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formerly Central European Journal of Mathematics

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Volume 14, Issue 1 (Jan 2016)


Convolutions of harmonic right half-plane mappings

YingChun Li
  • College of Mathematics, Honghe University, Mengzi 661199, Yunnan, China
  • Email:
/ ZhiHong Liu
  • Corresponding author
  • College of Mathematics, Honghe University, Mengzi 661199, Yunnan, China
  • School of Mathematics and Econometrics, Hunan University, Changsha 410082, Hunan, China
  • Email:
Published Online: 2016-10-20 | DOI: https://doi.org/10.1515/math-2016-0069


We first prove that the convolution of a normalized right half-plane mapping with another subclass of normalized right half-plane mappings with the dilatation z(a+z)/(1+az) is CHD (convex in the horizontal direction) provided a=1 or 1a0. Secondly, we give a simply method to prove the convolution of two special subclasses of harmonic univalent mappings in the right half-plane is CHD which was proved by Kumar et al. [1, Theorem 2.2]. In addition, we derive the convolution of harmonic univalent mappings involving the generalized harmonic right half-plane mappings is CHD. Finally, we present two examples of harmonic mappings to illuminate our main results.

Keywords: Harmonic univalent mappings; Harmonic convolution; Generalized right half-plane mappings

MSC 2010: 30C45; 58E20

1 Introduction and main results

Assume that f=u+iv is a complex-valued harmonic function defined on the open unit disk U={zC:|z|<1}, where u and v are real harmonic functions in U. Such function can be expressed as f=h+g¯, where h(z)=z+n=1anznandg(z)=n=1bnzn

are analytic in U,h and g are known as the analytic part and co-analytic part of f, respectively. A harmonic mapping f=h+g¯ is locally univalent and sense-preserving if and only if the dilatation of f defined by ω(z)=g(z)/h(z), satisfies ω(z)<1 for all zU.

We denote by SH the class of all harmonic, sense-preserving and univalent mappings f=h+g¯ in U, which are normalized by the conditions h(0)=g(0)=0 and h(0)=1. Let SH0 be the subset of all fSH in which g(0)=0. Further, let KH,CH (resp. KH0,CH0) be the subclass 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 every line parallel to the real axis has a connected intersection with Ω.

For harmonic univalent functions f(z)=h(z)+g(z)¯=z+n=2anzn+n=1bn¯z¯n

and F(z)=H(z)+G(z)¯=z+n=2Anzn+n=1Bn¯z¯n,

the convolution (or Hadamard product) of them is given by (fF)(z)=(hH)(z)+(gG)(z)¯=z+n=2anAnzn+n=1bnBn¯z¯n.

Many research papers in recent years have studied the convolution or Hadamard product of planar harmonic mappings, see [2-12]. However, corresponding questions for the class of univalent harmonic mappings seem to be difficult to handle as can be seen from the recent investigations of the authors [13-17]. In [1], Kumar et al. constructed the harmonic functions fa=ha+ga¯KH in the right half-plane, which satisfy the conditions ha+ga=z/(1z) with ωa(z)=(az)/(1az)(1<a<1) . By using the technique of shear construction (see [18]), we have ha(z)=11+az12z2(1z)2andga(z)=a1+az12z2(1z)2.(1)

Obviously, for a=0,f0(z)=h0(z)+g0(z)¯KH0 is the standard right half-plane mapping, where h0(z)=z12z2(1z)2andg0(z)=12z2(1z)2.(2)

Recently Dorff et al. studied the convolution of harmonic univalent mappings in the right half-plane (cf. [14, 15 They proved that:

([14, Theorem 5])

Let f1=h1+g1¯,f2=h2+g2¯SH0 with hi+gi=z/(1z) for i = 1,2. If f1 * f2 it is locally univalent and sense-preserving, then f1f2SH0 and convex in the horizontal direction.

([15, Theorem 3])

Let fn=h+g¯SH0 with h+g=z+(1z) and ω(z)=g(z)/h(z)=eiθzn(θR,nN+). If n = 1,2, then f0fnSH0 and is convex in the horizontal direction.

([15, Theorem 4])

Let f=h+g¯KH0 with h(z)+g(z)=z/(1z) and ω(z)=(z+a)/(1+az) with a(1,1). Then f0fSH0 and is convex in the horizontal direction.

We now begin to state the elementary result concerning the convolutions of f0 with the other special subclass harmonic mappings.

Let f=h+g¯SH0 with h(z)+g(z)=z/(1z) and ω(z)=z(z+a)/(1+az), then f0fSH0 and is convex in the horizontal direction for a=1or1a0.

The following generalized right half-plane harmonic univalent mappings were introduced by Muir [7]: Lc(z)=Hc(z)+Gc(z)¯=11+cZ1z+CZ(1z)2+11+cZ1zCZ(1z)2¯(zU;c>0).(3)

Clearly, L1(z)=f0(z), it was proved in [7] that Lc(U)={Re(ω)>1/(1+c)} for each c > 0. Moreover, if f=h+g¯SH, then the above representation gives that Lcf=h+czh1+c+gczg1+c.¯(4)

The following Cohn’s Rule is helpful in proving our main results.

([19, p.375])

Given a polynomial p(z)=p0(z)=an;0zn+an1;0zn1++a1;0z+a0;0(an;00)(5)

of degree n, let p(z)=p0(z)=znp(1/z¯)¯=an;0¯+an1;0¯z++a1;0¯zn1+a0;0¯zn(6)

Denote by r and s the number of zeros of p(z) inside the unit circle and on it, respectively. If a0,0<an,0, then p1(z)=an;0¯p(z)a0;0p(z)Z

is of degree n – 1 with r1 = r – 1 and s1 = s the number of zeros of p1 (z) inside the unit circle and on it, respectively.

It should be remarked that Cohn’s Rule and Schur-Cohn’s algorithm [19, p. 378] are important tools to prove harmonic mappings are locally univalent and sense-preserving. Some related works have been done on these topics, one can refer to [1, 3, 6, 13, 15, 16]. In [1], the authors proved the following result.

Let fa=ha+ga¯ be given by (1). If fn=h+g¯ is the right half-plane mapping given by h+g=z/(1z) with ω(z)=eiθzn(θR,nN+), then fafnSH is CHD for an2n+2,1.

In this paper, we will use a new method to prove the above theorem. The main difference of our work from [1] is that we construct a sequence of functions for finding all zeros of polynomials which are in U¯, and we will show that the dilatation of fa * fn satisfies |ω~1(z)|=|(gag)/(hah)<1 by using mathematical induction, which greatly simplifies the calculation compared with the proof of Theorem 2.2 in [1]. We also show that L_{c * fa is univalent and convex in the horizontal direction for 0<c2(1+a)/(1a), and derive the following theorem.

Let Lc=Hc+Gc¯ be harmonic mappings given by (3). If fa=ha+g¯a is the right half-plane mappings given by (1), then Lc * fa is univalent and convex in the horizontal direction for 0<c2(1+a)/(1a).

Recently, Liu and Li [6] defined a subclass of harmonic mappings defined by Pc(z)=Hc(z)Gc(z)¯=11+cCZ(1z)2+Z1z+11+cCZ(1z)2Z1z¯(zU;c>0).(7)

They proved the following result.

([6, Theorem 7])

Let Pc(z) be harmonic mappings defined by (7) and fn=h+g¯ with hg = z/(1–z) and dilatation ω(z)=eiθzn(θR,nN+). Then Pc * fn is univalent and convex in the horizontal direction for 0<c2/n.

Similar to the approach used in the proof of Theorem E, we get the following result.

Let Lc=Hc+Gc¯ be harmonic mappings given by (3) and fn=h+g¯ with h + g = z/(1–z) and dilatation ω(z)=eiθzn(θR,nN+). Then Lc * fn is univalent and convex in the horizontal direction for 0<c2/n.

In Theorem 1.2 and Theorem 1.3, let c = 1, clearly, L1 = f0, so Theorem 1.2 and Theorem 1.3 are generalization of Theorem C and Theorem B, respectively. It also explains why Theorem B does not hold for n ≥ 3, since c = 1, according to Theorem 1.3, it follows that 0<c2/n hold for n = 1,2.

2 Preliminary results

In this section, we will give the following lemmas which play an important role in proving the main results.

([15, Eq.(6)])

If f=h+g¯SH0 with h + g = z/(1–z) and dilatation ω(z)=g(z)/h(z), then the dilatation of f0 * f is given by ω¯(z)=zω2+[ω12ωz]+12ω1+[ω12ωz]+12ωz2.(8)

([1, Lemma 2.1])

Let fa=ha+ga¯ be defined by (1) and f=h+g¯SH0 be right half-plane mapping, where h + g = z/(1–z) with dilatation ω(z)=g(z)/h(z)(h(z)0,zU). Then ω~1, the dilatation of fa * f, is given by ω¯1(z)=2(az)ω(1+ω)+(a1)ωz(1z)2(1az)(1+ω)+(a1)ωz(1z).(9)

Let Lc=Hc+Gc¯ be defined by (3) and f=h+g¯SH0 be right half-plane mapping, where h + g = z/(1–z) with dilatation ω(z)=g(z)/h(z)(h(z)0,zU) . Then ω~2, the dilatation of Lc * f, is given by ω~2(z)=[(1c)(1+c)z]ω(1+ω)cωz(1z)[(1+c)(1c)z](1+ω)cωz(1z)(10)

By (4), we know that ω¯2(z)=(gczg)(h+czh).

Similar calculation as in the proof of [6, Lemma 7] gives (10). □

([20, Theorem 5.3])

A harmonic function f=h+g¯ locally univalent in U is a univalent mapping of U onto a domain convex in the horizontal direction if and only if hg is a conformal univalent mapping of U onto a domain convex in the horizontal direction.

(See [21])

Let f be an analytic function in U with f(0) = 0 and f(0)0, and let φ(z)=Z(1+zeiθ1)(1+zeiθ2)(11)

where θ1,θ2R. If Re(zf(z)(z))>0(zU).

Then f is convex in the horizontal direction.

Let Lc=Hc+Gc¯ be a mapping given by (3) and f=h+g¯ be the right half-plane mapping with h + g = z/(1–z) . If Lc * f is locally univalent, then LcfSH0 and is convex in the horizontal direction.

Recalling that Lc=Hc+Gc¯ and


Hence hg=1+c2(Hc+Gc)(hg)=1+c2(HchHcg+GchGcg),HcGc=(HcGc)(h+g)=(Hch+HcgGchGcg).

Thus HchGcg=1221+c(hg)+(HcGc).(12)

Next, we will show that 21+c(hg)+(HcGc) is convex in the horizontal direction. Letting φ(z)=z/(1z)2S, we have ReZφ21+c(hg)+(HcGc)=Rez21+c(h+g)(hgh+g)+(Hc+Gc)(HcGcHc+Gc)φ=Rez21+c(h+g)(1ω1+ω)+(Hc+Gc)(1ωc1+ωc)φ=21+cReZ(1z)2[r(z)+rc(z)]z(1z)2=21+cRe{r(z)+rc(z)}>0,

where r(z)=1ω(z)1+ω(z),rc(z)=1ωC(z)1+ωC(z). Therefore, by Lemma 2.5 and the equation (12), we know that HchGcg is convex in the horizontal direction.

Finally, since we assumed that Lc * f is locally univalent and HchGcg is convex in the horizontal direction, we apply Lemma 2.4 to obtain that Lcf=Hch+Gcg¯ is convex in the horizontal direction. □

3 Proofs of theorems

By Theorem A, we know that f0f is convex in the horizontal direction. Now we need to establish that f0f is locally univalent.

Substituting ω(z) = −z(z + a)/(1 + az) into (8) and simplifying, yields ω~(z)=Zz3+2+3a2z2+(1+a)z+a21+2+3a2z+(1+a)z2+a2z3=Zq(z)q(z)=Z(zA)(zB)(zC)(1A¯z)(1B¯z)(1C¯z),(13)

If a = 1, then q(z)=z3+52z2+2z+12=12(1+z)2(1+2z) has all its three zeros in U¯. By Cohn's Rule, so |ω~(z)|<1 for all zU.

If a = 0, it is clearly that |ω~(z)|=|z2|<1 for all zU.

If −l ≤ a < 0, we apply Cohn’s Rule to q(z)=z3+2+3a2z2+(1+a)z+a2. Note that |a2|<1, thus we get Q(z)=a3¯q(z)a0q(z)Z=4a24z2+2(2+2aa2)4a2z+4+2a3a24a2:=4a24q1(z).

Since 4+2a3a24a2=1+2a(1a)4a2<1(asla<0),

we use Cohn’s Rule on q1(z) again, we get q2(z)=q1(z)4+2a3a24a2q1(z)Z=4a(a1)(4+a2a2)(4a2)2Z+2+2aa24+a2a2.

Clearly, q2(z) has one zero at z0=2+2aa24+a2a2.

We show that |z0 ≤ 1, or equivalently, |2+2aa2|2|4+a2a2|2=(2+2aa2)2(4+a2a2)2=3(4a2)(a21)0.

Therefore, by Cohn’s Rule, q(z) has all its three zeros in U¯, that is A,B,CU¯ and so |ω~(z)|<1 for all zU.□

By Theorem A, it suffices to show that the dilatation of fafn satisfies |ω~(z)|<1 for all zU. Setting ω(z)=eiθzn in (9), we have ω~1(z)=zne2iθzn+1azn+12(2+ann)eiθz+12(n2aan)eiθ1az+12(2+ann)eiθzn+12(n2aan)eiθZn+1=zne2iθp(z)p(z),(14)

where p(z)=zn+1azn+12(2+ann)eiθz+12(n2aan)eiθ,(15)

and p(z)=zn+1p(1/z¯)¯.

Firstly, we will show that |ω~1(z)|<1 for a=n2n+2. In this case, substituting a=n2n+2 into (14), we have |ω~1(z)|=zne2iθzn+1n2n+2znn2n+2eiθz+eiθ1n2n+2zn2n+2eiθzn+eiθZn+1=zneiθeiθzn+1n2n+2eiθznn2n+2z+11n2n+2zn2n+2eiθzn+eiθZn+1=|zneiθ|<1.

Next, we will show that |ω~1(z)|<1 for n2n+2<a<1. Obviously, if z0 is a zero of p(z), then 1/z0¯ is a zero of p(z). Hence, ifA1, A2, ⋯, An + 1 are the zeros of p(z) (not necessarily distinct), then we can write ω~1(z)=zne2iθ(zA1)(1A1¯z)(zA2)(1A2¯z)(zAn+1)(1An+1¯z).

Now for Aj1,zAj1Aj¯z(j=1,2,,n+1) maps U onto U. It suffices to show that all zeros of (15) lie on U¯ for n2n+2<a<1. Since |a0,0|=|12(n2aan)eiθ|<|an+1,0|=1, then by using Cohn’s Rule on p(z), we get p1(z)=an+1;0¯p(z)a0;0p(z)Z=p(z)12(n2aan)eiθp(z)Z=(1a)(2+n)[2(1+a)(1a)n]4(znnn+2zn1+2n+2eiθ).

Since n2n+2<a<1, we have (1a)(2+n)[2(1+a)(1a)n]4>0. Let q1(z)=znnn+2zn1+2n+2eiθ, since |a0,1|=|2n+2eiθ|<1=|an,1|, by using Cohn’s Rule on q1(z) again, we obtain p2(z)=an;1¯q1(z)a0;1q1(z)Z=q1(z)2n+2eiθq1(z)Z=n(n+4)(n+2)2zn1n+2n+4zn2+2n+4eiθ.

Let q2(z)=zn1n+2n+4zn2+2n+4eiθ, then |a0,2|=2n+4<1=|an1,2|, we get p3(z)=an1;2¯q2(z)a0;2q2(z)Z=q2(z)2n+4eiθq2(z)Z=(n+2)(n+6)(n+4)2zn2n+4n+6zn3+2n+6eiθ.

Let q3(z)=zn2n+4n+6zn3+2n+6eiθ, then |a0,3|=2n+6<1=|an2,3|, we obtain p4(z)=an1;3¯q3(z)a0;3q3(z)Z=q3(z)2n+6eiθq3(z)Z=(n+4)(n+8)(n+6)2zn3n+6n+8zn4+2n+8eiθ.

By using this manner, we claim that Pk(z)=[n+2(k2)](n+2k)[n+2(k1)]2znk+1n+2(k1)n+2kznk+2n+2keiθ.(16)

where k = 2, 3, ⋯, n.

To prove the equation (16) is correct for all kN+(k2), it suffices to show Pk+1(z)=[n+2(k1)][n+2(k+1)](n+2k)2znkn+2kn+2(k+1)znk+1)+2n+2(k+1)eiθ.(17)

Let qk(z)=znk+1n+2(k1)n+2kznk+2n+2keiθ, then qk(z)=znk+1qk(1/z¯)¯=1n+2(k1)n+2kz+2n+2keiθznk+1. Since |a0,k|=|2n+2keiθ|=2n+2k<1=|ank+1,k|, by using Cohn’s Rule on qk(z), we deduce that Pk+1(z)=ank+1,k¯qk(z)a0,kqk(z)Z=qk(z)2n+2keiθqk(z)Z=[n+2(k1)][n+2(k+1)](n+2k)2znkn+2kn+2(k+1)zn(k+1)+2n+2(k+1)eiθ.

Setting n = k(k ≥ 2) in (16), we have Pn(z)=3n(3n4)(3n2)2(z3n22eiθ3n).

Then z0=3n22eiθ3n is a zero of pn(z), and |z0|=3n22eiθ3n|3n2|+|2eiθ|3n=3n2+23n=1.

So z0 lies inside or on the unit circle |z| = 1, by Lemma 1, we know that all zeros of (15) lie on U¯. This completes the proof.□

In view of Lemma 2.6, it suffices to show that Lcfa is locally univalent and sense-preserving. Substituting ω(z) = ωa(z) = (az)/(1 − az) into (10), we have ω~2(z)=[(1c)(1+c)z](az1az)(1+az1az)ca21)(1az)2z(1z)[(1+c)(1c)z](1+az1az)ca21)(1az)2z(1z)=[(1c)(1+c)z][(az)(1az)+(az)2](a21)cz(1z)[(1+c)(1c)z][(1az)2+(az)(1az)](a21)cz(1z)=z32+ac+2ac1+cz2+1+2a2c+ac1+cza1c)1+c12+ac+2ac1+cz+1+2a2c+ac1+cz2a1c)1+cz3(18)

Next we just need to show that |ω~2(z)|<1 for 0<c2(1+a)/(1a), where −1 < a < 1. We shall consider the following two cases.

Case 1. Suppose that a = 0. Then substituting a = 0 into (18) yields ω~2(z)=Zz22c1+cz+12c1+c12c1+cz+12c1+cz2=Z(z1)z12c1+c(1z)112c1+cz.

Then two zeros z1 = 1 and z2 = (1 − 2c)/(1 + c) of the above numerator lie in or on the unit circle for all 0 < c ≤ 2, so we have |ω~2(z)|<1.

Case 2. Suppose that a ≠ 0. From (18), we can write ω~2(z)=z32+ac+2ac1+cz2+1+2a2c+ac1+cza1c)1+c12+ac+2ac1+cz+1+2a2c+ac1+cz2a1c)1+cz3=p(z)p(z)=(zA)(zB)(zC)(1A¯z)(1B¯z)(1C¯z).

We will show that A,B,CU¯ for 0 < c ≤ 2(1 + a)/(1 − a). Applying Lemma 1 to p(z)=z32+ac+2ac1+cz2+1+2a2c+ac1+cza(1c)1+c.

Note that |a(1c)1+c|<1 for c > 0 and −1 < a < 1, we get p1(z)=a3¯p(z)a0p(z)Z=p(z)+a1c)1+cp(z)Z=(1+c+aac)(1+ca+ac)(1+c)2z2+2c6ac+c2+2a2a2ca2c2(1+c)2z+1c+6ac2c2a2a2c+2a2c2(1+c)2=(1+c+aac)(1+ca+ac)(1+c)2z2+2+c2aac1+c+aacZ+12c+a+2ac1+c+aac=(1+c+aac)(1+ca+ac)(1+c)2(z1)z1+a2c(1a)1+a+c(1a).

So p1(z) has two zeros z1=1 and z2=1+a2c(1a)1+a+c(1a) which are in or on the unit circle for 0 < c ≤ 2(1 + a)/(1 - a). Thus, by Lemma 1, all zeros of p(z) lie on U¯, that is A,B,CU¯ and so |ω~2(z)|<1 for all zU. □

4 Examples

In this section, we give interesting examples resulting from Theorem 1.1 and Theorem 1.2.

In Theorem 1.1, note that f=h+g¯SH0 with h + g = z/(1 - z) and the dilatation ω(z) = −z(a + z)/(1 + az) . By shearing h(z)+g(z)=1(1z)2andg(z)=ω(z)h(z).

Solving these equations, we get h(z)=1+az(1z)3(1+z)andg(z)=z(a+z)(1z)3(1+z).

Integration gives h(z)=12Z1z+1+a4Z(1z)2+1a8log1+z1z,(19)

and then, g(z)=12Z1z1+a4Z(1z)21a8log1+z1z(20)

By the convolutions, we have f0f=h0h+g0g¯=h(z)+zh(z)2+g(z)zg(z)2¯.

So that h0h=14Z1z+1+a8Z(1z)2+1a16log1+z1z+12z(1+az)(1z)3(1+z)

and g0g=14Z1z1+a8Z(1z)21a16log1+z1z+12z2(a+z)(1z)3(1+z).

Images of U under f0f for certain values of a are drawn in Figure 1 and Figure 2 (a)-(d) by using Mathematica. By Theorem 1.1, it follows that f0f is locally univalent and convex in the horizontal direction for a = −1, −0.5, 0,1, but it is not locally univalent for a = 0.2.

Fig. 1

lmages of U under f0f for a = 0.2.

Fig. 2

lmages of U under f0f for various values a.

In Theorem 1.2, by (1) and (4), we have Lcfa=11+cha(z)+czha(z)+11+cga(z)czga(z)¯=11+c11+az12z2(1z)2+cz(1az)(1+a)(1z)3+11+ca1+az12z2(1z)2cz(az)(1+a)(1z)3¯=Rez(1+c)(1z)+c(1a)z(1+z)(1+c)(1+a)(1z)3+iIm(1a1+a+c)z(1+c)(1z)2.

If we take a = 0.5, c = 6, in view of Theorem 1.2, we know that L6f0.5 is univalent and convex in the horizontal direction. The image of U under f0.5, L6 and L6f0.5 are shown in Figure 3 (a)-(c), respectively

Fig. 3

lmages of U under f0.5, L6 and L6f0.5.

If we take a = 0.5, c = 6.2, then L6.2f0.5=Re17.2z1z+3.110.8z(1+z)(1z)3+iIm13+6.27.2z(1z)2.

The image of U under L6.2f0.5 is shown in Figure 4 (a). Figure 4(b) is a partial enlarged view of Figure 4(a) showing that the images of two outer most concentric circles in U are intersecting and so L6.2f0.5 is not univalent.

Fig. 4

lmages of the unit disk U under L6.2f0.5.

This example shows that the condition 0 < c ≤ 2(1 + a)/(1 - a) in Theorem 1.2 is sharp.


The authors would like to thank the referees for their helpful comments. According to the hints of the referees the authors were able to improve the paper considerably.

The research was supported by the Project Education Fund of Yunnan Province under Grant No. 2015Y456, the First Bath of Young and Middle-aged Academic Training Object Backbone of Honghe University under Grant No. 2014GG0102.


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About the article

Received: 2016-03-22

Accepted: 2016-07-27

Published Online: 2016-10-20

Published in Print: 2016-01-01

Citation Information: Open Mathematics, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2016-0069.

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© 2016 Li and Liu, published by De Gruyter Open. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License. BY-NC-ND 3.0

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