγ4−3eiγ2+3−eiγ21−eiγ22=Reisin4γ−4sin2γ−4cosγsinγ+2αcos2γ−121−eiγ22=2α1−eiγ22cos2γ−12≥0,0≤α≤1.Hence, by the minimum principle for harmonic functions, we obtain(55)ReΨz=Re1−z2Kα′z>0,for all z∈U, and this yields the result.In addition, by Lemma1, we can deduce that the analytic function Kα=hα+gα is univalent and maps onto domain convex in the direction of the imaginary axis. Further, Theorem 3 yields that the harmonic mapping hα+g¯α belongs to the class SH and is convex in the direction of imaginary axis. Figure 8 gives an illustration of images of U under Kα for several values of α.Figure 8
Images of (a)K0U, (b) K1/3U, (c) K1/2U, and (d) K0.9U.
(a)(b)(c)(d)Let us begin by presenting our own dilatationω.Theorem 7.
Letfαj=hαj+g¯αj, for j=1,2, be two normalized harmonic mapping satisfying hαjz+gαjz=z1−αjz2/1−z2, 0≤αj≤1, and dilatation ωj=gj′/hj′ with ωj<1 in U. Then, the second complex dilatation ω of the mapping f=tfα1+1−tfα2, 0≤t≤1, is defined to be(56)ωz=1+z2ω1ω2+tω1+1−tω2+z2z2−3α1tω11+ω2+α21−t1+ω1ω21+z21+tω2+1−tω1+z2z2−3α1t1+ω2+α21−t1+ω1.Proof.
Sincef=tfα1+1−tfα2=thα1+1−thα2+tg¯α1+1−tg¯α2, it follows that the dilatation ω of the mapping f is given by(57)ω=tgα1′+1−tgα2′thα1′+1−thα2′=tω1hα1′+1−tω2hα2′thα1′+1−thα2′.
Making use of the mappinghαjz+gαjz=z1−αjz2/1−z2 and ωj=gαj′/hαj′, for j=1,2 , we obtain(58)hα1′z=1+z21+α1z2−31+ω11−z22,hα2′z=1+z21+α2z2−31+ω21−z22.
Replacing the above expressions withhα1′ and hα2′ in (57), it follows that(59)ω=tω11+z21+α1z2−3/1+ω11−z22+1−tω21+z21+α2z2−3/1+ω21−z22t1+z21+α1z2−3/1+ω11−z22+1−t1+z21+α2z2−3/1+ω21−z22=tω11+z2+α1z2z2−31+ω2+1−tω21+z2+α2z2z2−31+ω1t1+z2+α1z2z2−31+ω2+1−t1+z2+α2z2z2−31+ω1,which yields the results in (56) through some computation.
The next result clarifies that the necessary and sufficient condition for the convex combination offα1 and fα2 is in the class SH and convex in the direction of the imaginary axis if it is locally univalent and sense-preserving.Theorem 8.
Letfαj=hαj+g¯αj, for j=1,2, be two normalized harmonic mapping satisfying hαjz+gαjz=z1−αjz2/1−z2, 0≤αj≤1, and dilatation ωj=gαj′/hαj′ with ωj<1 in U. Then, the mappingf=tfα1+1−tfα2,0≤t≤1with dilatationωas given in (56), is inSHand maps into region convex in the direction of the imaginary axis, providedfis locally univalent and sense-preserving.Proof.
Define(60)f=h+g¯=thα1+1−thα2+tgα1+1−tgα2¯.
Let(61)h+g=thα1+gα1+1−thα2+gα2=thα1+gα1+1−thα2+gα2=tKα1+1−tKα2,where Kαj=hαj+gαj,j=1,2. Since the mapping fαj satisfies (49), it follows from (7) that(62)Re1−z2h′+g′=tRe1−z2Kα1′+1−tRe1−z2Kα2′>0,on U for all 0≤t≤1. Hence, according to Lemma 1, one can deduce that the mapping h+g is univalently convex in the direction of the imaginary axis. Also, If f=h+g¯ is sense-preserving locally univalent mapping in U, then in view of Theorem 3, f∈SH is univalent analytic mapping of U onto a region convex in the direction of imaginary axis.Theorem 9.
Forj=1,2, definefαj=hαj+g¯αjto be normalized sense-preserving harmonic mapping such thathαjz+gαjz=z1−αjz2/1−z2,0≤αj≤1withαj=α. Letωjbe the dilatation offαj(ωj<1inU). Ifω1≠ω2, then the mappingf=tfα1+1−tfα2,0≤t≤1, is inSHand is convex in the direction of the imaginary axis.Proof.
From Theorem8, it will be sufficient if we show that the dilatation of f given in (56) satisfies the condition ω<1 in U. Suppose that α1=α2=α in (56); then,(63)ω=1+z2ω1ω2+tω1+1−tω2+αz2z2−3ω1ω2+tω1+1−tω21+z21+tω2+1−tω1+αz2z2−31+tω2+1−tω1=ω1ω2+tω1+1−tω21+αz4+1−3αz21+tω2+1−tω11+z2+αz2z2−3=ω1ω2+tω1+1−tω21+tω2+1−tω1.
From the proof of Theorem3 in [21], it is clear that ω<1. Hence, we deduce that the mapping f is locally univalent and sense-preserving.
Next, we will consider one of the harmonic mappings involved in the linear combination induced by shearing analytic mappingz1−βz/1−z2 where β∈−1,1.Theorem 10.
Definefβ=hβ+g¯β, satisfying hβz+gβz=z1−βz/1−z2, −1≤β≤1, with dilatation ω1=gβ′/hβ′ lies in U and let fα=hα+g¯α, satisfying hαz+gαz=z1−αz2/1−z2, 0≤α≤1, with dilatation ω2=gα′/hα′ lies in U, be two normalized harmonic mapping. Then, the mappingfβ,α=tfβ+1−tfα,0≤t≤1is univalent and convex in the direction of the imaginary axis iffβ,αis locally univalent and sense-preserving.Proof.
Sincehαz+gαz=z1−αz2/1−z2 and hβz+hβz=z1−βz/1−z2, it follows that(64)h+g=thα+1−thβ+tgα+1−tgβ=thα+gα+1−thβ+gβ,which implies that(65)h′+g′=thα′+gα′+1−thβ′+gβ′.
Hence,(66)Re1−z2h′+g′=tRe1−z2hα′+gα′+1−tRe1−z2hβ′+gβ′,from (55). Since(67)Re1−z2hβ′+gβ′=1−z21+z2−2βRez1−z22>0,for all 0≤t≤1, we can easily see that(68)Re1−z2h′+g′>0.
In view of Theorem3, we deduce that the mapping f=h+g¯ is univalent and maps onto domain convex in the direction of the imaginary axis provided f is locally univalent and sense-preserving on U.
It is known that for0≤θ<π, the mapping fθ=hθ+g¯θ is called vertical strip mapping if it satisfies(69)hθz+gθz=12isinθlog1+eiθz1−eiθz.
Thus, Theorem10 can be expressed in terms of vertical strip mappings.Corollary 2.
Letfθ=hθ+g¯θ, where hθz+gθz=1/2isinθlog1+eiθz/1−eiθz with dilatation gθ′/hθ′ lies in U and fα=hα+g¯α, where hαz+gαz=z1−αz2/1−z2, 0≤α≤1 with dilatation gα′/hα′ lies in U, to be two normalized harmonic mapping. Then, the mappingfθ,α=tfθ+1−tfα,0≤t≤1, is univalent and convex in the direction of the imaginary axis iffθ,αis locally univalent and sense-preserving.



---

*Source: 6553600-2021-08-12.xml*