# Argument Properties for a Class of Analytic Functions Involving Libera Transform

Authors: Badr S. Alkahtani; Teodor Bulboacă; Rakesh Kumar; Rubayyi Alqahtani
Journal: Journal of Function Spaces
 (2016)

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## Abstract

The purpose of this paper is to find sufficient argument properties, such that the images of some subclasses of functions by the Libera transform have bounded arguments.


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## Body

## 1. Introduction

LetA denote the class of functions of the form (1)fz=z+∑n=2∞anzn, which are analytic in the open unit disk U=z∈C:z<1.A functionf∈A is said to be in the class S∗(α) of star-like functions of order α in U, if and only if it satisfies the condition (2)Re⁡zf′zfz>α,z∈U,α<1.A functionf∈A is said to be in the class K∗(α) of convex functions of order α in U, if and only if it satisfies the condition (3)Re⁡1+zf′′zf′z>α,z∈U,α<1.It is well-known thatS∗(α)⊂S∗, K(α)⊂K whenever 0≤α<1, where S∗≡S∗(0) and K≡K(0) represent, respectively, the class of starlike and convex (normalized) functions.Iff∈A satisfies the condition (4)Re⁡fzzf′z>β,z∈U,0≤β<1, then f is said to bestar-like of reciprocal orderβ, and we denote this class by NS∗(β).The above definition was recently discussed by Nunokawa et al. [1] and Ravichandran and Sivaprasad Kumar [2].Motivated by their works, we define a class of convex functions of reciprocal orderβ as follows.Definition 1.
Iff∈A satisfies the condition (5)Re⁡1+zf′′zf′z-1>β,z∈U,0≤β<1,then f is said to beconvex function of reciprocal orderβ, and we denote this class by NK(β).Definition 2 (see [3, 4]).
Suppose thatf and g are two analytic functions in U. We say that the function f is subordinate to g, written f(z)≺g(z), if there exists aSchwarz functionw, that is, a function w analytic in U, with w(0)=0 and w(z)<1, such that f(z)=g(w(z)), z∈U.It is well-known that (see, e.g., [3, 4]) if g is univalent in U, then the followingsubordination property holds:(6)fz≺gz⟺f0=g0,fU⊂gU.Remarks 1.
(1) In view of the fact that(7)Re⁡pz>β,z∈U,0≤β<1⇓Re⁡1pz=Re⁡pzpz2>0,z∈U, it follows that a star-like (convex) function of reciprocal order β is a star-like (convex) function. Thus, (8)NS∗β⊂S∗0,NKβ⊂K0,0≤β<1, and the equality holds in both cases if and only if β=0.
(2) Let0<β<1, and suppose that f∈A. Then (i)
(9)f∈NS∗β⟺zf′zfz-12β<12β,z∈U,(ii)
(10)f∈NKβ⟺1+zf′′zf′z-12β<12β,z∈U.
(3) We note that, for0≤β<1, we have (11)f∈NKβ⟺zf′z∈NS∗β.In the following, we give some examples of functions belonging to the class of star-like functions of reciprocal order and the class of convex functions of reciprocal order.Example 3.
(1) For0≤β<1, define the function fβ:U→C by (12)fβz=z1+1-2βz-21-β/1-2β,ifβ∈0,1∖12,ze-z,ifβ=12.
Sincefβ∈A and a simple computation shows that (13)Re⁡fβzzfβ′z=Re⁡1+1-2βz1-z>β,z∈U, we conclude that fβ∈NS∗(β).
(2) Using the third part of the Remarks1, we deduce that (14)φβ∈NKβ⟺fβz=zφβ′z∈NS∗β.Since we proved that fβ∈NS∗(β), and from the above differential equation we obtain (15)φβz=∫0z1+1-2βu-21-β/1-2βdu,ifβ∈0,1∖12,1-e-z,ifβ=12,it follows that φβ∈NK(β).Example 4.
(1) For1/2≤β<1, consider the function ψβ:U→C, defined by (16)ψβz=ze1-β/βz. This gives that (17)Re⁡ψβzzψβ′z=Re⁡ββ+1-βz=Re⁡hz,z∈U,where(18)hz=ββ+1-βz.
SinceRe⁡β≥1/2, it follows that h is analytic in U. Using the fact that h is a convex (not necessary normalized) function in U and h(U) is symmetric with respect to the real axis, we deduce that Re⁡h(z)>β for all z∈U; hence ψβ∈NS∗(β).
(2) From the above result, using the same reasons like in the second part of Example3, we obtain that (19)χβz=∫0ze1-β/βudu∈NKβ,12≤β<1.We mention that several authors have investigated the strongly star-like functions and the strongly convex functions (see [5–18]). In the present investigation we give some argument properties of analytic functions belonging to A, such that the images of these functions by the Libera transform have bounded arguments.


## 2. Main Results

The following lemma will be used to prove our main results.Lemma 5 (see [16]).
Letα0 be the solution of(20)απ=3π2-tan-1⁡γfor a suitable fixed γ>0, so that λ:U→C satisfies(21)αRe⁡λz1+αIm⁡λz≥γ,z∈U,and let(22)η≔ηα,γ=α+2πtan-1⁡γ,0<α≤α0.If p is analytic in U, p(0)=1 and (23)arg⁡pz+λzzp′z<π2η,z∈U,then(24)arg⁡pz<π2α,z∈U.Remark that in the article [19] the author considered some special situations improving many results with a lot of applications. Thus, in [19, Lemma 2], the author proved the next result.Letλ be a function defined on U satisfying (25)arg⁡λz-η<π3,z∈U,such  thatη=infz∈U⁡Re⁡λz-Im⁡λz3,and let(26)β′η=6+5η2+29+15η225η21/3·9-29+15η210 be such that 2β′(η)+η≥0. If p is analytic in U, p(0)=1, then (27)Repz+λzzp′z>β′η,z∈U⇓arg⁡pz<π3,z∈U.We emphasize that a special case of this lemma improves the conclusion of Lemma 1 from [19], and in the same article the author derived a number of interesting consequences of it.LetL:A→A defined by (28)Lfz=2z∫0zftdtbe the well-knownLibera transform [20].We first prove the following theorem, which is essential for proving our other results.Theorem 6.
Forg∈A suppose that the Libera transform G=Lg satisfies the condition (21), with(29)λz=2+zG′′zG′z-1,where 0<α≤α0 and α0 is defined by (20). If f∈A and F=Lf, then (30)arg⁡f′zg′z<π2ηα,γ=π2α+2πtan-1⁡γ,z∈U implies (31)arg⁡F′zG′z<π2α,z∈U.Proof.
Let us define the functionp by (32)pz=F′zG′z,z∈U. Suppose that there exists a number z0∈U, such that G′(z0)=0; since G′(0)=1, then z0∈U∖{0}. It follows that there exists a unique number n∈N and a unique function Q, analytic in U, with Q(z0)≠0, such that (33)G′z=z-z0nQz,z∈U. From the above relation we deduce that (34)zG′′zG′z=nzz-z0+zQ′zQz,z∈U∖z0, which implies that z0 is a simple pole for the function zG′′(z)/G′(z). Consequently, from (29) we obtain that λ(z0)=0, and using the fact that γ satisfies the inequality (21) for all z∈U it follows that (35)γ≤αRe⁡λz01+αIm⁡λz0=0, which contradicts the assumption γ>0.
Thus,G′(z)≠0 for all z∈U, which implies that the function p is analytic in U, and p(0)=1.
Now, a simple calculus shows that(36)f′zg′z=zF′′z+2F′zzG′′z+2G′z=pz+λzzp′z, and our result follows immediately from Lemma 5.Takingg(z)=z in Theorem 6, for the function λ defined by (29) we have λ(z)=1/2, and we obtain the following result.Corollary 7.
Suppose that the parametersα and γ satisfy the conditions (20) and (22), for a given γ≤α/2. If f∈A and F=Lf; then (37)f′z<π2ηα,γ=π2α+2πtan-1⁡γ,z∈U implies (38)arg⁡F′z<π2α,z∈U.Example 8.
Forf∈A and F=Lf, considering the special cases γ=1/3, γ=2-1 and γ=2-3, from the above corollary we obtain respectively the next implications:(1)
If2/3≤α≤4/3, then (39)f′z<π2ηα,13=π2α+13,z∈U⇓arg⁡F′z<π2α,z∈U.(2)
If2(2-1)≤α≤11/8, then (40)f′z<π2ηα,2-1=π2α+14,z∈U⇓arg⁡F′z<π2α,z∈U.(3)
If2(2-3)≤α≤17/12, then (41)f′z<π2ηα,2-3=π2α+16,z∈U⇓arg⁡F′z<π2α,z∈U.Theorem 9.
Forg∈A suppose that the Libera transform G=Lg∈NK(β), 0<β<1. Suppose that α0 is defined by (20), and for 0<α≤α0 let γ satisfy the condition(42)0<γ≤γ~≔γ~α,β=2αβ1+2β+β/1+βα2+1.If f∈A and F=Lf, then (43)arg⁡f′zg′z<π2ηα,γ=π2α+2πtan-1⁡γ,z∈U implies (44)arg⁡F′zG′z<π2α,z∈U.Proof.
Letp be the function defined by(45)pz=F′zG′z,z∈U.Similarly like in the proof of Theorem 6, we will prove that the function p is analytic in U. Supposing that there exists a number z0∈U, such that G′(z0)=0, since G′(0)=1, then z0∈U∖{0}. Hence, there exists a unique number n∈N and a unique function Q, analytic in U, with Q(z0)≠0, such that (46)G′z=z-z0nQz,z∈U. From here we deduce that (47)zG′′zG′z=nzz-z0+zQ′zQz,z∈U∖z0, which implies that z0 is a simple pole for the function zG′′(z)/G′(z). Now, using the fact that G=Lg∈NK(β), we obtain (48)0=Re⁡1+zG′′z0G′z0-1>β, which contradicts the assumption 0≤β.
Consequently,G′(z)≠0 for all z∈U, and this implies that the function p is analytic in U, with p(0)=1.
From (45) we easily get (49)f′zg′z=pz+λzzp′z, where the function λ is given by (29).
The assumptionG=Lg∈NK(β) is equivalent to (50)1+zG′′zG′z-1≺1+1-2βz1-z; hence there exists aSchwarz functionw such that (51)1+zG′′zG′z-1=1+1-2βwz1-wz,z∈U. By simple calculations, we get (52)2+zG′′zG′z-1=1+1-2βwz21-βwz,z∈U;that is,(53)λz=2+zG′′zG′z-1≺1+1-2βz21-βz≕ϕz.Since the circular transform ϕ maps the unit disk U onto the disk (54)Ω=w∈C:w-ω<r,withω=1+2β21+β,r=121+β, from the subordination property (6) we deduce that the subordination relation (53) is equivalent to (55)λz=uz+ivz∈Ω,z∈U.
With the above notation, the condition (21) becomes(56)αuz-αγvz-γ≥0,z∈U.
We will prove that ifγ satisfies the assumption (42), that is, 0<γ≤γ~, then the subordination (53) implies that the inequality (56) holds, that is, (57)Ω⊂Δ, where (58)Δ=w=u+iv∈C:αu-αγv-γ≥0. Since the sets Ω and Δ are symmetric with respect to the real axis, the above inclusion is equivalent to(59)Ω~⊂Δ~,where (60)Ω~=w=u+iv∈Ω:v≥0,Δ~=w=u+iv∈Δ:v≥0.
The line(d) containing the half-line (61)αu-αγv-γ=0,v≥0, which is a part of the boundary the set Δ~, has the equation (62)αu-αγv-γ=0. Thus, the line (d) intersects the real axis in the point γ/α and contains the point -i/α (see Figure 1). Thesmallest set Δ~ that includes the domain Ω~ is obtained for the case when the line (d), containing the point -i/α, is tangent to the upper boundary of Ω~, which is the half-circle (63)C=w=u+iv∈C:u-ω2+v2=r2,v≥0. In this case, as it is shown in the figure, the line (d) becomes the tangent line (dτ) to (C).
We will determine now the equation of the tangent line(dτ). If we consider the family (dm) of all the lines containing the point -i/α, with nonnegative angular coefficient, that is, (64)dm:v=mu-1α,m∈0,+∞, we will solve the system (65)u-ω2+v2=r2,v≥0v=mu-1α. It follows that (66)α21+m2u2-2ααω+mu+α2ω2-r2+1=0, and the line is tangent to the half-circle if and only if the discriminant of the above quadratic form is zero, that is, (67)α2ω2-r2m2-2αωm-α2r2+1=0. This last equation has the biggest positive root (68)m~=ω+rω2-r2α2+1αω2-r2; hence the tangent line to the half-circle (C), which contains the point -i/α, has the equation (69)dτ:v=m~u-1α. The tangent (dτ) intersects the real axis in the point 1/(αm~); hence we deduce that the inclusion (59) holds if and only if (70)0<γα≤1αm~,that is(71)0<γ≤1m~=αω2-r2ω+rω2-r2α2+1=2αβ1+2β+β/1+βα2+1=γ~α,β.
Thus, if the parameterγ satisfies the assumption (42), then the condition (21) of Lemma 5 is fulfilled, and our result follows from Theorem 6.Figure 1
The inclusionΩ~⊂Δ~.Puttingg(z)=z in Theorem 9, for β→1-, we get the next result.Corollary 10.
Suppose thatα0 is defined by (20), and for 0<α≤α0 let γ satisfy the condition (72)0<γ≤2α3+α2/2+1. If f∈A and F=Lf, then (73)f′z<π2ηα,γ=π2α+2πtan-1⁡γ,z∈U implies (74)arg⁡F′z<π2α,z∈U.Example 11.
Forf∈A and F=Lf, considering the special case γ=2-3, from Corollary 10 we deduce the next implication.
If(75)0.54552…≃-168+1083+1642-92647≤α≤1712, then (76)f′z<π2ηα,2-3=π2α+16,z∈U⇓arg⁡F′z<π2α,z∈U.



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