On Convolutions of Harmonic Univalent Mappings Convex in the Direction of the Real Axis∗

Home , Harmonic function

Journal of Applied Analysis and Computation Website:http://jaac-online.com/ Volume 6, Number 1, February 2016, 145–155 doi:10.11948/2016012

ON CONVOLUTIONS OF HARMONIC UNIVALENT MAPPINGS CONVEX IN THE DIRECTION OF THE REAL AXIS∗

Zhi-Gang Wang1,†, Zhi-Hong Liu2,3 and Ying-Chun Li3

Abstract In this paper, we show that convolutions of some planar harmonic mappings which convex in the direction of the real axis are also convex in the same direction. Furthermore, by means of the Mathematica software, we present an example to illuminate the main result. Keywords Harmonic univalent mapping, convolution, shear construction. MSC(2010) 58E20, 30C45.

1. Introduction

A complex-valued harmonic function f in the open unit disk D = {z ∈ C : |z| < 1} is given by f = h + g, where h and g are analytic functions in D. As usual, h is called the analytic part of f, and g is called the co-analytic part of f. The Jacobian 0 2 0 2 of the mapping f = h + g is given by Jf = |h | − |g | . A necessary and suﬃcient condition for f to be locally univalent and sense-preserving in D is that |g0| < |h0|, or equivalently, the dilatation ω = g0/h0 has the property |ω| < 1 in D for h0 6= 0 (see [3] or [11]). We denote SH by the class of harmonic, sense-preserving and univalent mappings in D, normalized by the conditions f(0) = 0 and fz(z) = 1. Thus, a harmonic mapping in the class SH can be expressed as f = h + g, where

∞ ∞ X n X n h(z) = z + anz and g(z) = bnz . n=2 n=1

0 Let SH be the subclass of SH whose members satisfy the additional condition 0 0 0 fz(0) = 0. Also, let KH and CH be the subclasses of SH whose image domains are convex and close-to-convex, respectively. A domain Ω ⊂ C is said to be convex in the direction γ, if for all a ∈ C, the set Ω ∩ {a + teiγ : t ∈ R} is either connected or empty. In particular, a domain is convex in the horizontal direction (CHD), if every line parallel to the real axis has a connected intersection with the domain. The shear construction is essential to the present work as it allows one to study harmonic functions through their related analytic functions, it produces a univalent harmonic function that maps D to a

†Corresponding author. Email address: [email protected] (Z.-G. Wang) 1School of Mathematics and Computing Science, Hunan First Normal University, Changsha 410205, Hunan, China 2School of Mathematics and Econometrics, Hunan University, Changsha 410082, Hunan, China 3College of Mathematics, Honghe University, Mengzi 661199, Yunnan, China ∗The authors were supported by National Natural Science Foundation of China (11301008, 11226088) and the Aid Program for Science and Technology Innovative Research Team in Higher Educational Institution of Hunan Province. 146 Z. Wang, Z. Liu & Y. Li region which is CHD. This construction relies on the following result due to Clunie and Sheil-Small [3]. Theorem A. A harmonic function f = h + g locally univalent in D is a univalent mapping of D onto a domain convex in the direction of the real axis if and only if h − g is a conformal univalent mapping of D onto a domain convex in the direction of the real axis. 0 A function f = h + g ∈ SH is called a CHD mapping if f maps D onto a CHD 0 domain. We denote all such CHD mappings by SCHD. Clearly, we know that 0 0 SCHD ⊂ CH . For two harmonic functions given by

∞ ∞ X n X n f(z) = h(z) + g(z) = z + anz + bnz n=2 n=2 and ∞ ∞ X n X n F (z) = H(z) + G(z) = z + Anz + Bnz , n=2 n=2 their convolution f ∗ F is deﬁned by

∞ ∞ X n X n f ∗ F = h ∗ H + g ∗ G = z + anAnz + bnBnz . n=2 n=2 For some recent investigations involving harmonic mappings and related topics, one can refer to [1,2,4,5,7–9,12–24,26,29–32]. In particular, Dorﬀ [8] and Dorﬀ et al. [9] derived convolutions involving right half-plane mappings, and obtained the following results, respectively. 0 0 Theorem B. Let f1 = h1 + g1 ∈ KH , f2 = h2 + g2 ∈ KH with hk + gk = z/(1 − z) 0 for k = 1, 2. If f1 ∗f2 is locally univalent and sense-preserving, then f1 ∗f2 ∈ SCHD. 0 0 0 Theorem C. Let f = h + g ∈ KH with h + g = z/(1 − z) and ω = g /h = iθ n + 0 e z (n ∈ Z ; θ ∈ R). If n = 1, 2, then f0 ∗ f ∈ SCHD, where

z − 1 z2 − 1 z2 f = h + g = 2 + 2 . 0 0 0 (1 − z)2 (1 − z)2

We remark that

z − 1 z2 − 1 z2 z z f (z) = 2 + 2 = < + i= 0 (1 − z)2 (1 − z)2 1 − z (1 − z)2 is the well-known right half-plane mapping. In this paper, we consider the harmonic mapping fc(z) = hc + gc satisﬁes the condition z h − g = c c 1 − z with the dilatation ω = z, by applying the shearing technique, we have

z − 1 z2 1 z z h = 2 = + , c (1 − z)2 2 (1 − z)2 1 − z On convolutions of harmonic univalent mappings 147 and

1 z2 1 z z g = 2 = − , c (1 − z)2 2 (1 − z)2 1 − z that is, z z f (z) = < + i= . (1.1) c (1 − z)2 1 − z

If a function F is analytic in D with F (0) = 0, then 1 h (z) ∗ F (z) = (zF 0(z) + F (z)) , c 2 and 1 g (z) ∗ F (z) = (zF 0(z) − F (z)) . c 2

In what follows, we consider the image domain fc(D). For convenience, we write z z f (z) = < + i= = u + iv. c (1 − z)2 1 − z

Let z = eiθ ∈ ∂D. Then z eiθ 2 cos θ − 2 1 u = < = < = = − , (1 − z)2 (1 − eiθ)2 (2 − 2 cos θ)2 2 − 2 cos θ and

z eiθ sin θ v = = = = = . 1 − z 1 − eiθ 2 − 2 cos θ

Thus, we get

1 v2 = − u + . 4

Since the point z = 0 is mapped into fc(0) = 0, we ﬁnd that

1 f ( ) = u + iv : v2 > − u + . c D 4

The images of concentric circles inside D under the harmonic mapping fc(z) are shown in Figure 1, which implies that fc(z) is a CHD mapping and not a right half-plane mapping. Recently, Nagpal and Ravichandran [25, Theorems 2.2 and 2.3] gave the radii of √ √ 1 q 1 convexity and starlikeness of fc(z) are 2 − 3 and 3 3 (37 − 8 10), respectively. In this paper, we aim at deriving convolutions of fc(z) and some special harmonic 0 0 mappings in the class SCHD are also belonging to the class SCHD. Furthermore, we present an example to illustrate the result with the aid of Mathematica software. 148 Z. Wang, Z. Liu & Y. Li

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Figure 1. Image of fc

2. Preliminaries

The following lemmas will be required in the proof of our main result.

Lemma 2.1. (see [27]) Let f be an analytic function in D with f(0) = 0 and f 0(0) 6= 0, and let z ϕ(z) = , (2.1) (1 + zeiθ1 )(1 + zeiθ2 ) where θ1, θ2 ∈ R. If zf 0(z) < > 0 (z ∈ ), ϕ(z) D then f is convex in the direction of the real axis. Lemma 2.2. (see [6, Cohn’s Rule]) Given a polynomial

n f(z) = a0 + a1z + ··· + anz of degree n, let

∗ n n f (z) = z f(1/z) = an + an−1z + ··· + a0z .

Denote by p and s the number of zeros of f inside the unit circle and on it, respectively. If |a0| < |an|, then

a f(z) − a f ∗(z) f (z) = n 0 1 z is of degree n − 1 with p1 = p − 1 and s1 = s the number of zeros of f1 inside the unit circle and on it, respectively. On convolutions of harmonic univalent mappings 149

Lemma 2.3. (see [10]) Let ϕ and G be analytic in D with ϕ0(0) = G(0) = 0. If ϕ is convex and G is starlike, then for each function F analytic in D and satisfying <(F (z)) > 0, we have (ϕ ∗ FG)(z) < > 0 (z ∈ ). (ϕ ∗ G)(z) D

0 Lemma 2.4. Let f1 = h1 + g1 ∈ SCHD with h1 − g1 = z/(1 − z), f2 = h2 + g2 ∈ 0 SCHD with 1 1 + zeiα h − g = log 2 2 2i sin α 1 + ze−iα

0 for π/2 ≤ α < π. If f1 ∗ f2 is locally univalent, then f1 ∗ f2 ∈ SCHD. Proof. Let

F1 = (h1 − g1) ∗ (h2 + g2) = h1 ∗ h2 + h1 ∗ g2 − h2 ∗ g1 − g1 ∗ g2, and

F2 = (h1 + g1) ∗ (h2 − g2) = h1 ∗ h2 − h1 ∗ g2 + h2 ∗ g1 − g1 ∗ g2. Thus, we have 1 h ∗ h − g ∗ g = (F + F ). (2.2) 1 2 1 2 2 1 2 1 Next, we shall prove that 2 (F1 + F2) is CHD. Since 1 1 + zeiα h − g = log , 2 2 2i sin α 1 + ze−iα we know that 0 0 0 0 0 0 0 h2 + g2 zF1 = (h1 − g1) ∗ [z (h2 + g2)] = (h1 − g1) ∗ z(h2 − g2) 0 0 h2 − g2 z z 1 + ω2 zp2(z) = ∗ iα −iα = iα −iα , 1 − z (1 + ze )(1 + ze ) 1 − ω2 (1 + ze )(1 + ze ) where

1 + ω2 p2(z) = 1 − ω2 satisﬁes the condition <(p2(z)) > 0. Thus, we get

0 ! zF1 < z = < (p2(z)) > 0. (2.3) (1+zeiα)(1+ze−iα) In what follows, we consider

0 0 0 0 0 0 0 h1 + g1 zF2 = [z (h1 + g1)] ∗ (h2 − g2) = z(h1 − g1) 0 0 ∗ (h2 − g2) h1 − g1 0 0 1 + ω1 zp1(z) = z(h1 − g1) ∗ (h2 − g2) = 2 ∗ (h2 − g2), 1 − ω1 (1 − z) 150 Z. Wang, Z. Liu & Y. Li where

1 + ω1 p1(z) = 1 − ω1 satisﬁes the condition <(p1(z)) > 0. Using the fact that z Ψ(z) ∗ = zΨ0(z) (1 − z)2 and h2 − g2 is convex, by Lemma 2.3, we have

0 ! (h − g ) ∗ p (z) z ! zF2 2 2 1 (1−z)2 < z = < 0 0 (1+zeiα)(1+ze−iα) z(h2 − g2) (2.4) z ! (h2 − g2) ∗ p1(z) (1−z)2 = < z > 0. (h2 − g2) ∗ (1−z)2

Combining (2.3) with (2.4), we get

0 ! z(F1 + F2) < z > 0, (1+zeiα)(1+ze−iα) by Lemma 2.1, we know that F1 + F2 is convex in the direction of the real axis. Thus, by Theorem A, we get the desired result of Lemma 2.4.

3. Main result

We now give the main result below. 0 Theorem 3.1. Let fn = hn + gn ∈ SCHD with 1 1 + zeiα h − g = log n n 2i sin α 1 + ze−iα

iθ n 0 for π/2 ≤ α < π and ωn = e z . If n = 1, 2, then fc ∗ fn ∈ SCHD, where fc is given by (1.1).

Proof. By Lemma 2.4, we only need to prove that fc ∗ fn = Hn + Gn are locally univalent. By noting that

1 1 + zeiα h − g = log n n 2i sin α 1 + ze−iα

0 0 00 00 0 0 and gn = ωnhn, we have gn = ωnhn + ωnhn. Therefore, we know that (g ∗ g )0 (zg0 − g )0 zg00 ω h00 + ω0 h0 ω(z) = c n = n n = n = z n n n n . (3.1) e 0 0 0 0 00 0 00 (hc ∗ hn) (zhn + hn) 2hn + zhn 2hn + zhn Moreover, we observe that

0 1 hn = iα −iα , (3.2) (1 − ωn)(1 + ze )(1 + ze ) On convolutions of harmonic univalent mappings 151 and 0 2 00 2(cos α + z)(1 − ωn) − ωn(1 + 2z cos α + z ) hn = − 2 iα 2 −iα 2 . (3.3) (1 − ωn) (1 + ze ) (1 + ze ) By substituting (3.2) and (3.3) into (3.1), yields

2 1 0 1 0 1+z cos α ω − ωn − ω z + ω ω(z) = z n 2 n 2 n cos α+z . (3.4) e 1+z cos α 1 0 1+z cos α 1 0 2 cos α+z − ωn − 2 ωnz cos α+z + 2 ωnz

iθ Now, we consider the case ω1 = e z, then

e2iθz2 − eiθz − 1 eiθz + 1 eiθz 1+z cos α ω(z) = z 2 2 cos α+z e 1+z cos α iθ 1 iθ 1+z cos α 1 iθ 2 cos α+z − e z − 2 e z cos α+z + 2 e z z3 + cos α − 1 e−iθ z2 + 1 e−iθ = ze2iθ 2 2 1 iθ 1 iθ 3 1 + cos α − 2 e z + 2 e z p(z) = ze2iθ . p∗(z)

∗ 3 1 ∗ Note that p (z) = z p(1/z), if z0 is one zero of p(z), then is one zero of p (z). z0 Thus, we have

(z + A)(z + B)(z + C) ω(z) = ze2iθ , e (1 + Az)(1 + Bz)(1 + Cz) where α ∈ [π/2, π) and θ ∈ [−π, π]. It is suﬃcient to show that A, B, C ∈ D. We apply Cohn’s Rule to 1 1 p(z) = z3 + cos α − e−iθ z2 + e−iθ. 2 2

1 −iθ 1 By noting that 2 |e | = 2 < 1, we obtain a p(z) − a p∗(z) 3 1 1 1 p (z) = 3 0 = z2 + cos α − e−iθ z − e−iθ cos α − eiθ . 1 z 4 2 2 2

Since 1 −iθ 1 iθ 1 1 3 − e cos α − e ≤ + | cos α| < , 2 2 4 2 4 we use Cohn’s Rule on p1(z) again, then

3 p (z) + 1 e−iθ cos α − 1 eiθ p∗(z) p (z) = 4 1 2 2 1 2 z 3 3 z2 + cos α − 1 e−iθ z − 1 e−iθ cos α − 1 eiθ = 4 4 2 2 2 z 1 e−iθ cos α − 1 eiθ 3 + (cos α − 1 e−iθ)z − 1 e−iθ(cos α − 1 eiθz2) + 2 2 4 2 2 2 z 9 1 1 3 1 1 1 2 = − | cos α − eiθ|2 z + cos α − e−iθ + e−iθ cos α − eiθ 16 4 2 4 2 2 2 152 Z. Wang, Z. Liu & Y. Li has one zero at

2 − 3 cos α − 1 e−iθ − 1 e−iθ cos α − 1 eiθ z = 4 2 2 2 0 9 1 1 iθ 2 16 − 4 | cos α − 2 e | − 3 cos α + 3 e−iθ − 1 e−iθ cos2 α + 1 cos α − 1 eiθ = 4 8 2 2 8 9 1 1 iθ 1 −iθ 16 − 4 cos α − 2 e cos α − 2 e − cos α − 2e−iθ cos2 α + 3 e−iθ − 1 eiθ = 2 2 . 2 − cos2 α + cos α cos θ

By noting that

2 2 2 −iθ 2 3 −iθ 1 iθ 2 − cos α + cos α cos θ − cos α + 2e cos α − e + e 2 2 =4 − 4 cos2 α + cos4 α + 4 cos α cos θ − 2 cos3 α cos θ + cos2 α cos2 θ 3 1 3 1 − cos α + 2e−iθ cos2 α − e−iθ + eiθ cos α + 2eiθ cos2 α − eiθ + e−iθ 2 2 2 2 =4 − 4 cos2 α + cos4 α + 4 cos α cos θ − 2 cos3 α cos θ + cos2 α cos2 θ 7 1 3 − 4 1 + cos4 α + cos3 α cos θ − cos2 α + cos2 α cos2 θ − cos α cos θ − cos2 θ 4 2 4 =3 − cos4 α − 2 cos3 α cos θ − cos2 α cos2 θ + cos2 α + cos α cos θ + cos2 θ 2 =3 1 − cos2 α)(cos α + cos θ ≥ 0, which shows that |z0| ≤ 1. Thus, by Cohn’s Rule, we know that f exists three zeros in D, that is A, B, C ∈ D, and so |ωe(z)| < 1 for z ∈ D. iθ 2 iθ 2 Finally, we consider the case ω2 = e z , by substituting ω2 = e z into (3.4), we have

eiθz3 + 1+z cos α ! ω(z) = z2eiθ cos α+z = z2eiθ, e 1+z cos α iθ 3 cos α+z + e z which implies that |ωe(z)| < 1. The proof of Theorem 3.1 is completed. Remark 3.1. The range of the dilatation function ωe(z) in Theorem 3.1 is not n contained in D for n ≥ 3. To check this, we choose ωn(z) = z and substitute it into (3.4), yields

zn+1 + ( n − 1)z + n 1+z cos α ω(z) = zn 2 2 cos α+z = znR(z). e 1+z cos α n n 1+z cos α n n+1 cos α+z − ( 2 − 1)z cos α+z + 2 z

It is a simple calculation to see that R(eiα) = 1 and 1/ R(1/z) = R(z). So R(z) maps the closed unit disk |z| ≤ 1 onto itself. Hence R can be written as a ﬁnite Blaschke product of order n + 1. However, n/2 is product of the module of zeros of R in the unit disk D. This means that there exists a point z0 ∈ D such that |ωe(z0)| > 1 for n ≥ 3. Thus, the restriction of n = 1, 2 in Theorem 3.1 becomes necessary for our result. Finally, we give an example to illuminate the main result. On convolutions of harmonic univalent mappings 153

Example 3.1. In Theorem 3.1, by setting f1 = h1 + g1 with

1 1 + iz h − g = log 1 1 2i 1 − iz and ω1 = z, we get

1 1 − i 1 + i h = − log(1 − z) + log(1 + iz) + log(1 − iz), 1 2 4 4 and

1 1 + i 1 − i g = − log(1 − z) + log(1 + iz) + log(1 − iz). 1 2 4 4

Consider the function

F1 = fc ∗ f1 = H1 + G1, we have 1 H = h ∗ h = (zh0 + h ) 1 c 1 2 1 1 1 z 1 1 − i 1 + i = − log(1 − z) + log(1 + iz) + log(1 − iz) , 2 (1 + z2)(1 − z) 2 4 4 and 1 G = g ∗ g = (zg0 − g ) 1 c 1 2 1 1 1 z2 1 1 + i 1 − i = + log(1 − z) − log(1 + iz) − log(1 − iz) . 2 (1 + z2)(1 − z) 2 4 4

Thus, we obtain

F1 =<(H1 + G1) + i=(H1 − G1) z + z2 1 1 + iz =< + log 2(1 + z2)(1 − z) 4i 1 − iz z 1 1 + i= − log(1 − z) + log 1 + z2 . 2(1 + z2) 2 4

The images of concentric circles inside D under the harmonic mappings fc and f1 are shown in Figure 1 and Figure 2, respectively. The images of these concentric circles under the convolution map fc ∗ f1 = F1 are shown in Figure 3.

Acknowledgments

The authors would like to thank the referees for their valuable comments and sug- gestions, which essentially helped to improve the quality of this paper. 154 Z. Wang, Z. Liu & Y. Li

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Figure 2. Image of f1 Figure 3. Image of fc ∗ f1

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