# Convolutions of Harmonic Functions with Certain Dilatations

Authors: Om P. Ahuja; Jay M. Jahangiri
Journal: International Journal of Mathematics and Mathematical Sciences
 (2017)

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## Abstract

The convolution of harmonic functions, unlike the analytic case, proved to be very challenging. In this paper, we introduce dilatation conditions that guarantee the convolution of two harmonic functions to be locally one-to-one, sense-preserving, and close-to-convex harmonic in the unit disk.


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## Body

## 1. Introduction

LetA denote the class of functions that are analytic in the open unit disc D≔{z∈C:|z|<1} and let A′ be the subclass of A consisting of functions h with the normalization h(0)=h′(0)-1=0. We let K(α) denote the class of functions h∈A′ so that(1)Re1+zh′′zh′z>α;-12≤α;z∈D.Consider the family of complex-valued harmonic functionsf=u+iv defined in D, where u and v are real harmonic in D. Such functions can be expressed as f=h+g¯, where h∈A and g∈A. Clunie and Sheil-Small in their remarkable paper [1] explored the functions of the form f=h+g¯ that are locally one-to-one, sense-preserving, and harmonic in D. By Lewy’s Theorem (see [2] or [1]), a necessary and sufficient condition for the harmonic function f=h+g¯ to be locally one-to-one and sense-preserving in D is that its Jacobian Jf=|h′|2-|g′|2 is positive or equivalently, if and only if h′(z)≠0 in D and the second complex dilatation ω of f satisfies |ω|=|g′/h′|<1 in D.In an interesting article, Bshouty and Lyzzaik [3] proved the following.Theorem 1.
Letf=h+g¯ be a harmonic mapping of D, with h′(0)≠0, that satisfies g′(z)=zh′(z) and h∈K(-1/2) for all z∈D. Then f is a univalent close-to-convex mapping.A simply connected proper subdomain ofC is said to be close-to-convex if its complement in C is the union of closed half-lines with pairwise disjoint interiors. Consequently, a univalent analytic or harmonic function f:D→C is said to be close-to-convex if f(D) is close-to-convex (e.g., see Clunie and Sheil-Small [1] or Bshouty and Lyzzaik [3]).Ruscheweyh and Sheil-Small in a striking article [4] proved that the Hadamard product or convolution of two analytic convex functions is also convex analytic and that the convolution of an analytic convex function and an analytic close-to-convex function is close-to-convex analytic in the unit disk D. Ironically, these results could not be extended to the harmonic case, since the convolution of harmonic functions, unlike the analytic case, proved to be very challenging. The purpose of the present paper is to introduce dilatation conditions that guarantee the convolution of two harmonic functions to be locally one-to-one, sense-preserving, and close-to-convex harmonic in the unit disk D. In other words, we extend Theorem 1 to the convolution of two harmonic functions f1=h1+g1¯ and f2=h2+g2¯ with certain dilatations, where h1∗h2∈K(α).The operator∗ stands for the convolution or Hadamard product of two power series h1(z)=∑n=1∞anzn and h2(z)=∑n=1∞bnzn given by h1(z)∗h2(z)=(h1∗h2)(z)=∑n=1∞anbnzn. Similarly, the convolution of two harmonic functions f1=h1+g1¯ and f2=h2+g2¯ is given by f1∗f2=h1∗h2+g1∗g2¯.In regard to the convolution of harmonic univalent functions, Clunie and Sheil-Small [1] proved the following.Theorem 2.
Ifϕ∈K(0) and if f is convex harmonic in D, then their convolution (ϕ+εϕ¯)∗f(|ε|≤1) is close-to-convex harmonic in D.A mappingf:D→C is called convex harmonic if f(D) is a convex domain.The convexity condition for the functionϕ in Theorem 2 cannot be compromised as it is demonstrated in the following.Example 3.
Set(2)fz=hz+gz¯=z-1/2z21-z2-1/2z21-z2¯and consider the starlike analytic function ϕ(z)=z+zn/n; n≥2 in D. Letting ε=0 in Theorem 2, we observe that the harmonic convolution (3)ϕz+0·ϕz¯∗hz+gz¯=z+n+12nznis not even univalent in D.In an attempt to investigate the possibilities of improving the required convexity condition forϕ, the authors in [5] proved the following.Theorem 4.
Letϕ∈K(0) and h∈K(0). Also let w be a Schwarz function. Then the convolution function (4)ϕz+εϕz¯∗hz+∫0zwth′tdt¯;ε=1,is close-to-convex harmonic in D.Theorem4 for ϕ(z)=z/(1-z) and ε=1 yields a theorem given by Bshouty et al. ([6], Theorem 2). From what is said above, especially Example 3, one wonders if there are other conditions that guarantee the close-to-convexity of the convolution of two harmonic functions. In the following theorem, we find such conditions.Theorem 5.
Leth1∈A′ and h2∈A′ so that h1∗h2∈K(α), where K(α) is given by inequality (1). If one of the following conditions hold(i)
α≥-1/2; g1(z)=zh1(z) and g2′(z)=zh2′(z),(ii)
α≥0; g1′(z)=znh1′(z) and g2′(z)=znh2′(z), where n∈N,
then the convolution functionF(z)=h1(z)∗h2(z)+g1(z)∗g2(z)¯ is locally one-to-one, sense-preserving, and close-to-convex harmonic in D.Since the convolution of two convex analytic functions is also convex (see Ruscheweyh and Sheil-Small [4]), an obvious consequence of the above theorem would be as follows.Corollary 6.
Leth1∈K(0) and h2∈K(0) and set g1(z)=znh1′(z) and g2′(z)=znh2′(z). Then the convolution function F(z)=h1(z)∗h2(z)+g1(z)∗g2(z)¯ is locally one-to-one, sense-preserving, and close-to-convex harmonic in D.


## 2. Preliminary Lemmas and Proof of Theorem5

To prove our Theorem5, we shall need the following three lemmas, the first of which is a celebrated result by Clunie and Sheil-Small [1] and the second one is given by Kaplan [7]. The third lemma which is on subordination is a modification of a result given by Miller and Mocanu (e.g., see [8] Lemma 1 or [9]). For functions p and q, where p(0)=q(0)=0, we write p≺q (i.e., p is subordinate to q) if there exists an analytic function ω with ω(0)=0 and |ω(z)|<1 so that p(z)=q(ω(z)) in D.Lemma 7.
(i) Ifg and h are analytic in D so that g′(0)<h′(0) and if h+εg is close-to-convex analytic in D for each ε(|ε|=1), then the function f=h+g¯ is close-to-convex harmonic in D.
(ii) Ifh and g are analytic in D so that h∈K(0) and if f=h+g¯ is locally univalent in D, then the function f=h+g¯ is close-to-convex harmonic in D.Lemma 8.
A necessary and sufficient condition for the analytic functionh:D→C to be close-to-convex is that h′ is nonvanishing in D and(5)∫θ1θ2Re1+reiθh′′reiθh′reiθdθ>-π;θ1<θ2,0<r<1.Lemma 9.
IfRp(z)>0 and q(z) is analytic in D, then q(z)+zp(z)q′(z)≺z implies q(z)≺z.Proof of Theorem5.
Proof of Part (i). The convolution function F(z)=h1(z)∗h2(z)+g1(z)∗g2(z)¯=H(z)+G(z)¯ is locally univalent and sense-preserving since (6)G′zH′z=1/zg1z∗g2′z1/zh1z∗h2′z=g1z∗zg2′zh1z∗zh2′z=zh1z∗z2h2′zh1z∗zh2′z=z<1.
ObviouslyG′(0)<H′(0); therefore, in view of Lemma 7, it suffices to prove that Tλ(z)=H(z)-λG(z) for |λ|=1 is close-to-convex analytic in D.
We note that(7)Tλ′z=H′z-λG′z=H′z-λ1/zg1z∗g2′z=H′z-λh1z∗zh2′z=H′z-λz1/zh1z∗h2′z=H′z-λzH′z=1-λz·H′z,and Tλ′′(z)=-λH′(z)+1-λzH′′(z).
We also observe thatTλ is nonvanishing in D since H′(0)≠0. Therefore, (8)Re1+zTλ′′zTλ′z=Re1+z-λH′z+1-λzH′′z1-λzH′z=Re1+λzλz-1+zH′′zH′reiθ.
Now, by Lemma8 and inequality (5) for θ1<θ2<θ1+2π and 0<r<1, it suffices to show that(9)∫θ1θ2Re1+reiθTλ′′reiθTλ′reiθdθ>-π.For z∈D, one may verify (also see Bshouty and Lyzzaik [3] p. 770) that (10)Rezz-1=12-121-z21-z2.For z=reiθ, replacing z by λz and letting ζ=λ¯r yield (11)Reλzλz-1=12-121-ζ2eiθ-ζ2=12-12Pζθ,where Pζ(θ) is the Poisson Kernal. It then follows that (12)∫θ1θ2Reλreiθλreiθ-1dθ=∫θ1θ2Re12-12·Pζθdθ=θ2-θ12-12∫θ1θ2Pζθdθ>θ2-θ12-12∫02πPζθdθ=θ2-θ12-π.On the other hand, since H(z)=h1(z)∗h2(z)∈K(-1/2), we obtain (13)∫θ1θ2Re1+reiθH′′reiθH′reiθdθ>θ1-θ22.Therefore, in view of the required condition (9), we get (14)∫θ1θ2Re1+reiθTλ′′reiθTλ′reiθdθ=∫θ1θ2Re1+λreiθλreiθ-1dθ+∫θ1θ2RereiθH′′reiθH′reiθdθ>θ2-θ12-π+θ1-θ22=-π.Proof of Part (ii). In view of Lemma 7, it suffices to show that the convolution function F(z)=h1(z)∗h2(z)+g1(z)∗g2(z)¯=H(z)+G(z)¯ is locally univalent and sense-preserving in D. In other words, we need to show that (15)G′zH′z=g1z∗g2z′h1z∗h2z′<1.
Using the Hadamard product properties of power series, we have(16)1>zn=znh1′∗h2′h1′∗h2′=g1′∗g2′h1′∗h2′=g1∗zg2′′h1∗zh2′′=z1/zg1∗g2′′z1/zh1∗h2′′=zg1∗g2′′zh1∗h2′′=g1∗g2′+zg1∗g2′′zh1∗h2′′=g1∗g2′h1∗h2′+zg1∗g2′′h1∗h2′h1∗h2′zh1∗h2′′=g1∗g2′h1∗h2′+zh1∗h2′′+zg1∗g2′′h1∗h2′-zg1∗g2′h1∗h2′′h1∗h2′zh1∗h2′′=g1∗g2′zh1∗h2′′+zg1∗g2′′h1∗h2′-zg1∗g2′h1∗h2′′h1∗h2′zh1∗h2′′=g1∗g2′h1∗h2′+zzh1∗h2′′g1∗g2′′h1∗h2′-g1∗g2′h1∗h2′′h1∗h2′=g1∗g2′h1∗h2′+zh1∗h2′zh1∗h2′′g1∗g2′′h1∗h2′-g1∗g2′h1∗h2′′h1∗h2′2=g1∗g2′h1∗h2′+zh1∗h2′zh1∗h2′′g1∗g2′h1∗h2′′.
Therefore,(17)g1∗g2′h1∗h2′+zh1∗h2′zh1∗h2′′g1∗g2′h1∗h2′′≺z.On the other hand, since h1∗h2∈K(0), z(h1∗h2)′ is starlike or (18)Rezzh1∗h2′′zh1∗h2′=Rezh1∗h2′′h1∗h2′>0.Thus, in view of Lemma 9, (g1(z)∗g2(z))′/(h1(z)∗h2(z))′≺z or G′(z)/H′(z)=[g1(z)∗g2(z)]′/[h1(z)∗h2(z)]′<1.Remark 10.
It is left as an open problem whether Theorem5(i) can be extended to the case g1(z)=znh1(z) and g2′(z)=znh2′(z) if n>1.



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