# Convolutions of Harmonic Mappings Convex in the Horizontal Direction

Authors: Serkan Çakmak; Elif Yaşar; Sibel Yalçın
Journal: Journal of Function Spaces
 (2021)

---

## Abstract

In this paper, we establish some results concerning the convolutions of harmonic mappings convex in the horizontal direction with harmonic vertical strip mappings. Furthermore, we provide examples illustrated graphically with the help of Maple to illuminate the results.


---

## Body

## 1. Introduction

For real-valued harmonic functionsu and v in the open unit disk E=z∈ℂ:z<1, the complex-valued continuous function f=u+iv is said to be harmonic and can be expressed as f=h+g¯, where h and g are analytic in E. Let H be the class of harmonic mappings f=h+g¯ normalized by h0=g0=h′0−1=0 and have the following power series representations:
(1)hz=z+∑m=2∞amzm,gz=∑m=1∞bmzm.We callh the analytic part and g the coanalytic part of f, respectively. The Jacobian of f=h+g¯ is given by Jf=h′2−g′2. Lewy’s theorem [1] implies that f∈H is locally univalent and sense-preserving if and only if Jf>0 in E. The condition Jf>0 is equivalent to that dilatation ωz=g′z/h′z satisfying ωz<1 for all z∈E (see [2, 3]).We denote bySH the class of all harmonic, sense-preserving, and univalent mappings f=h+g¯ in E, which are normalized by the condition h0=g0=0 and h′0=1. Let SH0 be the subset of all f∈SH in which g′0=0. Further, let KH,ℂH (resp., KH0,ℂH0) be the subset of SH (resp., SH0) whose images are convex and close-to-convex domains. A domain Ω is said to be convex in the horizontal direction (CHD) if the intersection of Ω with each horizontal line is connected (or empty). A function f=h+g¯∈SH is said to be a CHD mapping if f maps E onto a CHD domain. Let SCHD be the subset of ℂH which consist of CHD mappings. The following basic theorem of Clunie and Sheil-Small [2] is known as shear construction that constructs harmonic mappings with prescribed dilatations onto a domain convex in one direction.Theorem 1 (see [2]).
A locally univalent harmonic mappingf=h+g¯ in E is a univalent mapping of E onto a domain convex in a direction ϕ if and only if h−e2iϕg is a conformal univalent mapping of E onto a domain convex in the direction of ϕ.Letf∗F=h∗H+g∗G¯ be the convolution of two harmonic functions f=h+g¯ and F=H+G¯ where the operator ∗ is convolution (or Hadamard product) of two power series.There are several research papers in recent years which investigate the convolution of harmonic univalent functions. In particular, Dorff [4] and Dorff et al. [5] studied the convolution of harmonic univalent mappings in the right half-plane. For some recent investigations involving convolution of harmonic mappings, we refer the reader to [6–13].LetFa=Ha+Ga¯ sheared by Ha−Ga=z/1−z with the dilatation ωa=a+z/1+az, where a∈−1,1. Using shear construction of Clunie and Sheil-Small [2], we have
(2)Haz=1/1−az−1/2z21−z2=12z1−z+1+a1−az1−z2,(3)Gaz=a/1−az+1/2z21−z2=12−z1−z+1+a1−az1−z2.It is clear that by settinga=0 in (2) and (3), we obtain F0=H0+G0¯ which satisfy the conditions H0−G0=z/1−z and ωz=z, studied by Liu and Li [8]. Wang et al. [14] also studied convolutions of this mapping. Note that Fa is a CHD mapping.Recently, Liu and Li [8] introduced the following generalized harmonic univalent mappings:
(4)Pδz=Hδz+Gδz¯=11+δδz1−z2+z1−z+11+δδz1−z2−z1−z¯,where δ>0 and z∈E. Obviously, P1z=F0z. If f=h+g¯∈SH, then
(5)Pδ∗f=δzh′+h1+δ+δzg′−g1+δ¯.Also,Pδz maps E onto the domain u+iv:v2>−2δ/1+δu+1/1+δ2,δ>0 which is a CHD domain. Very recently, Yasar and Ozdemir [15] studied convolutions of these generalized harmonic mappings.Letfγ=hγ+gγ¯∈SCHD0 with
(6)hγ−gγ=12isinγlog1+zeiγ1+ze−iγ,where π/2≤γ<π.In this paper, we investigate the conditions under which the convolutions of harmonic mappingsPδ, fγ, and Fa with prescribed dilatations are univalent and CHD provided that the convolutions are locally univalent and sense-preserving. Furthermore, we provide two examples illustrated graphically with the help of Maple to illuminate our results.


## 2. Preliminary Results

Lemma 2 (see [16]).
Letf be an analytic fuction in E with f0=0 and f′0≠0 and let
(7)φz=z1+zeiθ11+zeiθ2,where θ1,θ2∈ℝ. If
(8)Rezf′zφz>0,then f is convex in the horizontal direction.Lemma 3 (see [17]).
Letφ and G be analytic in E with φ′0=G0=0. If φ is convex and G is starlike, then for each function F analytic in E and satisying ReFz>0, we have
(9)Reφ∗FGzφ∗Gz>0z∈E.Lemma 4 ([18], Cohn’s rule).
Given a polynomial(10)pz=p0z=ak,0zk+ak−1,0zk−1+⋯+a1,0z+a0,0ak,0≠0of degree k, let
(11)p∗z=p0∗z=zkp1z¯¯=ak,0¯+ak−1,0¯z+⋯+a1,0¯zk−1+a0,0¯zkak,0≠0.
Denote byr and s the number of zeros of pz inside the unit circle and on it, respectively. If a0,0<ak,0, then
(12)p1z=ak,0¯pz−a0,0p∗zzis of degree k−1 with r1=r−1 and s1=s the number of zeros of p1z inside the unit circle and on it, respectively.Lemma 5.
LetPδ=Hδ+Gδ¯ be defined by (4) and fγ=hγ+gγ¯ be defined by (6) with dilatation ω=gγ′/hγ′. Then the dilatation of Pδ∗fγ is given by
(13)ω~z=ω1−ωδ−1−2zcosγ−δ+1z2+δzω′1+2zcosγ+z21−ωδ+1+2zcosγ−δ−1z2+δzω′1+2zcosγ+z2.Proof.
Sincehγ−gγ=1/2isinγlog1+zeiγ/1+ze−iγπ/2≤γ<π and gγ′=ωhγ′, then gγ′′=ω′hγ′+ωhγ′′. We immediately get
(14)hγ′=11−ω1+zeiγ1+ze−iγ,(15)hγ′′=−2cosγ+z1−ω−ω′1+2zcosγ+z21−ω21+zeiγ21+ze−iγ2.
From (4), we have
(16)ω~z=Gδ∗gγ′Hδ∗hγ′=δzgγ′−gγ′δzhγ′+hγ′=δ−1gγ′+δzgγ′′δ+1hγ′+δzhγ′′=δ−1ωhγ′+δzω′hγ′+ωhγ′′δ+1hγ′+δzhγ′′=ω1−ωδ−1−2zcosγ−δ+1z2+δzω′1+2zcosγ+z21−ωδ+1+2zcosγ−δ−1z2+δzω′1+2zcosγ+z2.☐Lemma 6.
LetPδ=Hδ+Gδ¯be defined by (4) andfγ=hγ+gγ¯be defined by (6). IfPδ∗fγis locally univalent and sense-preserving, thenPδ∗fγis univalent and convex in the horizontal direction.Proof.
Let(17)F1=Hδ−Gδ∗hγ+gγ=Hδ∗hγ+Hδ∗gγ−Gδ∗hγ−Gδ∗gγ,F2=Hδ+Gδ∗hγ−gγ=Hδ∗hγ−Hδ∗gγ+Gδ∗hγ−Gδ∗gγ.
Thus,(18)Hδ∗hγ−Gδ∗gγ=12F1+F2.
By Theorem1, we need to prove that 1/2F1+F2 is convex in the horizontal direction. Since
(19)hγ−gγ=12isinγlog1+zeiγ1+ze−iγπ2≤γ<π,we have
(20)zF1′=Hδ−Gδ∗zhγ′+gγ′=Hδ−Gδ∗zhγ′−gγ′hγ′+gγ′hγ′−gγ′=2z1+δ1−z∗z1+zeiγ1+ze−iγ1+ωγ1−ωγ=2zp1z1+δ1+zeiγ1+ze−iγ,where p1z=1+ωγ/1−ωγ satisfies the condition Rep1z>0. Thus, we have
(21)RezF1′2z/1+δ1+zeiγ1+ze−iγ=Rep1z>0.
Now, we consider(22)zF2′=zHδ′+Gδ′∗hγ−gγ=zHδ′−Gδ′Hδ′+Gδ′Hδ′−Gδ′∗hγ−gγ=zHδ′−Gδ′1+ωδ1−ωδ∗hγ−gγ=2zp2z1+δ1−z2∗hγ−gγ,where p2z=1+ωδ/1−ωδ satisfies the condition Rep2z>0. Using the fact that
(23)ψz∗z1−z2=zψ′zand hγ−gγ is convex, by Lemma 3, we have
(24)RezF2′z/1+zeiγ1+ze−iγ=Rehγ−gγ∗p2z2z/1+δ1−z2zhγ′−gγ′=Rehγ−gγ∗p2z2z/1+δ1−z2hγ−gγ∗z/1−z2>0.☐Finally, using Lemma2, we obtain that F1+F2 is convex in the horizontal direction.Lemma 7.
Letfγ=hγ+gγ∈SCHD0 be given by (6) with dilatation ω=gγ′/hγ′andFa=Ha+Ga¯ be a mapping defined by (2) and (3). Then the dilatation of Fa∗fγ is given by
(25)W~z=2ω1−ωa−1−azcosγ−z2+1+azω′1+2zcosγ+z221−ω1+1−azcosγ−az2+1+azω′1+2zcosγ+z2.Proof.
From (2) and (3), we have
(26)W~z=Ga∗gγ′Ha∗hγ′=1+azgγ′−1−agγ′1+azhγ′+1−ahγ′=2agγ′+1+azgγ′′2hγ′+1+azhγ′′=2aωhγ′+1+azω′hγ′+ωhγ′′2hγ′+1+azhγ′′.
Using (14) and (15), then we obtain the dilatation of Fa∗fγ as follows:
(27)W~z=2ω1−ωa−1−azcosγ−z2+1+azω′1+2zcosγ+z221−ω1+1−azcosγ−az2+1+azω′1+2zcosγ+z2.☐Lemma 8 ([14], Lemma 2.4).
LetFa=Ha+Ga¯ be a mapping defined by (2), (3) and fγ=hγ+gγ¯∈SCHD0 be defined by (6). If Fa∗fγ is locally univalent and sense-preserving, then Fa∗fγ is univalent and convex in the horizontal direction.Lemma 9 ([19], Gauss-Lucas theorem).
LetTz be a nonconstant polynomial with complex coefficients. Then, the zeros of the derivative T′z are contained in the convex hull of the set of the zeros of Tz.Lemma 10.
Let(28)qz=zk+1+2k+1cosγk+2zk+kk+2zk−1−2k+2e−iθz−2cosγk+2e−iθbe a complex polynomial of degree k+1, where θ∈ℝ,k∈ℕ+, and π/2≤γ<π. Then, all zeros of qz lie in the closed unit disk z≤1.Proof.
Note thatqz=1/k+2T′z, where
(29)Tz=zk−e−iθ1+2zcosγ+z2.
It is obvious that the roots ofzk−e−iθ lie on the unit circle. Also, −cosγ±isinγ which are the roots of 1+2zcosγ+z2 lie on the unit circle as well. Hence, the result follows from Lemma 9.☐


## 3. Main Results

Theorem 11.
LetPδ=Hδ+Gδ¯∈SCHD be a mapping given by (4) and fγ=hγ+gγ¯∈SCHD0 be given by (6) with the dilatation ωk=gγ′/hγ′=eiθzkθ∈ℝ,k∈ℕ+. Then Pδ∗fγ is univalent and convex in the horizontal direction.Proof.
By Lemma6, we need to prove that the dilatation ω~ of Pδ∗fγ satisfies ω~<1 for all z∈E. Substituting ω=eiθzk in (13), we yield
(30)ω~z=eiθzk1−eiθzkδ−1−2zcosγ−δ+1z2+δkeiθzk1+2zcosγ+z21−eiθzkδ+1+2zcosγ−δ−1z2+δkeiθzk1+2zcosγ+z2=e2iθzktzt∗z,where
(31)tz=zk+2+2cosγ1+δzk+1+1−δ1+δzk+δk−1−11+δe−iθz2+2kδ−1cosγ1+δe−iθz+δ1+k−11+δe−iθ,(32)t∗z=1+2cosγ1+δz+1−δ1+δz2+δk−1−11+δeiθzk+2kδ−1cosγ1+δeiθzk+1+δ1+k−11+δeiθzk+2.
If we substituteδ=2/k into (30), then tz/t∗z=e−iθ, and it is clear that ω~<1 for all z∈E. Now, we need to show that ω~<1 for 0<δ<2/k. Obviously, if z0 is a zero tz, then 1/z0¯ is zero of t∗z. Then, we may write
(33)ω~z=e2iθzkz+A1z+A2⋯z+Ak+21+A1¯z1+A2¯z⋯1+Ak+2¯z.
Using Lemma4, we only need to show that all zeros of (31) lie in the closed unit disk for 0<δ<2/k. Since a0,0=δ1+k−1/1+δe−iθ=δ1+k−1/1+δ<ak+2,0=1 for 0<δ<2/k, thus we have
(34)t1z=ak+2,0¯tz−a0,0t∗zz=δk+22−kδ1+δ2zk+1+2k+1cosγk+2zk+kk+2zk−1−2k+2e−iθz−2cosγk+2e−iθ.
By Lemma10, we know that all zeros of
(35)qz=zk+1+2k+1cosγk+2zk+kk+2zk−1−2k+2e−iθz−2cosγk+2e−iθlie inside the closed disk. Then, by Cohn’s rule, tz given by (31) has all its zeros in the closed unit disk. The proof is complete.☐Theorem 12.
LetFa be a mapping given by (2) and fγ=hγ+gγ∈SCHD0 be a mapping given by (6) with the dilatation ωk=gγ′/hγ′=eiθzkθ∈ℝ,k∈ℕ+. Then, Fa∗fγ is univalent and convex in the horizontal direction for −1<a≤2−k/2+k.Proof.
By Lemma8, we need to prove that Fa∗fγ is locally univalent and sense-preserving, i.e., the dilatation W~ of Fa∗fγ satisfies ∣W~z∣<1 for all z∈E. Substituting ω=eiθzk in (25),
(36)W~z=e2iθzk×zk+2+1−acosγzk+1−azk+1+ak−2/2e−iθz2+k−1+ak+1cosγe−iθz+k+2a+k/2e−iθ1+1−acosγz−az2+1+ak−2/2eiθzk+k−1+ak+1cosγeiθzk+1+k+2a+k/2eiθzk+2=e2iθzkuzu∗z, where
(37)uz=zk+2+1−acosγzk+1−azk+1+ak−2/2e−iθz2+k−1+ak+1cosγe−iθz+k+2a+k/2e−iθ,u∗z=1+1−acosγz−az2+1+ak−2/2eiθzk+k−1+ak+1cosγeiθzk+1+k+2a+k/2eiθzk+2.
If we substitutea=2−k/2+k into (36), we yield
(38)W~z=e2iθzkzk+2+2k/k+2cosγzk+1−2−k/k+2zk−2−k/k+2e−iθz2+2k/k+2cosγe−iθz+e−iθ1+2k/k+2cosγz−2−k/k+2z2−2−k/k+2eiθzk+2k/k+2cosγeiθzk+1+eiθzk+2=eiθzk.
Hence,∣W~z∣=eiθzk<1.
Next, we will show that∣W~z∣<1 for all −1<a<2−k/2+k. If z0 is a zero of uz, then 1/z0¯ is zero of u∗z; hence,
(39)W~z=e2iθzkuzu∗z=e2iθzkz+A1z+A2⋯z+Ak+21+A1¯z1+A2¯z⋯1+Ak+2¯z.
By Lemma4, we need to show that all zeros of uz lie inside or on the unit disk for −1<a<2−k/2+k. Since
(40)a0,0=k+2a+k2e−iθ<1=ak+2,0for−1<a<2−k2+k,from (12), we have
(41)u1z=ak+2,0¯uz−a0,0u∗zz=−k+21+ak+2a+k−24zk+1+2k+1cosγk+2zk+kk+2zk−1−2k+2e−iθzk−2cosγk+2e−iθ=−k+21+ak+2a+k−24qz,where
(42)qz=zk+1+2k+1cosγk+2zk+kk+2zk−1−2k+2e−iθz−2cosγk+2e−iθ.
Becausek+21+ak+2a+k−2/4≠0 for −1<a<2−k/2+k, it follows that both u1z and qz have the same zeros. By Lemma 10, we know that all zeros of qz lie inside the closed unit disk. Then, by Cohn’s rule, we know that all zeros uz lie inside or on the boundary of the unit disk. The proof is completed.☐Theorem 13.
LetF0=H0+G0¯∈SCHD0 be a harmonic mapping with H0−G0=z/1−z and dilatation G0′z/H0′z=z. Let fπ/2=hπ/2+gπ/2¯∈SCHD be a mapping defined by (6) with γ=π/2 and dilatation ωμz=μ+z2/1+μz2,−1<μ<1. Then the mapping F0∗fπ/2 is univalent and convex in the horizontal direction.Proof.
Sincefπ/2=hπ/2+gπ/2¯∈SCHD is a mapping defined by (6) with γ=π/2, we have
(43)hπ/2z−gπ/2z=12ilog1+iz1−iz.
Therefore, we know that(44)W~z=G0∗gπ/2′H0∗hπ/2′=zgπ/2′−gπ/2′zhπ/2′+hπ/2′=zgπ/2′′2hπ/2′+zhπ/2′′=zωμhπ/2′′+ωμ′hπ/2′2hπ/2′+zhπ/2′′.
Substituting(45)hπ/2′z=1ωμ1+z2,hπ/2′′z=ωμ′1+z2−2zωμ1−ωμ21+z22,into (44) yields
(46)W~z=zωμ2−ωμ−1/2ωμ′z+1/2ωμ′1/z1/z−ωμ−1/2ωμ′z1/z+1/2ωμ′z2.
Settingωμz=μ+z2/1+μz2 in the above equation, we get W~z=z2, and hence, ∣W~z∣<1 for all z∈E.☐Example 14.
Supposefγ=hγ+gγ¯∈SCHD0be given by ((6)). If we setγ=π/2andω1=−z3then by shear construction of Clunie and Sheil-Small [2], we have(47)hγz=16log1+z−i4log1+iz1−iz+14log1+z2−13log1−z+z2,gγz=16log1+z+i4log1+iz1−iz+14log1+z2−13log1−z+z2.
Recall that, iff=h+g¯∈SH, then
(48)Pδ∗f=δzh′+h1+δ+δzg′−g1+δ¯.
So, we have(49)Pδ∗fγ=11+δδzhγ′z+hγz+11+δδzgγ′z−gγz¯=11+δδz1+z31+z2+16log1+z−i4log1+iz1−iz+14log1+z2−13log1−z+z2+11+δ−δz41+z31+z2−16log1+z−i4log1+iz1−iz¯−14log1+z2+13log1−z+z2¯=Re11+δδz1−z31+z31+z2−i2log1+iz1−iz+iIm11+δδz1+z2+13log1+z+12log1+z2−23log1−z+z2.Now, in view of Theorem11, if we set the parameter δ=2/3, then Pδ∗fγ is univalent and CHD. Also, if we choose δ=3/4, then Pδ∗fγ is not guaranteed to be univalent. The images of ∣z∣=r<1 under fΠ/2,P2/3,P2/3∗fΠ/2 and P3/4∗fΠ/2 are shown in Figures 1–4.Example 15.
Supposefγ=hγ+gγ¯∈SCHD0 be given by (6). If we set γ=2π/3 and ω2=z4, then calculations lead to
(50)hγz=112log1+z+14log1+z21−z−16log1−z+z2−i36log1+ze2π/3i1+ze−2π/3i,gγz=112log1+z+14log1+z21−z−16log1−z+z2+i36log1+ze2π/3i1+ze−2π/3i.
Iff=h+g¯∈SH, then
(51)Fa∗f=121+azh′1−a+h+121+azg′1−a−g¯.
So, we have(52)Fa∗fγ=121+azhγ′1−a+hγ+121+azgγ′1−a−gγ¯=121+az1−a1−z+z21−z4+112log1+z+14log1+z21−z−16log1−z+z2−i36log1+ze2π/3i1+ze−2π/3i+121+az51−a1−z+z21−z4−112log1+z−14log1+z21−z¯+16log1−z+z2−i36log1+ze2π/3i1+ze−2π/3i¯=Re121+az1+z41−a1−z+z21−z4−i33log1+ze2π/3i1+ze−2π/3i+iIm121+az1−a1−z+z2+16log1+z+12log1+z21−z−13log1−z+z2.Figure 1
Image offΠ/2.Figure 2
Image ofP2/3.Figure 3
Image ofP2/3∗fΠ/2.Figure 4
Image ofP3/4∗fΠ/2.Now, if we set the parametera=−1/3, in view of Theorem 11, Fa∗fγ is univalent and CHD. If we choose a=−1/4, then Fa∗fγ is not guaranteed to be univalent (see Figures 5–8).Figure 5
Image off2Π/3.Figure 6
Image ofF−1/3.Figure 7
Image ofF−1/3∗f2Π/3.Figure 8
Image ofF−1/4∗f2Π/3.



---

*Source: 2949573-2021-09-06.xml*