Convolutions of harmonic right half-plane mappings

YingChun Li; ZhiHong Liu

doi:10.1515/math-2016-0069

Artikel Open Access

Convolutions of harmonic right half-plane mappings

YingChun Li und ZhiHong Liu

Veröffentlicht/Copyright: 20. Oktober 2016

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Abstract

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 −1≤a≤0. 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={z∈C:|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=1∞anznandg(z)=∑n=1∞bnzn

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 z∈U.

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 f∈SH 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=2∞anzn+∑n=1∞bn¯z¯n

and

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

the convolution (or Hadamard product) of them is given by

(f∗F)(z)=(h∗H)(z)+(g∗G)(z)¯=z+∑n=2∞anAnzn+∑n=1∞bnBn¯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/(1−z) with ωa(z)=(a−z)/(1−az)(−1<a<1) . By using the technique of shear construction (see [18]), we have

ha(z)=11+az−12z2(1−z)2andga(z)=a1+az−12z2(1−z)2.(1)

Obviously, for a=0,f0(z)=h0(z)+g0(z)¯∈KH0 is the standard right half-plane mapping, where

h0(z)=z−12z2(1−z)2andg0(z)=−12z2(1−z)2.(2)

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

Theorem A

Theorem A ([14, Theorem 5])

Letf1=h1+g1¯,f2=h2+g2¯∈SH0withhi+gi=z/(1−z)fori = 1,2. Iff₁ * f₂it is locally univalent and sense-preserving, thenf1∗f2∈SH0and convex in the horizontal direction.

Theorem B

Theorem B ([15, Theorem 3])

Let fn=h+g¯∈SH0withh+g=z+(1−z)andω(z)=g′(z)/h′(z)=eiθzn(θ∈R,n∈N+). Ifn = 1,2, thenf0∗fn∈SH0and is convex in the horizontal direction.

Theorem C

Theorem C ([15, Theorem 4])

Let f=h+g¯∈KH0withh(z)+g(z)=z/(1−z)andω(z)=(z+a)/(1+az)witha∈(−1,1). Thenf0∗f∈SH0and 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.

Theorem 1.1

Letf=h+g¯∈SH0withh(z)+g(z)=z/(1−z)andω(z)=−z(z+a)/(1+az), thenf0∗f∈SH0and is convex in the horizontal direction fora=1or−1≤a≤0.

The following generalized right half-plane harmonic univalent mappings were introduced by Muir [7]:

Lc(z)=Hc(z)+Gc(z)¯=11+cZ1−z+CZ(1−z)2+11+cZ1−z−CZ(1−z)2¯(z∈U;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

Lc∗f=h+czh′1+c+g−czg′1+c.¯(4)

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

Cohn’s Rule

Cohn’s Rule ([19, p.375])

Given a polynomial

p(z)=p0(z)=an;0zn+an−1;0zn−1+⋯+a1;0z+a0;0(an;0≠0)(5)

of degreen, let

p∗(z)=p0∗(z)=znp(1/z¯)¯=an;0¯+an−1;0¯z+⋯+a1;0¯zn−1+a0;0¯zn(6)

Denote byrandsthe number of zeros ofp(z) inside the unit circle and on it, respectively. Ifa0,0<an,0, then

p1(z)=an;0¯p(z)−a0;0p∗(z)Z

is of degreen – 1 withr₁ = r – 1 ands₁ = sthe number of zeros ofp₁ (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.

Theorem D

Letfa=ha+ga¯be given by (1). Iffn=h+g¯is the right half-plane mapping given byh+g=z/(1−z)withω(z)=eiθzn(θ∈R,n∈N+), thenfa∗fn∈SHisCHDfora∈n−2n+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 f_a * f_n satisfies |ω~1(z)|=|(ga∗g)′/(ha∗h)′<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 * f_a is univalent and convex in the horizontal direction for 0<c≤2(1+a)/(1−a), and derive the following theorem.

Theorem 1.2

LetLc=Hc+Gc¯be harmonic mappings given by (3). Iffa=ha+g¯ais the right half-plane mappings given by (1), thenL_c * f_ais univalent and convex in the horizontal direction for0<c≤2(1+a)/(1−a).

Recently, Liu and Li [6] defined a subclass of harmonic mappings defined by

Pc(z)=Hc(z)−Gc(z)¯=11+cCZ(1−z)2+Z1−z+11+cCZ(1−z)2−Z1−z¯(z∈U;c>0).(7)

They proved the following result.

Theorem E

Theorem E ([6, Theorem 7])

LetP_c(z) be harmonic mappings defined by (7) andfn=h+g¯withh – g = z/(1–z) and dilatationω(z)=eiθzn(θ∈R,n∈N+). ThenP_c * f_nis univalent and convex in the horizontal direction for0<c≤2/n.

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

Theorem 1.3

LetLc=Hc+Gc¯be harmonic mappings given by (3) andfn=h+g¯withh + g = z/(1–z) and dilatationω(z)=eiθzn(θ∈R,n∈N+). ThenL_c * f_nis univalent and convex in the horizontal direction for0<c≤2/n.

Remark 1.4

In Theorem 1.2 and Theorem 1.3, letc = 1, clearly, L₁ = f₀, so Theorem 1.2 and Theorem 1.3 are generalization of TheoremCand TheoremB, respectively. It also explains why TheoremBdoes not hold forn ≥ 3, sincec = 1, according to Theorem 1.3, it follows that0<c≤2/nhold forn = 1,2.

2 Preliminary results

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

Lemma 2.1

Lemma 2.1 ([15, Eq.(6)])

Iff=h+g¯∈SH0withh + g = z/(1–z) and dilatationω(z)=g′(z)/h′(z), then the dilatation off₀ * fis given by

ω¯(z)=−zω2+[ω−12ω′z]+12ω′1+[ω−12ωz]+12ωz2.(8)

Lemma 2.2

Lemma 2.2 ([1, Lemma 2.1])

Letfa=ha+ga¯be defined by (1) andf=h+g¯∈SH0be right half-plane mapping, whereh + g = z/(1–z) with dilatationω(z)=g′(z)/h′(z)(h′(z)≠0,z∈U). Thenω~1, the dilatation off_a * f, is given by

ω¯1(z)=2(a−z)ω(1+ω)+(a−1)ω′z(1−z)2(1−az)(1+ω)+(a−1)ωz(1−z).(9)

Lemma 2.3

LetLc=Hc+Gc¯be defined by (3) andf=h+g¯∈SH0be right half-plane mapping, whereh + g = z/(1–z) with dilatationω(z)=g′(z)/h′(z)(h′(z)≠0,z∈U) . Thenω~2, the dilatation ofL_c * f, is given by

ω~2(z)=[(1−c)−(1+c)z]ω(1+ω)−cω′z(1−z)[(1+c)−(1−c)z](1+ω)−cω′z(1−z)(10)

Proof

By (4), we know that

ω¯2(z)=(g−czg′)′(h+czh′)′.

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

Lemma 2.4

Lemma 2.4 ([20, Theorem 5.3])

A harmonic functionf=h+g¯locally univalent inUis a univalent mapping ofUonto a domain convex in the horizontal direction if and only ifh – gis a conformal univalent mapping ofUonto a domain convex in the horizontal direction.

Lemma 2.5

Lemma 2.5 (See [21])

Letfbe an analytic function inUwithf(0) = 0 andf′(0)≠0, and let

φ(z)=Z(1+zeiθ1)(1+zeiθ2)(11)

whereθ1,θ2∈R. If

Re(zf′(z)(z))>0(z∈U).

Thenfis convex in the horizontal direction.

Lemma 2.6

LetLc=Hc+Gc¯be a mapping given by (3) andf=h+g¯be the right half-plane mapping withh + g = z/(1–z) . IfL_c * fis locally univalent, thenLc∗f∈SH0and is convex in the horizontal direction.

Proof

Recalling that Lc=Hc+Gc¯ and

Hc+Gc=2z(1+c)(1−z),h+g=Z1−z.

Hence

h−g=1+c2(Hc+Gc)∗(h−g)=1+c2(Hc∗h−Hc∗g+Gc∗h−Gc∗g),Hc−Gc=(Hc−Gc)∗(h+g)=(Hc∗h+Hc∗g−Gc∗h−Gc∗g).

Thus

Hc∗h−Gc∗g=1221+c(h−g)+(Hc−Gc).(12)

Next, we will show that 21+c(h−g)+(Hc−Gc) is convex in the horizontal direction. Letting φ(z)=z/(1−z)2∈S∗, we have

ReZφ21+c(h′−g′)+(Hc′−Gc′)=Rez21+c(h′+g′)(h′−g′h+g′)+(Hc′+Gc′)(Hc′−Gc′Hc′+Gc′)φ=Rez21+c(h′+g′)(1−ω1+ω)+(Hc′+Gc′)(1−ωc1+ωc)φ=21+cReZ(1−z)2[r(z)+rc(z)]z(1−z)2=21+cRe{r(z)+rc(z)}>0,

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

Finally, since we assumed that L_c * f is locally univalent and Hc∗h−Gc∗g is convex in the horizontal direction, we apply Lemma 2.4 to obtain that Lc∗f=Hc∗h+Gc∗g¯ is convex in the horizontal direction. □

3 Proofs of theorems

Proof of Theorem 1.1

By Theorem A, we know that f₀ ∗ f is convex in the horizontal direction. Now we need to establish that f₀ ∗ f 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(z−A)(z−B)(z−C)(1−A¯z)(1−B¯z)(1−C¯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 z∈U.

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

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=4−a24z2+2(2+2a−a2)4−a2z+4+2a−3a24−a2:=4−a24q1(z).

Since

4+2a−3a24−a2=1+2a(1−a)4−a2<1(as−l≤a<0),

we use Cohn’s Rule on q₁(z) again, we get

q2(z)=q1(z)−4+2a−3a24−a2q1∗(z)Z=4a(a−1)(4+a−2a2)(4−a2)2Z+2+2a−a24+a−2a2.

Clearly, q₂(z) has one zero at

z0=−2+2a−a24+a−2a2.

We show that |z₀ ≤ 1, or equivalently,

|2+2a−a2|2−|4+a−2a2|2=(2+2a−a2)2−(4+a−2a2)2=3(4−a2)(a2−1)≤0.

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

A new proof of Theorem D

By Theorem A, it suffices to show that the dilatation of f_a ∗ f_n satisfies |ω~(z)|<1 for all z∈U. Setting ω(z)=eiθzn in (9), we have

ω~1(z)=−zne2iθzn+1−azn+12(2+an−n)e−iθz+12(n−2a−an)e−iθ1−az+12(2+an−n)eiθzn+12(n−2a−an)eiθZn+1=−zne2iθp(z)p∗(z),(14)

where

p(z)=zn+1−azn+12(2+an−n)e−iθz+12(n−2a−an)e−iθ,(15)

and

p∗(z)=zn+1p(1/z¯)¯.

Firstly, we will show that |ω~1(z)|<1 for a=n−2n+2. In this case, substituting a=n−2n+2 into (14), we have

|ω~1(z)|=−zne2iθzn+1−n−2n+2zn−n−2n+2e−iθz+e−iθ1−n−2n+2z−n−2n+2eiθzn+eiθZn+1=−zneiθeiθzn+1−n−2n+2eiθzn−n−2n+2z+11−n−2n+2z−n−2n+2eiθzn+eiθZn+1=|−zneiθ|<1.

Next, we will show that |ω~1(z)|<1 for n−2n+2<a<1. Obviously, if z₀ is a zero of p(z), then 1/z0¯ is a zero of p^∗(z). Hence, ifA₁, A₂, ⋯, A_{n + 1} are the zeros of p(z) (not necessarily distinct), then we can write

ω~1(z)=−zne2iθ(z−A1)(1−A1¯z)(z−A2)(1−A2¯z)⋯(z−An+1)(1−An+1¯z).

Now for Aj≤1,z−Aj1−Aj¯z(j=1,2,⋯,n+1) maps U onto U. It suffices to show that all zeros of (15) lie on U¯ for n−2n+2<a<1. Since |a0,0|=|12(n−2a−an)e−iθ|<|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(n−2a−an)e−iθp∗(z)Z=(1−a)(2+n)[2(1+a)−(1−a)n]4(zn−nn+2zn−1+2n+2e−iθ).

Since n−2n+2<a<1, we have (1−a)(2+n)[2(1+a)−(1−a)n]4>0. Let q1(z)=zn−nn+2zn−1+2n+2e−iθ, since |a0,1|=|2n+2e−iθ|<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+2e−iθq1∗(z)Z=n(n+4)(n+2)2zn−1−n+2n+4zn−2+2n+4e−iθ.

Let q2(z)=zn−1−n+2n+4zn−2+2n+4e−iθ, then |a0,2|=2n+4<1=|an−1,2|, we get

p3(z)=an−1;2¯q2(z)−a0;2q2∗(z)Z=q2(z)−2n+4e−iθq2∗(z)Z=(n+2)(n+6)(n+4)2zn−2−n+4n+6zn−3+2n+6e−iθ.

Let q3(z)=zn−2−n+4n+6zn−3+2n+6e−iθ, then |a0,3|=2n+6<1=|an−2,3|, we obtain

p4(z)=an−1;3¯q3(z)−a0;3q3∗(z)Z=q3(z)−2n+6e−iθq3∗(z)Z=(n+4)(n+8)(n+6)2zn−3−n+6n+8zn−4+2n+8e−iθ.

By using this manner, we claim that

Pk(z)=[n+2(k−2)](n+2k)[n+2(k−1)]2zn−k+1−n+2(k−1)n+2kzn−k+2n+2ke−iθ.(16)

where k = 2, 3, ⋯, n.

To prove the equation (16) is correct for all k∈N+(k≥2), it suffices to show

Pk+1(z)=[n+2(k−1)][n+2(k+1)](n+2k)2zn−k−n+2kn+2(k+1)zn−⋅k+1)+2n+2(k+1)e−iθ.(17)

Let qk(z)=zn−k+1−n+2(k−1)n+2kzn−k+2n+2ke−iθ, then qk∗(z)=zn−k+1qk(1/z¯)¯=1−n+2(k−1)n+2kz+2n+2ke−iθzn−k+1. Since |a0,k|=|2n+2ke−iθ|=2n+2k<1=|an−k+1,k|, by using Cohn’s Rule on q_k(z), we deduce that

Pk+1(z)=an−k+1,k¯qk(z)−a0,kqk∗(z)Z=qk(z)−2n+2ke−iθqk∗(z)Z=[n+2(k−1)][n+2(k+1)](n+2k)2zn−k−n+2kn+2(k+1)zn−(k+1)+2n+2(k+1)e−iθ.

Setting n = k(k ≥ 2) in (16), we have

Pn(z)=3n(3n−4)(3n−2)2(z−3n−2−2e−iθ3n).

Then z0=3n−2−2e−iθ3n is a zero of p_n(z), and

|z0|=3n−2−2e−iθ3n≤|3n−2|+|2e−iθ|3n=3n−2+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.□

Proof of Theorem 1.2

In view of Lemma 2.6, it suffices to show that L_c ∗ f_a is locally univalent and sense-preserving. Substituting ω(z) = ω_a(z) = (a − z)/(1 − az) into (10), we have

ω~2(z)=[(1−c)−(1+c)z](a−z1−az)(1+a−z1−az)−c⋅a2−1)(1−az)2z(1−z)[(1+c)−(1−c)z](1+a−z1−az)−c⋅a2−1)(1−az)2z(1−z)=[(1−c)−(1+c)z][(a−z)(1−az)+(a−z)2]−(a2−1)cz(1−z)[(1+c)−(1−c)z][(1−az)2+(a−z)(1−az)]−(a2−1)cz(1−z)=−z3−2+a−c+2ac1+cz2+1+2a−2c+ac1+cz−a⋅1−c)1+c1−2+a−c+2ac1+cz+1+2a−2c+ac1+cz2−a⋅1−c)1+cz3(18)

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

Case 1. Suppose thata = 0. Then substitutinga = 0 into (18) yields

ω~2(z)=−Zz2−2−c1+cz+1−2c1+c1−2−c1+cz+1−2c1+cz2=−Z(z−1)z−1−2c1+c(1−z)1−1−2c1+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 thata ≠ 0. From (18), we can write

ω~2(z)=−z3−2+a−c+2ac1+cz2+1+2a−2c+ac1+cz−a⋅1−c)1+c1−2+a−c+2ac1+cz+1+2a−2c+ac1+cz2−a⋅1−c)1+cz3=−p(z)p∗(z)=−(z−A)(z−B)(z−C)(1−A¯z)(1−B¯z)(1−C¯z).

We will show thatA,B,C∈U¯for 0 < c ≤ 2(1 + a)/(1 − a). Applying Lemma 1 to

p(z)=z3−2+a−c+2ac1+cz2+1+2a−2c+ac1+cz−a(1−c)1+c.

Note that|−a(1−c)1+c|<1forc > 0 and −1 < a < 1, we get

p1(z)=a3¯p(z)−a0p∗(z)Z=p(z)+a⋅1−c)1+cp∗(z)Z=(1+c+a−ac)(1+c−a+ac)(1+c)2z2+−2−c−6ac+c2+2a2−a2c−a2c2(1+c)2z+1−c+6ac−2c2−a2−a2c+2a2c2(1+c)2=(1+c+a−ac)(1+c−a+ac)(1+c)2z2+−2+c−2a−ac1+c+a−acZ+1−2c+a+2ac1+c+a−ac=(1+c+a−ac)(1+c−a+ac)(1+c)2(z−1)z−1+a−2c(1−a)1+a+c(1−a).

Sop₁(z) has two zerosz1∗=1andz2∗=1+a−2c(1−a)1+a+c(1−a)which are in or on the unit circle for 0 < c ≤ 2(1 + a)/(1 - a). Thus, by Lemma 1, all zeros ofp(z) lie onU¯, that isA,B,C∈U¯and so|ω~2(z)|<1for allz∈U. □

4 Examples

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

Example 4.1

In Theorem 1.1, note thatf=h+g¯∈SH0withh + g = z/(1 - z) and the dilatationω(z) = −z(a + z)/(1 + az) . By shearing

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

Solving these equations, we get

h′(z)=1+az(1−z)3(1+z)andg′(z)=−z(a+z)(1−z)3(1+z).

Integration gives

h(z)=12Z1−z+1+a4Z(1−z)2+1−a8log1+z1−z,(19)

and then,

g(z)=12Z1−z−1+a4Z(1−z)2−1−a8log1+z1−z(20)

By the convolutions, we have

f0∗f=h0∗h+g0∗g¯=h(z)+zh′(z)2+g(z)−zg′(z)2¯.

So that

h0∗h=14Z1−z+1+a8Z(1−z)2+1−a16log1+z1−z+12z(1+az)(1−z)3(1+z)

and

g0∗g=14Z1−z−1+a8Z(1−z)2−1−a16log1+z1−z+12z2(a+z)(1−z)3(1+z).

Images ofUunderf₀ ∗ ffor certain values of a are drawn in Figure 1 and Figure 2 (a)-(d) by using Mathematica. By Theorem 1.1, it follows thatf₀ ∗ fis locally univalent and convex in the horizontal direction fora = −1, −0.5, 0,1, but it is not locally univalent fora = 0.2.

$Fig. 1 lmages of U${\mathbb{U}}$ under f0 ∗ f for a = 0.2.$

Fig. 1

lmages of U under f₀ ∗ f for a = 0.2.

$Fig. 2 lmages of U${\mathbb{U}}$ under f0 ∗ f for various values a.$

Fig. 2

lmages of U under f₀ ∗ f for various values a.

Example 4.2

In Theorem 1.2, by (1) and (4), we have

Lc∗fa=11+cha(z)+czha′(z)+11+cga(z)−czga′(z)¯=11+c11+az−12z2(1−z)2+cz(1−az)(1+a)(1−z)3+11+ca1+az−12z2(1−z)2−cz(a−z)(1+a)(1−z)3¯=Rez(1+c)(1−z)+c(1−a)z(1+z)(1+c)(1+a)(1−z)3+iIm(1−a1+a+c)z(1+c)(1−z)2.

If we takea = 0.5, c = 6, in view of Theorem 1.2, we know thatL₆ ∗ f_0.5is univalent and convex in the horizontal direction. The image ofUunderf_0.5, L₆andL₆ ∗ f_0.5are shown in Figure 3 (a)-(c), respectively

$Fig. 3 lmages of U${\mathbb{U}}$ under f0.5, L6 and L6 ∗ f0.5.$

Fig. 3

lmages of U under f_0.5, L₆ and L₆ ∗ f_0.5.

If we takea = 0.5, c = 6.2, then

L6.2∗f0.5=Re17.2z1−z+3.110.8z(1+z)(1−z)3+iIm13+6.27.2z(1−z)2.

The image ofUunderL_6.2 ∗ f_0.5is 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 inUare intersecting and soL_6.2 ∗ f_0.5is not univalent.

$Fig. 4 lmages of the unit disk U${\mathbb{U}}$ under L6.2 ∗ f0.5.$

Fig. 4

lmages of the unit disk U under L_6.2 ∗ f_0.5.

Remark 4.3

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

Acknowledgement

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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Received: 2016-3-22

Accepted: 2016-7-27

Published Online: 2016-10-20

Published in Print: 2016-1-1

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License.