Convolutions of harmonic convex mappings

# Convolutions of harmonic convex mappings

Michael Dorff Department of Mathematics, Brigham Young University, Provo, Utah, 84602, U.S.A. Maria Nowak Department of Mathematics, Maria Curie-Sklodowska University, 20-031 Lublin, Poland  and  Magdalena Wołoszkiewicz Department of Mathematics, Maria Curie-Sklodowska University, 20-031 Lublin, Poland
February 2, 2009
###### Abstract.

The first author proved that the harmonic convolution of a normalized right half-plane mapping with either another normalized right half-plane mapping or a normalized vertical strip mapping is convex in the direction of the real axis. provided that it is locally univalent. In this paper, we prove that in general the assumption of local univalency cannot be omitted. However, we are able to show that in some cases these harmonic convolutions are locally univalent. Using this we obtain interesting examples of univalent harmonic maps one of which is a map onto the plane with two parallel slits.

###### Key words and phrases:
Harmonic Mappings, Convoutions, Univalence
30C45

## 1. Introduction

Let be the unit disk. We consider the family of complex-valued harmonic functions defined in , where and are real harmonic in . Such functions can be expressed as , where

 h(z)=z+∞∑n=2anznandg(z)=∞∑n=1bnzn

are analytic in . The harmonic function is locally one-to-one and sense- preserving in if and only if

 |g′(z)|<|h′(z)|,∀z∈\D.

In such a case, we say that is locally univalent and satisfies the dilatation condition. Let be the class of complex-valued, harmonic, sense-preserving, univalent functions in , normalized so that , , and . Let and be the subclasses of mapping onto convex and close-to-convex domains, respectively. We will deal with mappings that are convex in one direction.

For analytic functions and , their convolution (or Hadamard product) is defined as . In the harmonic case, with

 f =h+ ¯¯¯g=z+∞∑n=2anzn+∞∑n=1 ¯¯bn ¯¯¯zn   and F =H+ ¯¯¯¯G=z+∞∑n=2Anzn+∞∑n=1 ¯¯¯¯Bn ¯¯¯zn,

define the harmonic convolution as

 f∗F=h∗H+ ¯¯¯¯¯¯¯¯¯¯¯¯g∗G=z+∞∑n=2anAnzn+∞∑n=1 ¯¯¯¯¯¯¯¯¯¯¯¯bnBn ¯¯¯zn,

There have been some results about harmonic convolutions of functions (see [2], [4], [6], and [10]). For the convolution of analytic functions, if , then . Also, the right half-plane mapping, , acts as the convolution identity. In the harmonic case, there are infinitely many right half-plane mappings and the harmonic convolution of one of these right half-plane mappings with a function may not preserve the properties of , such as convexity or even univalence (see [4] for an example).

In [4], the following theorems were proved:

###### Theorem A.

Let be right half-plane mappings. If satisfies the dilatation condition, then and is convex in the direction of the real axis.

###### Theorem B.

Let be a right half-plane mapping and be a vertical strip mapping. If satisfies the dilatation condition, then and is convex in the direction of the real axis.

In section 2, we generalize Theorem A for harmonic mappings onto slanted half-planes given by

Next, we deal mainly with the convolution of the canonical harmonic right half-plane mapping (see [2]) given by

 (1) f0(z)=h0(z)+¯¯¯¯¯¯¯¯¯¯¯¯g0(z)=z−12z2(1−z)2−¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯12z2(1−z)2

with harmonic mappings that are either right half-planes or strip mappings. We show that if the dilatation of is (), then is locally univalent. However, we give examples when local univalency fails for . Also, we provide some results about univalency in the case the dilatation of is . Finally, we give examples of univalent harmonic maps obtained by way of convolutions.

## 2. Convoluting slanted half-plane mappings

In [1], [5], and [7], explicit descriptions are given for half-plane and strip mappings. Specifically, the collection of functions that map onto the right half-plane, , have the form

 h(z)+g(z)=z1−z

and those that map onto the vertical strip, , where , have the form

 h(z)+g(z)=12isinαlog(1+zeiα1+ze−iα).

Now, we first prove a generalization of Theorem A for the slanted half-plane, , , described in the introduction. Let denote the class of harmonic functions that map onto . In the case when we get the normalized class of harmonic functions that map onto the right half-plane .

###### Lemma 1.

If , then

 h(z)+e−2iγg(z)=z1−zeiγ,z∈\D.
###### Proof.

If , then , which means that . In other words, . Since is a convex harmonic function, it follows from a convexity theorem by Clunie and Sheil-Small [2] that the function is convex in the direction , and so is univalent. It is also clear, that maps onto which implies result. ∎

###### Theorem 2.

If , , and is locally univalent in , then is convex in the direction

###### Proof.

Let

 F1 =(h1+e−2iγ1g1)∗(h2−e−2iγ2g2), and F2 =(h2+e−2iγ2g2)∗(h1−e−2iγ1g1).

Then

 12(F1+F2)=h1∗h2−e−2i(γ1+γ2)g1∗g2.

The shearing theorem of Clunie and Sheil-Small [2] establishes that it is sufficient to show that the function is convex in the direction , or equivalently, that is convex in the direction of the real axis. A result by Royster and Ziegler [9] shows that is convex in the real direction, if , where with some . Thus, if we show this last condition, we are done.

By Lemma 1,

 zF′1(z)=z1−zeiγ1∗[z(h2−e−2iγ2g2)′(z)].

Furthermore,

 z(h2−e−2iγ2g2)′(z) =zh′2(z)−e−2iγ2g′2(z)h′2(z)+e−2iγ2g′2(z)(h′2(z)+e−2iγ2g′2(z)) =zp2(z)(1−eiγ2z)2,

where for all . Consequently,

 zF′1(z) =z1−zeiγ1∗zp2(z)(1−eiγ2z)2 =e−iγ1zeiγ11−zeiγ1∗zp2(z)(1−eiγ2z)2 =zp2(zeiγ1)(1−ei(γ1+γ2)z)2

Analogously,

 zF′2(z)=zp1(zeiγ2)(1−ei(γ1+γ2)z)2,

where for all . Thus

 Re⎛⎜ ⎜ ⎜⎝ei(γ1+γ2)(zF′1(z)+zF′2(z))zei(γ1+γ2)(1−ei(γ1+γ2)z)2⎞⎟ ⎟ ⎟⎠=Re(p1(zeiγ2)+p2(zeiγ1))>0.

## 3. Convoluting f0 with right half-plane mappings

In Theorem A, Theorem B, and Theorem 2, we require that the resulting convolution function satisfy the dilatation condition

 |ω(z)|=∣∣∣g′(z)h′(z)∣∣∣<1,∀z∈\D.

When is this not a necessary assumption? In the rest of the paper we establish cases of these theorems for which this assumption can be omitted.

The following result about the number of zeros of polynomials in the disk is helpful in proving the next several theorems.

###### Cohn’s Rule.

([3] or see [8], p 375) Given a polynomial

 f(z)=a0+a1z+⋯+anzn

of degree , let

 f∗(z)=zn ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯f(1/ ¯¯¯z)= ¯¯¯¯¯an+ ¯¯¯¯¯¯¯¯¯¯an−1z+⋯+ ¯¯¯¯¯a0zn.

Denote by and the number of zeros of inside the unit circle and on it, respectively. If , then

 f1(z)= ¯¯¯¯¯anf(z)−a0f∗(z)z

is of degree with and the number of zeros of inside the unit circle and on it, respectively.

The main result of this section is the following.

###### Theorem 3.

Let with and and . If , then and is convex in the direction of the real axis.

###### Proof.

Let the the dilatation of be given by . By Theorem A and by Lewy’s Theorem, we just need to show that .

First, note that if is analytic in and , then from eq. (1)

 (2) h0(z)∗F(z)=12[F(z)+zF′(z)] g0(z)∗F(z)=12[F(z)−zF′(z)].

Also, since , we know .

Hence

 (3) ˜ω(z)=−zg′′(z)2h′(z)+zh′′(z)=−zω′(z)h′(z)−zω(z)h′′(z)2h′(z)+zh′′(z).

Using and , we can solve for and in terms of and :

 h′(z)= 1(1+ω(z))(1−z)2 h′′(z)= 2(1+ω(z))−ω′(z)(1−z)(1+ω(z))2(1−z)3.

Substituting these formulas for and into the equation for , we derive:

 (4) ˜ω(z)=−zω′(z)h′(z)−zω(z)h′′(z)2h′(z)+zh′′(z)=−zω2(z)+[ω(z)−12ω′(z)z]+12ω′(z)1+[ω(z)−12ω′(z)z]+12ω′(z)z2.

Now, consider the case in which . Then eq. (4) yields

 ˜ω(z)= −ze2iθ(z2+12e−iθz+12e−iθ)(1+12eiθz+12eiθz2).

If we denote , then . In such a situation, if is a zero of , then is a zero of . Hence,

 ˜ω(z)= −ze2iθ(z+A)(z+B)(1+ ¯¯¯¯Az)(1+ ¯¯¯¯Bz).

By Cohn’s Rule we have

has one zero at . So, has two zeros, namely and , with .

Next, consider the case in which . In this case,

 |˜ω(z)|=∣∣z2∣∣∣∣∣z3+e−iθ1+eiθz3∣∣∣=|z|2<1.

###### Remark 1.

If we assume the hypotheses of the previous theorem with the exception of making , then for each we can find a specific such that . For example, if is odd, let and then eq. (4) yields

 ˜ω(z)=−znzn+1+(n2−1)z−n21+(n2−1)zn−n2zn+1.

It suffices to show that for some point , . Let . Then

 (5) ˜ω(z0)=(nn+1)n(nn+1)n+1−(n2−1)(nn+1)−n21−(n2−1)(nn+1)n−(n2)(nn+1)n+1=1+[(n+1n)n−(nn+1)n+1]+[1−nn+1](n2−1)+(n2)(nn+1)−(n+1n)n.

Note that . Also, since and is an increasing series converging to . Thus, if is odd, . If , it is easy to compute that . Now, if is even, let and . This simplifies to the same given eq. (5) and the argument above also holds for . If , .

###### Theorem 4.

Let with and with . Then and is convex in the direction of the real axis.

###### Proof.

Using eq. (4) with , where , we have

 ˜ω(z)= = −zf(z)f∗(z) = −z(z+A)(z+B)(1+ ¯¯¯¯Az)(1+ ¯¯¯¯Bz).

Again using Cohn’s Rule,

 f1(z)= ¯¯¯¯¯a2f(z)−a0f∗(z)z=(a+3)(1−a)4z+(1+3a)(1−a)4.

So has one zero at which is in the unit circle since . Thus, , . ∎

Next, we provide some examples.

###### Example 1.

Let , where with . Then

 h1 =14log(1+z1−z)+12z1−z g1 =−14log(1+z1−z)+12z1−z.

Consider . Using eq. (2) we have

 H1 =h0∗h1=12[h1(z)+zh′1(z)]=18log(1+z1−z)+34z−14z3(1−z)2(1+z) G1 =g0∗g1=12[g1(z)−zg′1(z)]=−18log(1+z1−z)+14z−12z2−14z3(1−z)2(1+z),

and from eq. (4)

 ˜ω(z)=−z(2z2+z+1z2+z+2).

We show that maps the unit disk onto the domain whose boundary consists of the four half-lines given by and (see Figure 2). In doing so, we use a similar argument to that used by Clunie and Sheil-Small in Example 5.4 [2]. We have

 F1(z)=Re⎛⎝z−12z2−12z3(1+z)(1−z)2⎞⎠+iIm(14ln(1+z1−z)+12z(1−z)2).

Applying the transformation , , we get

 F1(z) =18(3ξ−2−ξξ2+η2)+i4(arctanηξ+ξη).

Observe first that the positive real axis is mapped monotonically onto the whole real axis. Next we find the images of the level curves

 arctanηξ+ξη=c,ξ,η>0.

The polar coordinates equations of these level curves are

 (6) θ+r2sinθcosθ=c,0<θ<π2.

Hence

 ξ =√(c−θ)cotθ η =√(c−θ)tanθ,0<θ

Fix . Then the image of the curve given in (6) under is

 F1(z) =18(3√(c−θ)cotθ−2−√sinθcos3θc−θ)+i4c =u(c,θ)+i4c.

If , then and one finds easily that and . The intermediate value property implies that in this case the image of the level curve under is the entire horizontal line If , then and . So in this case the images of the level curves are horizontal half-lines . This means that images of the level curves under fill the domain whose boundary consists of the real axis and two half-lines and . Finally, our claim follows from the fact that the range of is symmetric with respect to the real axis.

The images of concentric circles inside under the harmonic maps and under are shown in Figure 1. The images of these concentric circles under the convolution map are shown in Figure 2.

###### Example 2.

Let be the harmonic mapping in the disk such that and . One can find that

 h2(z) =18ln(1+z1−z)+12z1−z+14z(1−z)2, g2(z) =−18ln(1+z1−z)+12z1−z−14z(1−z)2

and the image of under is the right half-plane, . We note here that if and if Next let

 F2=h0∗h2+¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯g0∗g2=H2+¯¯¯¯¯¯G2.

By eq. (2)

 H2(z) =12[18ln(1+z1−z)+12z1−z+14z(1−z)2+z(1−z)3(1+z)], G2(z) =12[−18ln(1+z1−z)+12z1−z−14z(1−z)2+z3(1−z)3(1+z)]

and

 ˜ω2(z)=G′2(z)H′2(z)=z2

Analysis similar to that in Example 1 can be used to show that maps the disk onto the plane minus two half-lines given by , . We have

 F2(z)=Re(12z(2−z+z3)(1−z)3(1+z))+iIm(18ln(1+z1−z)+68z(1−z)2),

which under the same transformation as in Example 1 becomes

 F2(z)=116(ξ3−3ξη2+4ξ−4−ξξ2+η2)+i8(arctanηξ+3ξη).

Analogously, we find that the images of the level curves

 θ+32r2sin2θ=c,0<θ<π2

are

 F2(z) =116[√13(c−θ)cotθ((c−θ)(13cotθ−tanθ)+4−3sin2θ2(c−θ))−4]+i8c =u(c,θ)+i8c.

If (or , respectively), then (or , respectively) and . This means that the images of the level curves are entire horizontal lines. If , then