# On the Largest Disc Mapped by Sum of Convex and Starlike Functions

Authors: Rosihan M. Ali; Naveen Kumar Jain; V. Ravichandran
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

For a normalized analytic functionf defined on the unit disc 𝔻, let ϕ(f,f′,f′′;z) be a function of positive real part in 𝔻, ψ(f,f′,f′′;z) need not have that property in 𝔻, and χ=ϕ+ψ. For certain choices of ϕ and ψ, a sharp radius constant ρ is determined, 0<ρ<1, so that χ(ρz)/ρ maps 𝔻 onto a specified region in the right half-plane.


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## Body

## 1. Introduction

Let𝒜be the class of functionsfanalytic in𝔻={z∈ℂ:|z|<1}and normalized byf(0)=0=f′(0)-1. Let𝒮be its subclass consisting of univalent functions. For two analytic functionsfandg, the functionfis subordinate tog, writtenf(z)≺g(z), if there is an analytic self-mapw:𝔻→𝔻withw(0)=0satisfyingf(z)=g(w(z)). Given an analytic functionpwithp(0)=1andRep(z)>0in𝔻, denote by𝒮𝒯(p)and𝒞𝒱(p)the subclasses of𝒜consisting, respectively, offsatisfyingzf′(z)/f(z)≺p(z)and1+zf′′(z)/f′(z)≺p(z).For various choices ofp, these classes reduce to well-known subclasses of starlike and convex functions. For instance, withp(z)=(1+(1-2α)z)/(1-z),0≤α<1, then𝒮𝒯(α)and𝒞𝒱(α)are, respectively, the subclasses consisting of starlike functions of orderαand convex functions of orderα. The classes𝒮𝒯:=𝒮𝒯(0)and𝒞𝒱:=𝒞𝒱(0)are the familiar subclasses of𝒮of starlike and convex functions. Forp(z)=(1+(1-2β)z)/(1-z),β>1, ℳ(β)=𝒮𝒯(p)is the class of functionsf∈𝒜satisfying
(1)ℳ(β):={f∈𝒜:Re(zf′(z)f(z))<β(z∈𝔻)}
studied by Uralegaddi et al. [1]. Various subclasses ofℳ(β)have been investigated in [2–5]. Forp(z)=((1+z)/(1-z))α,0<γ≤1, the class𝒮𝒮𝒯(γ):=𝒮𝒯(p)is the class of strongly starlike functions of orderγ. The class𝒮ℒ:=𝒮𝒯(1+z)introduced by Sokół and Stankiewicz [6] consists of functionsf∈𝒜satisfying
(2)|(zf′(z)f(z))2-1|<1(z∈𝔻).Thus, a functionf∈𝒜is in the class𝒮ℒifzf′(z)/f(z)lies in the region bounded by the right-half of the lemniscate of Bernoulli given by|w2-1|<1. Results related to the class𝒮ℒcan be found in [3, 7–9].In investigating the class𝒰𝒞𝒱of uniformly convex functions, Rønning [10] introduced a class𝒮𝒫of parabolic starlike functions. These are functionsf∈𝒜satisfying
(3)Re(zf′(z)f(z))>|zf′(z)f(z)-1|(z∈𝔻).It is important to keep in mind that the qualifier “parabolic” refers to the geometry of the image of𝔻under the mapzf′(z)/f(z); that is, the domain necessarily lies in a parabolic region of thew-plane. It does not convey the interpretation that the functionfmaps the disk𝔻onto a parabolic region. This terminology of parabolic starlike functions is however widely accepted and used by authors. Ali and Ravichandran [11] recently surveyed works on uniformly convex and parabolic starlike functions.This paper finds radius estimates for classes of functions in𝒜. The radius of a propertyPin a given set of functionsℳ[12, page 119] is the largest numberRsuch that every function in the setℳhas the propertyPin each disc𝔻r={z∈ℂ:|z|<r}for everyr<R. For example, the Koebe functionk(z)=z/(1-z)2, which maps𝔻onto the domainℂ∖{w∈ℝ:w≤-1/4}, is starlike but not convex. However,kmaps the disc𝔻ronto a convex domain for everyr≤2-3. Indeed, every univalent functionf∈𝒮maps𝔻ronto a convex domain forr≤2-3[13, Theorem  2.13, page 44]. This number is known as the radius of convexity for𝒮.It is known that𝒞𝒱⊆𝒮𝒯(1/2)⊆{f∈𝒜:Re(f(z)/z)>1/2,z∈𝔻}. The functiong(z)=z/(1-z)is convex and therefore starlike of order1/2; it is clear that the function
(4)ϕ(z):=zg′(z)g(z)
has real part greater than 1/2. Now the function
(5)ψ(z):=z2g′′(z)g(z)
takes values inℂ∖{w∈ℝ:w≤-1/2}, and therefore does not have positive real part for allz∈𝔻. Their sum
(6)ϕ(z)+ψ(z)=zg′(z)g(z)+z2g′′(z)g(z)=zg′(z)g(z)(1+zg′′(z)g′(z))
takes values in{w:=x+iy∈ℂ:y2>-x/2}and therefore the sumϕ+ψdoes not have positive real part in𝔻. This motivates us to determine the largest radiusρsuch that
(7)Re(z2g′′(z)g(z)+zg′(z)g(z))>α(|z|≤ρ).More generally, letϕ=ϕ(f,f′,f′′;z)andψ=ψ(f,f′,f′′;z)be functions satisfyingReϕ>0in𝔻, whileReψneed not necessarily be positive in the whole unit disc𝔻.For certain choices ofϕandψ, a sharp radius constantρis determined,0≤ρ<1, so that whenever|z|<ρ, the sumϕ+ψtakes values in specified regions in the complex plane. The results obtained are shown to reduce those of Singh and Paul [14] in certain special cases.


## 2. Main Results

Forϕ(z):=ϕ(f,f′,f′′;z)=zf′(z)/f(z)andψ(z):=ψ(f,f′,f′′;z)=z2f′′(z)/f(z), withf∈𝒮𝒯(1/2), several radius results for the sumϕ+ψto be in certain regions in the complex plane are obtained in the following result.Theorem 1.
Letf∈𝒮𝒯(1/2); letχ:𝔻→ℂbe defined by
(8)χ(z)=z2f′′(z)f(z)+zf′(z)f(z),χi(z)=χ(ρiz),i=1,2,…,6.
Then(a)
Reχ1(z)>α,  0≤α<1, whereρ1is given by
(9)ρ1={7-16α11-16α+82-4α,0≤α≤2+13182(1-α)1+2α+1+8α,2+1318<α<1.(b)
|χ22(z)-1|<1, whereρ2is given by
(10)ρ2=2(2-1)1+22+1+82≃0.112903.(c)
Reχ3(z)<β, β>1, whereρ3is given by
(11)ρ3=2(β-1)1+2β+1+8β.(d)
|χ4(z)-1|<1-α,  0≤α<1, whereρ4is given by
(12)ρ4=2(1-α)5-2α+17-8α.(e)
|arg(χ5(z))|<γπ/2,0<γ≤1whereρ5=ρ5(γ)∈(0,1)is the root of the equation inr:
(13)(1+2(1-r2)t0)4t0-(1+t0-r2t0)2-(1+t0(-1-3r2+2(1-r2)2t0))tan(πγ2)=0,t0=5-r2+9-10r2+17r48(1-r4).In particular,(14)ρ5(14)≃0.131522,ρ5(12)≃0.266747,ρ5(34)≃0.409049,ρ5(1)≃0.560097.(f)
Also|χ6(z)-1|<Reχ6(z), whereρ6≃0.23605∈(0,1)is the root of the following equation inr:
(15)2(1+9r2)t0-1+(-5+26r2-21r4)t02+4(1-r2)2(1+3r2)t03-4(1-r2)4t04=0,andt0∈(1/(1+r)2,1/(1-r)2)is the root of the equation int:
(16)2(1+9r2)+2(-5+26r2-21r4)t+12(1-r2)2(1+3r2)t2-16(1-r2)4t3=0.
Each radius constantρiis sharp.For two analytic functionsf,g∈𝒜, their convolution or Hadamard product, denoted byf*g, is defined by(f*g)(z):=z+∑n=2∞‍anbnzn. The following results are needed in the sequel.Lemma 2 ([15, Lemma  2.7, page 126; Lemma  3.5, page  130]).
Iff∈𝒞𝒱andg∈𝒮𝒯, orfandgbelong to𝒮𝒯(1/2), then
(17)f*gFf*g(𝔻)⊂co¯(F(𝔻))
for any functionFanalytic in𝔻, whereco¯(F(𝔻))denotes the closed convex hull ofF(𝔻).Lemma 3 ([7, Lemma2.2, page 6559]).
For0<a<2, letrabe given by
(18)ra={(1-a2-(1-a2))1/2,0<a≤22/32-a,22/3≤a<2,
and fora>0, letRabe given by
(19)Ra={2-a,0<a≤12,a,12≤a.
Then,(20){w:|w-a|<ra}⊆{w:|w2-1|<1}⊆{w:|w-a|<Ra}.Proof of Theorem1.
Leth:𝔻→ℂbe defined by
(21)h(z)=2(1-z)2-11-z.
First, for eachi=1,2,…,6, hi(z)=h(ρiz)will be shown to, respectively, satisfyReh1(z)>α,|h2(z)-1|<1,Reh3(z)<β,|h4(z)-1|<1-α,|argh5(z)|<γπ/2, and|h6(z)-1|<Reh6(z). Then, using Lemma 2,χiis deduced to satisfy the required condition.
As in [14], let
(22)11-z=Reiθ,
so that
(23)11+r≤R≤11-r(|z|=r),cosθ=1+R2-r2R22R.
(a) By (21), (22), and (23), it follows that
(24)Reh(z)=2R2cos2θ-Rcosθ=12-12(1+3r2)t+(1-r2)2t2(t≔R2)=:ϕ(t).
Case (i). Suppose that0≤α≤(2+13)/18.We assert thatminϕ(t)>αfor|z|<ρ1, where the minimum is taken over allt∈(1/(1+r)2,1/(1-r)2). Letr<ρ1. Then
(25)∂ϕ(t)∂t=-12(1+3r2)+2(1-r2)2t=0
ift=t0:=(1+3r2)/(4(1-r2)2),∂2ϕ(t0)/∂t2>0, and that forr≥4-13,
(26)1(1+r)2≤t0≤1(1-r)2.
Thus, for4-13≤r<ρ1,
(27)minϕ(t)=ϕ(t0)=7-22r2-r416(1-r2)2>α.
On the other hand, ifr<4-13, then it can be shown that
(28)minϕ(t)=ϕ(1(1+r)2)=1-r(1+r)2>13-3(5-13)2=2+1318>α.
Case (ii). For(2+13)/18<α<1, thenminϕ(t)>αin|z|<ρ1,t∈(1/(1+r)2,1/(1-r)2). Indeed forr<ρ1<4-13, as in the case (i),
(29)minϕ(t)=ϕ(1(1+r)2)=1-r(1+r)2>α.
The previously mentioned two cases show thatReh1(z)>αin𝔻. Figure 1 illustrates sharpness of the radiusρ1=5-2in the caseα=0.5.
(b) Forhgiven by (21), a calculation shows that
(30)|h(z)-1|=|2(1-z)2-11-z-1|=|z1-z+2z(1-z)2|≤r1-r+2r(1-r)2.
By Lemma3, the functionhsatisfies
(31)|h2(z)-1|<1
provided
(32)r1-r+2r(1-r)2≤2-1;
that is,
(33)2r2-(1+22)r+(2-1)≥0.
This inequality holds ifr≤ρ2. Figure 2 illustrates sharpness of the radiusρ2≃0.1129.
(c) From (30), it follows that
(34)Reh(z)≤1+r1-r+2r(1-r)2≤β
provided
(35)βr2-(1+2β)r-(1-β)≥0
holds, which occurs wheneverr≤ρ3. Sharpness of the radiusρ3=(4-13)/3in the caseβ=1.5is illustrated in Figure 3.
(d) Inequality (30) also yields
(36)|h(z)-1|≤r1-r+2r(1-r)2≤1-α
provided
(37)r2(α-2)+r(5-2α)+α-1≤0,
that is, whenr≤ρ4. Figure 4 illustrates sharpness of the radiusρ4=(4-13)/3in the caseα=0.5.
(e) For the functionhgiven by (21), it follows from (22) and (23) that
(38)argh(z)=arctan(2Rsin2θ-sinθ2Rcos2θ-cosθ)=arctan((4Rcosθ-1)sinθ2R(2cos2θ-1)-cosθ)=arctan((1+2(1-r2)R2)4R2-(1+(1-r2)R2)21-(1+3r2)R2+2(1-r2)2R4)=arctan((1+2(1-r2)t)4t-(1+(1-r2)t)21-(1+3r2)t+2(1-r2)2t2):=arctan(ϕ(t)),t∈(1/(1+r)2,1/(1-r)2). A calculation shows thatϕ′(t)=0where
(39)t=t0=5-r2+9-10r2+17r48(1-r4)∈(1(1+r)2,1(1-r)2).
Nowϕ′(t)>0fort<t0,ϕ′(t)<0ift>t0, and
(40)ϕ(1(1+r)2)=ϕ(1(1-r)2)=0.
Thus(41)maxϕ(t)=ϕ(t0)=(1+2(1-r2)t0)4t0-(1+(1-r2)t0)21-(1+3r2)t0+2(1-r2)2t02.
Evidently, (38) and (41) give
(42)|argh(z)|≤|arctan((1+2(1-r2)t0)4t0-(1+(1-r2)t0)21-(1+3r2)t0+2(1-r2)2t02)|≤γπ2
provided
(43)(1+2(1-r2)t0)4t0-(1+(1-r2)t0)2-(1+t0(-1-3r2+2(1-r2)2t0))tan(πγ2)≤0.
Figure5 illustrates sharpness of the radiusρ5≃0.266747in the caseγ=0.5.
(f) The inequality(44)|h(z)-1|<Reh(z)
holds if
(45)(2R2cos2θ-Rcosθ-1)2+(2R2sin2θ-Rsinθ)2<(2R2cos2θ-Rcosθ)2,or, witht=R2,(46)ϕ(t)≔2(1+9r2)t-1+(-5+26r2-21r4)t2+4(1-r2)2(1+3r2)t3-4(1-r2)4t4<0,t∈(1/(1+r)2,1/(1-r)2). Then,
(47)ϕ′(t)=2(1+9r2)+2(-5+26r2-21r4)t+12(1-r2)2(1+3r2)t2-16(1-r2)4t3.
Letr<ρ6. Since
(48)ϕ′(1(1+r)2)=4(-3+11r+6r2+7r3-r4)(1+r)2>0ifr>0.234722,ϕ′(1(1-r)2)=-4(3+11r-6r2+7r3+r4)(1-r)2<0,
there exists a uniquet0∈(1/(1+r)2,1/(1-r)2)such thatϕ′(t0)=0andmaxϕ(t)=ϕ(t0).
Thus,maxϕ(t)=ϕ(t0)<0for0.234722<r<ρ6. Whenr≤0.234722,
(49)maxϕ(t)=ϕ(1(1+r)2)=4(4r-1+r2)(1+r)2<0;
hence,ϕ(t)<0forr<ρ6.Figure 6 illustrates sharpness of the radiusρ6≃0.23605.
Next, considergi(z)=z/(1-ρiz)∈𝒞𝒱⊂𝒮𝒯(1/2),i=1,2,…,6. Then,
(50)f(z)*gi(z)hi(z)f(z)*gi(z)=f(z)*(z/(1-ρiz))(2ρiz/(1-ρiz)2+1/(1-ρiz))f(z)*(z/(1-ρiz))=f(z)*(2ρiz2/(1-ρiz)3+z/(1-ρiz)2)f(z)*(z/(1-ρiz))=ρi2z2f′′(ρiz)+ρizf′(ρiz)f(ρiz)=χ(ρiz)=χi(z).
Lemma2, together with (50) and the corresponding inequality for the functionhi, shows that each functionχisatisfies the required condition. For sharpness, consider the functionf0(z)=z/(1-z)∈𝒞𝒱⊂𝒮𝒯(1/2). Then,
(51)z2f0′′(z)+zf0′(z)f0(z)=2(1-z)2-11-z=h(z).
Sharpness of the numbersρiis now evident in view of the definitionh.Figure 1
Image of|z|≤5-2touchesRew=0.5.Figure 2
Image of|z|≤0.1129touches|w2-1|=1.Figure 3
Image of|z|≤(4-13)/3touchesRew=1.5.Figure 4
Image of|z|≤(4-13)/3touches|w-1|=0.5.Figure 5
Image of|z|≤0.266747touches|argw|=π/4.Figure 6
Image of|z|≤0.23605touches|w-1|=Rew.Forα=0, Theorem 1(a) reduces to the following corollary.Corollary 4 ([14, Theorem5, page 724]).
Iff∈𝒮𝒯(1/2), then
(52)Re(z2f′′(z)f(z)+zf′(z)f(z))>0
in|z|<ρ=82-11=0.56.The numberρis sharp.Theorem 5.
Letf∈𝒮𝒯(1/2)andχ:𝔻→ℂbe defined by
(53)χ(z)=f(z)z+f′(z),χi(z)=χ(ρiz),i=1,2,3.
Then(a)
Reχ1(z)>α,0≤α<1, whereρ1is given by
(54)ρ1={7-8α5-8α+421-α,0≤α<7+4718,4-2α2α-1+1+4α,7+4718≤α<1.(b)
|argχ2(z)|<γπ/2,0<γ≤1, whereρ2=ρ2(γ)∈(0,1)is the root of the equation
(55)(9+5r2(-2+r2)-2r2)×16r2+(3+3r2-9+5r2(r2-2))2+(9+5r2(r2-2)-11-11r4+5r2×××(2+9+5r2(r2-2)))tan(πγ2)=0.In particular,(56)ρ2(14)≃0.257136,ρ2(12)≃0.487998,ρ2(34)≃0.674274,ρ2(1)=75+42≃0.810465.(c)
Also|χ3(z)-1|<Reχ3(z), whereρ3≃0.44915is given by the equation inr(57)24r2t0-8+(-1+18r2-17r4)t02+2(1-r2)2(-1+3r2)t03-(1-r2)4t04=0andt0is given by the equation int(58)24r2+2(-1+18r2-17r4)t+6(1-r2)2(-1+3r2)t2-4(1-r2)4t3=0.
Each radius constantρiis sharp.Proof.
Let(59)h(z)=11-z+1(1-z)2(z∈𝔻).
Eachhi(z)=h(ρiz), i=1,2,3, is shown to, respectively, satisfyReh1(z)>α, |argh2(z)|<γπ/2, and|h3(z)-1|<Reh3(z). Then, it follows from Lemma 2 thatχisatisfies the required condition.
(a) We claim thatReh(z)>αin|z|<ρ1. By (22) and (23),
(60)Reh(z)=Rcosθ+R2cos2θ=1+12((1-3r2)t+(1-r2)2t2)(t≔R2):=ϕ(t).
Case (i). Suppose0≤α<(7+47)/18.In this case, it is shown thatminϕ(t)>αfor|z|<ρ1over alltin(1/(1+r)2,1/(1-r)2). Letr<ρ1. It can be verified that
(61)∂ϕ(t)∂t=12(1-3r2+2(1-r2)2t)=0
ift=t0=(3r2-1)/(2(1-r2)2),∂2ϕ(t0)/∂t2=(1-r2)2>0, and that forr≥7-2,
(62)1(1+r)2≤t0≤1(1-r)2.
Thus for7-2≤r<ρ1,
(63)minϕ(t)=ϕ(t0)=7-10r2-r48(1-r2)2>α.
On the other hand, ifr<7-2, then
(64)minϕ(t)=ϕ(1(1+r)2)=2+r(1+r)2.
Sinceg(r)=(2+r)/(1+r)2is a decreasing function in(0,7-2),
(65)minϕ(t)=(2+r)(1+r)2>2+7-2(1+7-2)2=7(7-1)2=47+718>α.
Case (ii). For(7+47)/18≤α<1, we prove thatminϕ(t)>αin|z|<ρ1,t∈(1/(1+r)2,1/(1-r)2).Letr<ρ1<7-2. As in Case (i), then
(66)minϕ(t)=ϕ(1(1+r)2)=(2+r)(1+r)2>α.
It is evident from the previous two cases thatReh1(z)>αin𝔻.Figure 7 shows that, forα=0.5, the radiusρ1=3/5is sharp.
(b) Leth(reit)=u+iv. Then,
(67)u=2(1+r2+r2cos2t)-r(5+r2)cost(1+r2-2rcost)2,v=r(3+r2-4rcost)sint(1+r2-2rcost)2.
By (67), it follows that
(68)argh(reit)=arctan(r(3+r2-4rcost)sint2(1+r2+r2cos2t)-r(5+r2)cost).
Letg:[-1,1]→ℝbe defined by
(69)g(x)=r(3+r2-4rx)1-x22-r(5+r2)x+4r2x2.
(The case-1-x2is similar.) A calculation shows that
(70)g′(x)=r(1+r2-2rx)(r(7+r2)-6x(1+r2)+4rx2)1-x2(2-r(5+r2)x+4r2x2)2.
Let(71)x0=3+3r2-9-10r2+5r44r∈(0,1].
Then,g′(x0)=0,g′(x)>0forx<x0, andg′(x)<0forx>x0. Thus,
(72)g(x)≤maxx∈[-1,1]g(x)=g(x0).
Now (68) and (72) show that
(73)|argh(reit)|≤γπ2
provided
(74)r(3+r2-4rx0)1-x022-r(5+r2)x0+4r2x02≤tan(γπ2);
that is,
(75)tan(γπ2)-r(3+r2-4rx0)1-x022-r(5+r2)x0+4r2x02≥0.
Thus,|argh2(z)|<γπ/2in𝔻.Figure 8 shows that, forγ=0.5, the radiusρ2=0.487998is sharp.
(c) Proceeding similarly as in part (a),(76)|h(z)-1|<Reh(z),
provided(77)24r2R2-8+(18r2-1-17r4)R4-2(1-r2)2(1-3r2)R6-(1-r2)4R8<0.
Letϕ:(1/(1+r)2,1/(1-r)2)→ℝbe defined by
(78)ϕ(t)=24r2t-8+(18r2-1-17r4)t2-2(1-r2)2(1-3r2)t3-(1-r2)4t4.
Now(79)ϕ′(t)=24r2+2(18r2-1-17r4)t-6(1-r2)2(1-3r2)t2-4(1-r2)4t3.
Letr<ρ3.Since
(80)ϕ′(1(1+r)2)=4(-3+7r+12r2+7r3+r4)(1+r)2>0ifr>0.27606,ϕ′(1(1-r)2)=4(-3-7r+12r2-7r3+r4)(1-r)2<0,
there exists a uniquet0∈(1/(1+r)2,1/(1-r)2)such thatϕ′(t0)=0andmaxϕ(t)=ϕ(t0).
Then, for0.27606<r<ρ3,
(81)maxϕ(t)=ϕ(t0)<0.
Whenr≤0.27606,
(82)maxϕ(t)=ϕ(1(1+r)2)=4(-3+r2)(1+r)2<0.
Evidently,ϕ(t)<0forr<ρ3and hence|h3(z)-1|<Reh3(z)in𝔻.Figure 9 shows that the radiusρ3≃0.44915is sharp.
Now, withg(z)=z∈𝒮𝒯(1/2),i=1,2,3,
(83)f(z)*zhi(z)f(z)*z=f(z)*z(1/(1-ρiz)+1/(1-ρiz)2)f(z)*z=f(ρiz)ρiz+f′(ρiz)=χ(ρiz)=χi(z).
Lemma2, together with (83) and the corresponding inequality for the functionhi, shows that each functionχisatisfies the required condition. For sharpness, consider the functionf0(z)=z/(1-z)∈𝒞𝒱⊂𝒮𝒯(1/2). Clearly
(84)f0(z)z+f0′(z)=11-z+1(1-z)2=h(z);
hence the fact that the numberρiis sharp follows from the definition ofh.Figure 7
Image of|z|≤3/5touchesRew=0.5.Figure 8
Image of|z|≤0.487998touches|argw|=π/4.Figure 9
Image of|z|≤0.44915touches|w-1|=Rew.Forα=0, Theorem 5(a) reduces to the following corollary.Corollary 6 ([14, Theorem3, page 722]).
Iff∈𝒮𝒯(1/2), then
(85)Re(f(z)z+f′(z))>0
in|z|<ρ=42-5≃0.81. The numberρis sharp.Theorem 7.
Letf∈𝒞𝒱,χ:𝔻→ℂbe defined by
(86)χ(z)=(1+zf′′(z)f′(z))+1f′(z),χi(z)=χ(ρiz),i=1,2.
Then,(a)
Reχ1(z)>α,0≤α<1, whereρ1=3-5+2α;(b)
|argχ2(z)|<γπ/2,0<γ≤1, whereρ2=ρ(γ)∈(0,1)is the root of the equation inr(87)2r21-x02(r-(3+r2)x0+2rx02)+tan(πγ2)×(2-r2-r4-4rx0+(6r2+2r4)x02-4r3x03)=0,andx0∈[-1,1]is the root of the equation
(88)6-5r2-4r4-r6+(-6r+15r3+7r5)x0+(-12-8r2-16r4)x02+(24r+16r3)x03-16r2x04=0.In particular,(89)ρ2(12)≃0.6355,ρ2(1)=3-5.The radii are sharp.Proof.
Let(90)h(z)=21-z+(1-z)2-1(z∈𝔻).
(a) We claim thatReh(z)>αin|z|<ρ1. By (22) and (23), it follows that
(91)Reh(z)=2Rcosθ-1+cos2θR2=1+t2+r4t2+2t3-2r2(t+t2+t3)2t2(t≔R2):=ϕ(t).
A calculation shows thatϕ′(t)=0ift=t0=1∈(1/(1+r)2,1/(1-r)2),ϕ′′(t0)=3-2r2>0, and that
(92)minϕ(t)=ϕ(1)=12(4-6r2+r4)>α
over allt∈(1/(1+r)2,1/(1-r)2)provided
(93)r4-6r2+4-2α>0.
This inequality reduces tor≤ρ1.Thus,Reh1(z)>αin𝔻.Figure 10 shows that, forα=0.5, the radiusρ1=3-6is sharp.
(b) Leth(reit)=u+iv. Then
(94)u=(2(1-rcost)+(r2(cos2t-sin2t)-2rcost)××(1+r2-2rcost))×(1+r2-2rcost)-1,v=2rsint(1-(1-rcost)(1+r2-2rcost))1+r2-2rcost.
By (94), it follows that
(95)argh(reit)=arctan(r2+r2(3+r2)cos2t-r3cos3t)-1(2r2(-r+cost(3+r2-2rcost))sint)=arctan×(2+2r2-r(4+3r2)costlhh+r2(3+r2)cos2t-r3cos3t)-1(-r+cost(3+r2-2rcost))).
Let(96)g(x)=(2r2(-r+x(3+r2-2rx))1-x2)×(2+2r2-r(4+3r2)xmmm+r2(3+r2)(2x2-1)-r3(4x3-3x))-1.
A calculation shows that there existsx0∈[0,1]such thatg′(x0)=0andg′′(x0)<0. Thus
(97)g(x)≤g(x0),x∈[-1,1].
By (95), (96), and (97), evidently
(98)|argh(reit)|≤γπ2
if
(99)2r2(-r+x0(3+r2-2rx0))1-x022-r2-r4-4rx0+(6r2+2r4)x02-4r3x03≤tan(γπ2);
that is,
(100)2r21-x02(r-(3+r2)x0+2rx02)+tan(γπ2)(2-r2-r4-4rx0+(6r2+2r4)x02-4r3x03)≥0.
Thus,|argh2(z)|<γπ/2in𝔻.Figure 11 shows that, forγ=0.5, the radiusρ2≃0.6335is sharp.
To conclude the proof, letg(z)=z/(1-ρiz)2∈𝒮𝒯. Then,
(101)f(z)*(z/(1-ρiz)2)hi(z)f(z)*(z/(1-ρiz)2)=f(z)*(z/(1-ρiz)2)(2/(1-ρiz)-1+(1-ρiz)2)f(z)*(z/(1-ρiz)2)=f(z)*(2z/(1-ρiz)3+z-z/(1-ρiz)2)f(z)*(z/(1-ρiz))=(1+ρizf′′(ρiz)f′(ρiz))+1f′(ρiz)=χ(ρiz)=χi(z).
As in the earlier proofs, Lemma2 together with (101) and the corresponding inequality for the functionhishows that the functionχisatisfies the required condition. For sharpness, considerf0(z)=z/(1-z)∈𝒞𝒱⊂𝒮𝒯(1/2). Then,
(102)(1+zf0′′(z)f0′(z))+1f0′(z)=21-z-1+(1-z)2=h(z).Figure 10
Image of|z|≤3-6touchesRew=0.5.Figure 11
Image of|z|≤0.6335touches|argw|=π/4.Forα=0, Theorem 7(a) reduces to the following result.Corollary 8 ([14, Theorem4, page 723]).
Iff∈𝒞𝒱, then
(103)Re((1+zf′′(z)f′(z))+1f′(z))>0
in|z|<ρ=3-5=(5-1)/2≃0.874032.The result is sharp.



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*Source: 682413-2013-12-09.xml*